diff --git a/apps/pdp10/tapes/diags/klad/README.md b/apps/pdp10/tapes/diags/klad/README.md index 5cffdf9b0..2721039a6 100644 --- a/apps/pdp10/tapes/diags/klad/README.md +++ b/apps/pdp10/tapes/diags/klad/README.md @@ -10,4 +10,5 @@ PDP-10 KLAD Diagnostic Tapes PCjs has archived selected files from [PDP-10 KLAD Diagnostics Sources](http://pdp-10.trailing-edge.com/klad_sources/index.html), including: -- [KA10 BASIC INSTRUCTION DIAGNOSTIC (MAINDEC-10-DAKAD-B-D)](dakad/) +- [KA10 Basic Instruction Diagnostic #1 (MAINDEC-10-DAKAA-B-D)](dakaa/) +- [KA10 Basic Instruction Diagnostic #4 (MAINDEC-10-DAKAD-B-D)](dakad/) diff --git a/apps/pdp10/tapes/diags/klad/dakaa/DAKAA.MAC b/apps/pdp10/tapes/diags/klad/dakaa/DAKAA.MAC new file mode 100644 index 000000000..cf610bb6a --- /dev/null +++ b/apps/pdp10/tapes/diags/klad/dakaa/DAKAA.MAC @@ -0,0 +1,1042 @@ +SUBTTL DIAGNOSTIC SECTION + +START: SETZM USER# ;CLEAR USER CONTROL WORD + JSP 0,.+1 ;GET FLAGS + TLNE USERF ;IN USER MODE? + SETOM USER ;YES, SET USER CONTROL WORD + SKIPN MONFLG ;SPECIAL USER MODE? + SETZM USER ;YES, CLEAR USER CONTROL WORD + SKIPN USER + JRST STARTA + SKIPL MONCTL + TTCALL 3,PGMNAM ;MENTION OUR NAME + JRST STARTA + +PGMNAM: ASCIZ/ +PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) [DAKAA] +/ + +;BASIC INSTRUCTION TEST (3) +;THE TEST IS DESIGNED FOR INITIAL DEBUGGING OF +;PROCESSOR HARDWARE AND TO DETECT (SOLID) FAILURES +;IN THE FIELD. + + +STARTA: JRST .+1 +SUBTTL TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS + +;NOTE: AN "*" IN THE COMMENT FIELD OF AN INSTRUCTION INDICATES THAT IT IS THE TESTED +;INSTRUCTION. +;********** + +;THIS TEST VERIFIES THAT SKIPA ALWAYS SKIPS THE NEXT INSTRUCTION + +A00=. +A12500: SKIPA ;*SKIPA SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + HALT ;IF PROGRAM HALTS, SKIPA DID NOT SKIP + +;IF PROGRAM HANGS UP ON SKIPA INSTRUCTION, CHECK AB PC EN [ABC], AB PC F/F [ABC], +;AB PC B [ABC OR AB], AB PC (FETCH) EN [ABC], IR SKIPS [IR3], IR SKIPX [IR1], +;IR 3XX [IR1], IR BITS 00 - 08 [IR1], F CYC ACT EN C [FI], FT6 F/F [F2], +;PC CLOCK EN [PCC], ET2 J F/F [E], PC CLK (ET2) EN [PCC], ADZ COND P [ADZ] AND +;ADZ COND R [ADZ], PC CLOCK A [PCC], PC CLOCK B [PCC] + +;IF PROGRAM HALTED ON HALT INSTRUCTION, CHECK ST1 COND [ST2] + +;********** + +;THIS TEST VERIFIES THAT JUMP NEVER JUMPS + +A15000: JUMP .+1 ;*JUMP SHOULD NEVER JUMP + SKIPA ;SKIP HALT INSTRUCTION IF JUMP IS SUCCESSFUL + HALT ;HALT IF JUMP FAILS + +;IF PROGRAM HANGS UP ON JUMP .+1 INSTRUCTION, THE JUMP INSTRUCTION +;MAY HAVE FAILED. CHECK ST1 COND [ST2], ST INST FET ST1 EN [ST2], ET2 D F/F [E], +;IR JUMPS [IR1], IR JUMPX [IR1] + +;********** +;THIS TEST VERIFIES THAT JUMP NEVER JUMPS + +A15100: JUMP .+2 ;*JUMP SHOULD NEVER JUMP + SKIPA ;SKIP HALT INSTRUCTION IF JUMP IS SUCCESSFUL + HALT ;HALT IF JUMP FAILS + +;AB PC EN [ABC] + +;********** + +;THIS TEST VERIFIES THAT JUMPA .+1 ALWAYS JUMPS TO THE NEXT INSTRUCTION + +A12700: JUMPA .+1 ;*JUMPA .+1 SHOULD NEVER JUMP + SKIPA ;SKIP HALT INSTRUCTION IF JUMPA IS SUCCESSFUL + HALT ;HALT IF JUMPA .+1 FAILED + +;********** + +;THIS TEST VERIFIES THAT JUMPA JUMPS TO THE LOCATION SPECIFIED BY E WHEN E=.+2 + +A12600: JUMPA .+2 ;*JUMPA .+2 SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + HALT ;PROGRAM HALTS HERE IF JUMPA .+2 FAILS + +;ST1 COND [ST2] + +;********** +;THIS TEST VERIFIES THAT SKIPA IS DATA INDEPENDENT. HENCE, IT ALWAYS SKIPS THE NEXT +;INSTRUCTION. +;THE MOVE [0] INSTRUCTION IS USED TO TEST THE DATA INDEPENDENCE OF SKIPA. + +A100: MOVE [0] ;PRESET LOCATION 0 TO ALL ZEROS + SKIPA ;*SKIPA SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST INSURES THAT SKIPA IS DATA INDEPENDENT. HENCE, IT ALWAYS SKIPS THE NEXT +;INSTRUCTION. +;THE MOVE [-1] INSTRUCTION IS USED TO TEST THE DATA INDEPENDENCE OF SKIPA. + +A200: MOVE [-1] ;PRESET LOCATION 0 TO ALL ONES + SKIPA ;*SKIPA SKOULD ALWAYS SKIP THE NEXT INSTRUCTION + STOP + +;COND P [ADZ] + +;********** +;THIS TEST VERIFIES THAT SKIPGE SKIPS THE NEXT INSTRUCTION WHEN C(E) +;IS ALL ZEROS. THE MOVE INSTRUCTION IS USED TO LOAD E WITH ALL ZEROS. + +A300: MOVE [0] ;PRESET E TO ALL ZEROS + SKIPGE ;*SKIPGE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;COND R [ADZ], AD00 S-A-1, AD AR + EN [ADAP], AD LT AR + EN [ADAP], +;AD RT AR + EN [ADAP], AD AR + EN C [ADAP], F CYC ACT EN A [F1] + +;********** + +;THIS TEST VERIFIES THAT SKIP IS DATA INDEPENDENT. HENCE, SKIP NEVER SKIPS. IT ALWAYS +;CONTINUES ON TO THE NEXT INSTRUCTION. + +A400: MOVE [-1] ;PRESET E TO ALL ONES + SKIP ;*SKIP SHOULD NEVER SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;PC CHANGE [PCC], PC CHANGE A [PCC], SEE MEMORY CONTROL DWGS [MC1-4] + +;********** +;THIS TEST VERIFIES THAT SKIPGE WILL NOT SKIP THE NEXT INSTRUCTION WHEN C(E) IS +;NEGATIVE. THE MOVE INSTRUCTION LOADS E WITH -1, A NEGATIVE NUMBER. + +A500: MOVE [-1] ;PRESET E WITH ALL ONES, A NEGATIVE NUMBER + SKIPGE ;*SKIPGE SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;IF MOVE INSTRUCTION FAILED, CHECK MOVX-: FCE [F1], -IR XXXI [IR1], -IR XXXM [IR1], +;IR HWTFWT [IR3], IR FWT [IR2], IR2XX [IR1] + +;IF SKIPGE FAILED, CHECK SKIPX: FCE [F1] + +;********** + +;THIS TEST VERIFIES THAT SKIPN WILL NOT SKIP THE NEXT INSTRUCTION WHEN +;C(E) IS ZERO. THE MOVE INSTRUCTION LOADS E WITH ALL ZEROS. + +A600: MOVE [0] ;PRESET E WITH ALL ZEROS + SKIPN ;*SKIPN SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT SHOULD SKIP +;THE NEXT INSTRUCTION WHEN C(E) IS NON-ZERO. THE MOVE INSTRUCTION LOADS +;E WITH -1, A NON-ZERO NUMBER. + +A700: MOVE [-1] ;PRESET E WITH ALL ONES, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPE IS DATA DEPENDENT. HENCE, IT SHOULD NOT +;SKIP THE NEXT INSTRUCTION WHEN C(E) IS NON-ZERO. THE MOVE INSTRUCTION LOADS E +;WITH -1, A NON-ZERO NUMBER. + +A1000: MOVE [-1] ;PRESET E WITH ALL ONES, A NON-ZERO NUMBER + SKIPE ;*SKIPE SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPE IS DATA DEPENDENT. HENCE, IT SHOULD SKIP +;THE NEXT INSTRUCTION WHEN C(E) IS ZERO. THE MOVE INSTRUCTION LOADS ALL +;ZEROS INTO E. + +A1100: MOVE [0] ;PRESET E TO ALL ZEROS + SKIPE ;*SKIPE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIP IS DATA INDEPENDENT. HENCE, IT NEVER SKIPS. IT ALWAYS +;CONTINUES ON TO THE NEXT INSTRUCTION. + +A1200: MOVE [0] ;PRESET E TO ALL ZEROS + SKIP ;*SKIP SHOULD NEVER SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NOT NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD +;E WITH ALL ZEROS, A NON-NEGATIVE NUMBER. + +A1300: MOVE [0] ;PRESET E T0 ZERO, A NON-NEGATIVE NUMBER + SKIPL ;*SKIPL SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD E WITH -1, +;A NEGATIVE NUMBER. + +A1400: MOVE [-1] ;PRESET E TO -1, A NEGATIVE NUMBER + SKIPL ;*SKIPL SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPG IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS GREATER THAN ZERO. THE MOVE INSTRUCTION IS USED TO LOAD +;E WITH A NON-ZERO POSITIVE NUMBER. + +A1500: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A NON-ZERO POSITIVE NUMBER + SKIPG ;*SKIPG SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPG IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NOT GREATER THAN ZERO. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH -1, A NUMBER NOT GREATER THAN ZERO. + +A1600: MOVE [-1] ;PRESET E TO -1, A NEGATIVE NUMBER + SKIPG ;*SKIPG SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NON-NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD E WITH +;377777777777, A NON-NEGATIVE NUMBER. + +A1700: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A NON-NEGATIVE NUMBER + SKIPL ;*SKIPL SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS ZERO. THE MOVE INSTRUCTION LOADS E WITH ALL ZEROS. + +A2000: MOVE [0] ;PRESET E TO ALL ZEROS + SKIPLE ;*SKIPLE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NEGATIVE NUMBER. THE MOVE INSTRUCTION IS USED TO LOAD +;E WITH 400000000000, A NEGATIVE NUMBER. + +A2100: MOVE [XWD 400000,0] ;PRESET E TO 400000000000, A NEGATIVE NUMBER. + SKIPLE ;*SKIPLE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A POSITIVE NUMBER. THE MOVE INSTRUCTION LOADS +;E WITH 377777777777, A POSITIVE NUMBER. + +A2200: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A POSITIVE NUMBER + SKIPLE ;*SKIPLE SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********* +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION ;WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD 777000000000, A NON-ZERO NUMBER INTO E. + +A2300: MOVE [XWD 777000,0] ;PRESET E WITH 777000000000, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD A NON-ZERO NUMBER, 377000000 INTO E. + +A2400: MOVE [XWD 377,0] ;PRESET E TO 377000000, A NON ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH A NON-ZERO NUMBER, 177000. + +A2500: MOVE [177000] ;PRESET E WITH 177000, A NON ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD 777, A NON-ZERO NUMBER INTO E. + +A2600: MOVE [777] ;PRESET E WITH 777, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH 600000, A NON-ZERO NUMBER. + +A2700: MOVE [600000] ;PRESET E WITK 600000, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH 400000000, A NON-ZERO NUMBER. + +A3000: MOVE [XWD 400,000] ;PRESET E WITH 400000000, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +SN=3100 + ZZ=0 + +A3100: REPEAT ^D36, +<;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST +;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE +;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO +;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). +;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +> +SUBTTL TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS + +;THIS TEST VERIFIES THAT MOVEI LOADS E INTO THE AC. SKIPG IS USED TO CHECK THAT +;THE AC WAS INDEED MODIFIED BY MOVEI. + +A3200: MOVEI [-1] ;*MOVEI SHOULD LOAD A POSITIVE NUMBER INTO AC0 + SKIPG ;SKIP HALT INSTRUCTION IF MOVEI LOADED AC CORRECTLY + STOP + +;IF PROGRAM HANGS UP ON MOVEI INSTRUCTION, CHECK AD FM +EN [ADFP], AD FM + EN A [ADFP], +;AD FM + F/F [ADFP], F CYC ACT EN C [F1] + +;********** + +;THIS TEST VERIFIES THAT SKIPL IS SENSITIVE TO THE DATA IN E AND NOT SENSITIVE TO +;THE DATA IN THE AC. HENCE, SKIPL WILL SKIP THE NEXT INSTRUCTION ONLY IF +;C(E) IS LESS THAN ZERO. E IS SET TO -1, A NEGATIVE NUMBER; AND +;THE AC IS SET TO 0 IN THIS TEST. THEREFORE SKIPL SHOULD SKIP. + +A3300: MOVE [0] ;SET THE AC TO ALL ZEROS + SKIPL [-1] ;*SKIPL SHOULD SKIP BECAUSE C(E) IS NEGATIVE + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPL IS SENSITIVE TO THE SIGN BIT IN E. SKIPL SHOULD +;SKIP THE NEXT INSTRUCTION WHEN AND ONLY WHEN THE SIGN BIT IS SET. IN THIS +;TEST, C(AC) IS PRESET TO ALL ZEROS, THEN SKIPL IS EXECUTED WITH C(E)=400000,,0. +;A FAILURE UCCURS IF SKIPL DOES NOT SKIP. + +A13100: MOVE [0] ;SET THE AC TO ALL ZEROS + SKIPL [400000,,0] ;*SKIPL SHOULD SKIP BECAUSE C(E) IS NEGATIVE + STOP + +;********** +;THIS TEST VERIFIES THAT THE CAM INSTRUCTION DOES NOT MODIFY C(AC). THE AC IS +;INITIALLY LOADED WITH ALL ZEROS; THEN CAM IS EXECUTED WITH C(E) NON-ZERO. +;THE AC IS THEN CHECKED FOR ALL ZEROS. THIS TEST FAILS IF THE AC WAS MODIFIED BY CAM. + +A3400: MOVE [0] ;PRESET THE AC TO ALL ZEROS + CAM [1234] ;*CAM SHOULD NOT MODIFY THE AC + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION SERVES AS A NO-OP. + ;IT IS A PROGRAM FILLER IN CASE CAM SKIPS.) + SKIPE ;SKIP HALT INSTRUCTION IF AC WAS NOT MODIFIED BY CAM + STOP + +;IF TEST HANGS ON CAM INSTRUCTION, AB PC EN [ABC], AB PC (FETCH) EN [ABC], +;IR CAXX [IR3] + +;IF TEST LOOPS ON CAM INSTRUCTION, MC INSTR FETCH EN [MC1], MC INST FET ET2 EN [MC1], +;ET2 J F/F [E], IR CAXX [IR3] + +;IF TEST HALTS AT HALT INSTRUCTION, SAC INH [ST2], IR CAXX [IR3], IR CAMX [IR1] + +;********** + +;THIS TEST VERIFIES THAT THE CAI INSTRUCTION DOES NOT MODIFY C(AC). THE AC IS +;INITIALLY LOADED WITH ALL ZEROS; THEN CAI IS EXECUTED WITH E NON-ZERO. THE AC IS +;THEN CHECKED FOR ALL ZEROS. THIS TEST FAILS IF THE AC WAS MODIFIED BY CAM. + +A3500: MOVE [0] ;PRESET THE AC TO ALL ZEROS + CAI [1234] ;*CAI SHOULD NOT MODIFY THE AC + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A PROGRAM FILLER + ;IN CASE CAI MODIFIES THE AC. IT SERVES AS A NO-OP) + SKIPE ;SKIP HALT INSTRUCTION IF CAI DID NOT MODIFY THE AC + STOP + +;********** +;THIS TEST VERIFIES THAT CAI DOES NOT MODIFY C(AC0) WHEN AN AC OTHER THAN AC0 IS +;SPECIFIED BY THE CAI INSTRUCTION. +;THIS IS THE FIRST TEST THAT USES AN AC OTHER THAN AC0. +;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, THE CAI INSTRUCTION IS EXECUTED WITH +;A NON-ZERO E. AC 0 IS THEN TESTED F0R ALL ZEROS. + +A3600: MOVE [0] ;LOAD AC0 WITH ALL ZEROS + CAI 17,[1234] ;*CAI SHOULD NOT MODIFY AC0 + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + ;CAI SKIPS THE NEXT INSTRUCTION. IT ACTS AS A NO-OP) + SKIPE ;SKIP HALT INSTRUCTION IF C(AC0) WAS NOT MODIFIED + STOP + +;********** + +;THIS TEST VERIFIES THAT CAM DOES NOT MODIFY C(AC0) WHEN AN AC OTHER THAN AC0 IS +;SPECIFIED BY THE CAM INSTRUCTION. +;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, THE CAM INSTRUCTION IS EXECUTED WITH +;A NON-ZERO E. AC 0 IS THEN TESTED F0R ALL ZEROS. + +A3700: MOVE [0] ;PRESET C(AC0) TO ALL ZEROS + CAM 17,[1234] ;*CAM SHOULD NOT MODIFY AC0 + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + ;CAM SKIPS THE NEXT INSTRUCTION. IT SERVES AS A NO-OP) + SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS NOT MODIFIED BY CAM + STOP + +;********** +;THIS TEST VERIFIES THAT MOVE WILL PLACE C(E) INTO THE AC. AC, E AND C(E) ARE NON-ZERO +;IN THIS TEST. THIS TEST PASSES IF FINAL C(AC) OF THE MOVE INSTRUCTION IS NON-ZERO. + +A4500: MOVE 1,[0] ;LOAD THE AC WITH ALL ZEROS + MOVE 1,[-1] ;*MOVE SHOULD PLACE ALL ONES IN THE AC + SKIPN 1 ;SKIP HALT INSTRUCTION IF MOVE MODIFIED THE AC + STOP + +;********** + +;THIS TEST VERIFIES THAT THE MOVE INSTRUCTION WILL PLACE C(E) INTO THE AC. +;AC, E ARE NON-ZERO AND C(E)=0. +;THIS TEST FAILS IF THE FINAL C(AC) IS NOT EQUAL TO ZERO, THE ORIGINAL C(E) + +A4600: MOVE 1,[0] ;*PLACE ALL ZEROS INTO THE AC + SKIPE 1 ;SKIP HALT INSTRUCTION IF MOVE PLACED ALL ZEROS INTO AC + STOP + +;IF JUMP .+1 OF PREVIOUS TEST FAILED, CHECK JUMPS: AD FM + EN [ADFP], IR JUMPS [IR1], +;IR JUMPX [IR1], AR AD EN [ARMA], AR AD EN A [ARMA], AR LT AD EN A [ARMA], +;AR LT ADD EN B [ARMA], AR RT AD EN A [ARMA], +;AR RT AD EN B [ARMA], AR A CLK INPUT [ARMA], +;ET2 C F/F [E], AR AD ET2 EN [ARMA], AR AD BR (ET2) B [ARMA] + +;********** +;THIS TEST VERIFIES THAT SOS DOES NOT MODIFY C(AC) WHEN THE AC IS AC0. +;FIRST, AC0 IS LOADED WITH ALL ZEROS AND E IS LOADED WITH A POSITIVE NUMBER; +;THEN, SOS IS EXECUTED. SOS SHOULD SUBTRACT ONE FROM C((E) +;AND EXECUTE THE NEXT INSTRUCTION. IT SHOULD NOT MODIFY AC0. +;THE TEST PASSES IF AC0 WAS NOT MODIFIED. + +A4000: MOVE [0] ;PRESET AC0 TO ALL ZEROS + MOVE 7,[1234] ;PRESET E WITH A POSITIVE NUMBER + SOS 7 ;*SOS SHOULD NOT MODIFY C(AC0) + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + ;SOS SKIPS THE NEXT INSTRUCTION. IT SERVES AS A NO-OP) + SKIPE ;SKIP THE HALT INSTRUCTION IF C(AC0) WAS NOT MODIFIED + STOP + +;********** +SN=4100 + AC=0 + +A4100: REPEAT 4, +<;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL PLACE THE CONTENTS OF E INTO THE AC +;WHEN THE AC IS NOT AC0. THIS TEST IS PERFORMED FOR AC1, AC2, AC4 AND AC10. +;FIRST, THE AC IS LOADED WITH ALL ZEROS; THEN, THE SKIP INSTRUCTION IS EXECUTED +;WITH AC1, AC2, AC4 OR AC10. C(E) IS EQUAL TO -1. THE SKIP INSTRUCTION SHOULD CAUSE +;THE AC TO CONTAIN -1. THE AC IS THEN CHECKED FOR +;NON-ZERO CONTENTS. THE TEST FAILS IF C(AC)=0. +;THIS TEST ALSO VERIFIES THAT 'SKIP AC,[-1]' DOES NOT CLOBBER AC0. + +SN=SN+1 + AC=AC+AC + IFE AC, + MOVE 0,[0] ;PRESET AC0 TO ALL ZEROS + MOVE AC,[0] ;PRESET THE AC TO ALL ZEROS + SKIP AC,[-1] ;*SKIP SHOULD PLACE -1 INTO THE AC + SKIPN AC ;SKIP THE HALT INSTRUCTION IF THE AC IS NON-ZERO + STOP + SKIPE 0 ;VERIFY THAT AC0 DID NOT GET CLOBBERED BY SKIP + STOP + +;********** +> +SUBTTL TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS + +;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL NOT MODIFY THE CONTENTS OF AC0. +;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, +;THE SKIP INSTRUCTION IS EXECUTED WITH C(E)=-1 AND AC0. AC0 IS THEN TESTED FOR ALL +;ZEROS. THE TEST FAILS IF AC0 WAS MODIFIED BY THE SKIP INSTRUCTION. + +A4200: MOVE [0] ;PRESET AC0 TO ALL ZEROS + SKIP [-1] ;*SKIP SHOULD NOT MODIFY AC0 + SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS NOT MODIFIED BY SKIP + STOP + +;AC=0: SAC INH, IR BIT 9, 10, 11 OR 12 S-A-1 + +;********** + +;THIS TEST VERIFIES THAT MOVSS DOES NOT MODIFY C(AC0) WHEN AC0 IS SPECIFIED AS THE AC. +;FIRST, E IS LOADED WITH ALL ONES AND THE AC (AC0) +;IS LOADED WITH ALL ZEROS; THEN, MOVSS IS EXECUTED. MOVSS SHOULD NOT MODIFY AC0. +;AC0 IS THEN CHECKED TO INSURE THAT IT WAS NOT MODIFIED BY MOVSS. + +A4300: MOVE 7,[-1] ;LOAD E OF THE MOVSS INSTRUCTION WITH ALL ONES + MOVE [0] ;PRESET AC0 TO ALL ZEROS + MOVSS 7 ;*MOVSS SHOULD NOT MODIFY AC0 + SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS INTACT + STOP + +;MOVX & AC=0: SAC INH [ST2], IR XXXS [IR1] + +;********** +;THIS TEST VERIFIES THAT HRRS DOES NOT MODIFY C(AC0) WHEN AC0 IS THE SPECIFIED AC. +;FIRST, E IS LOADED WITH ALL ONES AND AC0 IS LOADED WITH ALL ZEROS; +;THEN, HRRS IS EXECUTED WITH AC=AC0. AC0 IS THEN CHECKED FOR ALL ZEROS. +;IF AC0 WAS MODIFIED BY THE HRRS INSTRUCTION, THE TEST FAIL. + +A4400: MOVE 7,[-1] ;PRESET E OF THE HRRS INSTRUCTION TO ALL ONES + MOVE [0] ;PRESET AC0 TO ALL ZEROS + HRRS 7 ;*HRRS SHOULD NOT MODIFY THE AC WHEN THE AC IS AC0 + SKIPE ;SKIP THE HALT INSTRUCTION IF AC0 WAS INTACT + STOP + +;(HXXXS&AC=0): SAC INH [ST2], IR HWTFWT [IR3], IR XXXS [IR1], IR HWT [IR1], +;IR BITS S-A-1 + +;********** +;THIS TEST VERIFIES THAT THE JUMP INSTRUCTION DOES NOT MODIFY C(AC). +;THE AC IS FIRST LOADED WITH ALL ZEROS; +;THEN JUMP .+1 IS EXECUTED. THE AC IS THEN CHECKED TO INSURE THAT IT WAS NOT MODIFIED +;BY JUMP. THE TEST FAILS IF THE AC WAS MODIFIED + +A4700: MOVE [0] ; PRESET THE AC TO ALL ZEROS + JUMP .+1 ;*JUMP SHOULD NOT MODIFY THE AC + SKIPE ;SKIP HALT INSTRUCTION IF AC WAS INTACT + STOP + +;********** + +;THIS TEST VERIFIES THAT MOVS WILL NOT PICK UP ANY ONES WHEN SWAPPING A WORD +;OF ALL ZEROS. FIRST, E IS LOADED WITH A WORD OF ALL ZEROS; +;THEN, MOVS IS EXECUTED. THE AC IS THEN CHECKED TO INSURE THAT THE RESULT DOES NOT +;CONTAIN ANY ONES. + +A5000: MOVE [0] ;PRESET E WITH ALL ZEROS + MOVS ;*MOVS SHOULD RESULT WITH AN AC 0F ALL ZEROS + SKIPE ;SKIP HALT INSTRUCTION IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT A MOVS EXECUTED ON AN E CONTAINING A NON-ZERO NUMBER +;RESULTS IN A NON-ZERO NUMBER IN THE AC. FIRST A NON-ZERO WORD +;IS LOADED INTO E; THEN, MOVS IS EXECUTED. THE AC IS CHECKED FOR A NON-ZERO RESULT. +;IF C(AC)=0, THE TEST FAILS. + +A5100: MOVE [XWD 0,-1] ;LOAD E WITH A NON-ZERO VALUE + MOVS ;*MOVS SHOULD RESULT IN C(AC) NON-ZERO + SKIPN ;SKIP HALT INSTRUCTION IF C(AC) OF MOVS IS NON-ZERO + STOP + +;********** +;THIS TEST VERIFIES THAT A MOVS EXECUTED ON AN E CONTAINING A NON-ZERO NUMBER +;RESULTS IN A NON-ZERO NUMBER IN THE AC. FIRST A NON-ZERO WORD +;IS LOADED INTO E; THEN, MOVS IS EXECUTED. THE AC IS CHECKED FOR A NON-ZERO +;RESULT. IF C(AC)=0, THE TEST FAILS. + +A5200: MOVE [XWD -1,0] ;LOAD E OF MOVS WITH A NON-ZERO NUMBER + MOVS ;*MOVS SHOULD RESULT WITH C(AC) NON-ZERO + SKIPN ;SKIP HALT INSTRUCTION IF C(AC) OF MOVS IS NON-ZERO + STOP + +;********** + +;THIS TEST VERIFIES THAT THE MOVS INSTRUCTION SWAPS BITS 0 AND 188 OF E +;AND STORES THEM IN THE AC UNLESS AR BIT 0 IS S-A-0. FIRST, A WORD +;WITH BIT 0 SET TO A ONE AND BIT 18 CLEARED IS LOADED INTO E. THEN, MOVS +;IS EXECUTED. THE AC IS THEN CHECKED FOR A 0 IN BIT 0, WHICH IS THE +;CORRECT RESULT. + +A5300: MOVE [XWD -1,0] ;LOAD E OF MOVS WITH 777777000000 + MOVS ;*MOVS SHOULD RESULT WITH AC BIT 0 EQUAL TO 0 + SKIPG ;SKIP HALT INSTRUCTION IF BIT 0 OF AC IS 0 + STOP + +;MOVSX: AR SWAP [ARMB], AR LT AR RT A [ARMB], AR LT AR RT B [ARMB], +;AR RT AR LT A [ARMB], AR RT AR LT B [ARMB], ET2 A F/F [E], IR MOVSX [IR1], +;IR MOV(E,S)X [IR1] + +;********** + +;THIS TEST VERIFIES THAT THE MOVS INSTRUCTION SWAPS BITS 0 AND 18 OF E AND +;STORES THEM IN THE AC UNLESS AR BIT 0 IS S-A-1. FIRST, A WORD WITH BIT 0 +;SET TO A ONE AND BIT 18 CLEARED IS LOADED INTO E. THEN, MOVS IS EXECUTED. +;THE AC IS CHECKED FOR A ZERO IN BIT ZERO, WHICH IS THE CORRECT RESULT. + +A5400: MOVE [XWD 0,-1] ;LOAD E OF MOVS WITH 777777 + MOVS ;*MOVS SHOULD RESULT WITH BIT 0 OF THE AC SET + SKIPL ;PASS IF AC BIT 0 IS SET + STOP + +;********** +SN=5500 + ZZ=0 + +A5500: REPEAT ^D18, +<;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, +;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. +;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES +;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT +;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [0] ;CLEAR AC + MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + STOP + +;********** +> +SN=5600 + +A5600: REPEAT ^D18, +<;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL +;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS +;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS +;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT +;HALF OF THE AC IS TESTED WITH A 1. + +SN=SN+1 + ZZ=ZZ+ZZ + MOVE [0] ;CLEAR AC + MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + STOP + +;********** +> +SUBTTL TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) + +;THIS TEST VERIFIES THAT SETM LOADS AC FROM E. THE AC IS FIRST SET +;TO ZERO; THEN, SETM IS EXECUTED WITH ALL ONES IN E. BIT 0 OF THE AC +;IS THEN CHECKED FOR A 1. + +A5700: MOVE [0] ;PRELOAD AC WITH ALL ZEROS + SETM [-1] ;*SETM SHOULD LOAD AC WITH ALL ONES + SKIPL ;PASS TEST IF AC BIT 0 IS A ONE + STOP +;FCE [F1], IR BOOLE [IR1], IRXX(0,5,12,17) [IR2] + +;********** + +;THIS TEST VERIFIES THAT SETM LOADS AC FROM E. THE AC IS FIRST SET TO +;ALL ONES; THEN, SETM IS EXECUTED WITH ALL ZEROS IN E. THE AC IS CHECKED +;FOR ALL ZEROS + +A6000: MOVE [-1] ;PRESET AC TO ALL ONES + SETM [0] ;*SETM SHOULD LOAD AC WITH ALL ONES + SKIPE ;PASSES TEST IF C(AC) = 0 + STOP + +;AD AR + EN [ADAP], AD AR + EN C[ADAP], +;F CYC ACT EN A[F1], IR BOOLE AD A +F[IR2], +;IR BOOLE [IR1], IRXX(1-3,6,11,15,16) [IR2] + +;********** + +;THIS TEST VERIFIES THAT SETZ LOADS TH AC WITH ALL ZEROS. FIRST, A WORD +;OF ALL ZEROS IS LOADED INTO THE AC VIA A MOVE INSTC. THEN, SETZ IS +;EXECUTED. THE AC IS CHECKED FOR ALL ZEROS. +;IF ANY BITS IN THE AC ARE SET, THE TEST FAILS. + +A6100: MOVE [0] ;RESET AC TO ALL ZEROS + SETZ ;*SETZ SHOULD CLEAR THE AC + SKIPE ;PASS TEST IF C(AC) = 0 + STOP + +;AD MASK GEN EN [ADC1], AD MASK GEN F/F'S [ADC1] +;AD MASK EN B; F CYC ACT EN B[F1]; IRXX00 [IR1]; +;IR BOOLE [IR1] + +;********** +;THIS TEST VERIFIES THAT SETZ LOADS THE AC WITH ALL ZEROS. FIRST, +;THE AC IS SET TO ALL ONES; THEN, SETZ IS EXECUTED. THE AC IS THEN +;CHECKED FOR ALL ZEROS. IF ANY BITS IN THE AC ARE SET, THE TEST FAILS. + +A6200: MOVE [-1] ;PRESET AC TO ALL ONES + SETZ . ;*SETZ SHOULD CLEAR THE AC + SKIPE ;PASS TEST IF C(AC) = 0 + STOP + +;AR AD EN [ARMA], AR AD ET2 EN [ARMA], IR BOOLE [IR] + +;********** + +;THIS TEST VERIFIES THAT THE AND INSTRUCTION OPERATING ON AN AC +;CONTAINING ALL ZEROS AND AN E CONTAINING ALL ZEROS RESULTS IN AN AC +;OF ALL ZEROS. THE AC AND E ARE THE SAME IN THIS TEST. +;AC, E ARE LOADED WITH ZEROS; THEN, AND IS EXECUTED. THE AC IS THEN +;CHECKED FOR ALL ZEROS. IF ANY BITS IN THE AC ARE SET BY THE AND INSTRUCTION, +;THE TEST FAILS. + + +A6300: MOVE [0] ;PRESET AC, E TO ALL ZEROS + AND ;*AND SHOULD RESULT IN C(AC) = 0 + SKIPE ;PASS TEST IF C(AC) = 0 + STOP + +;********** +SN=6400 + ZZ=0 + +A6400: REPEAT ^D36, +<;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION +;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH +;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X +;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. +;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, +;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS +;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + SETZ ;PRESET E TO ALL ZEROS + MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + ;WHERE X VARIES FROM 0 THRU 35 + AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + SKIPE 1 ;FAIL IF C(AC) NON-ZERO + STOP + SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + STOP +;********** +> +SN=6500 + ZZ=0 + +A6500: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF +;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE +;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS +;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED +;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + ;POSITION WITH ITSELF SHOULD RESULT WITH + ;C(AC) NONZERO + SKIPN ;PASS TEST IF C(AC) NONZERO + STOP + +;********** +> +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE +;C(AC)=0 AND C(E)=0 AND E OUTSIDE OF THE AC RANGE IS ALL ZEROS IN THE AC. +;FIRST, THE AC IS CLEARED; THEN, THE C(AC) IS ANDED WITH C(E) WHEN +;E GREATER THAN 20 AND C(E)=0. THE RESULT SHOULD BE C(AC)=0. +;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A6600: MOVE [0] ;PRESET AC TO ALL ZEROS + AND [0] ;*AND OF TWO WORDS OF ALL ZEROS + ;SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(AC)=0 +;AND C(E) IS ALL ONES IS AN AC CONTAINING ALL ZEROS. FIRST, THE AC IS +;CLEARED; THEN THE C(AC) IS ANDED WITH AN E CONTAINING ALL ONES, WHERE +;E IS BEYOND THE AC RANGE. THE RESULT SHOULD BE C(AC)=0. +;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A6700: MOVE [0] ;PRESET AC TO ALL ZEROS + AND [-1] ;*AND OF TWO WORDS WHERE C(AC)=0 SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;AD FM+EN [ADFP],AD FM+EN A [ADFP], +;F CYC ACT EN C [F1],IR BOOLE AD FM+F [IR2], +;IR BOOLE [IR1],IRXX(1,4,5,11,13,16) [IR2] + +;********** +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(E)=0 +;AND C(AC) IS ALL ONES IS AN AC CONTAINING ALL ZEROS. FIRST, THE +;AC IS SET TO ALL ONES; THEN, THE AC IS ANDED WITH AN E CONTAINING ALL +;ZEROS, WHERE E IS BEYOND THE AC RANGE. THE RESULT SHOULD BE C(AC)=0. +;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A7000: MOVE [-1] ;PRESET AC TO ALL ONES + AND [0] ;*AND OF TWO WORDS WHERE C(E)=0 SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(E) IS +;ALL ONES AND C(AC) IS ALL ONES IS AN AC CONTAINING ALL ONES. FIRST, THE +;AC IS SET TO ALL ONES; THEN, C(AC) IS ANDED WITH AN E CONTAINING ALL +;ONES. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +;IF C(AC) IS ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A7100: MOVE [-1] ;PRELOAD AC WITH ALL ONES + AND [-1] ;*AND OF TWO WORDS OF ALL ONES SHOULD + ;RESULT IN C(AC)=ALL ONES + SKIPN ;PASS TEST IF C(AC) IS NON ZERO + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHERE CONTENTS IS NON-ZERO. +;FIRST, THE AC IS CLEARED, THEN, XOR IS EXECUTED WHERE C(E) IS NON-ZERO. +;THE AC IS THEN CHECKED FOR NON-ZERO CONTENTS. +A7200: MOVE [0] ;CLEAR AC + XOR [707070707070] ;*XOR WITH C(AC)=0 AND C(E)=7070707070 SHOULD + ;RESTULT IN C(AC)=707070707070 + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +SN=7300 + ZZ=0 + +A7300: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS +;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED +;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. +;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [0] ;CLEAR AC + XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + ;RESULT IN C(AC) NONZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +> +SN=7400 + ZZ=0 + +A7400: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE +;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS +;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE +;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E +;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + ;RESULT IN C(AC) NON-ZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], +;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], +;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], +;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + +;********** +> +SN=7500 + ZZ=0 + +A7500: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION +;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, +;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT +;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME +;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST +;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY +;INDICATE AN ILLEGAL CARRY GENERATION. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +> +;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION WHERE +;C(AC)=C(E)=0 IS AN AC CONTAINING ALL ZEROS; FIRST THE AC IS CLEARED; THEN +;XOR IS EXECUTED WITH C(E)=0. THE RESULT IN THE AC IS CHECKED FOR ALL ZEROS + +A7600: MOVE [0] ;INITIALIZE AC TO ALL ZEROS + XOR [0] ;*XOR WITH C(AC)=C(E)=0 SHOULD RESULT INC(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE +;C(AC)=0 AND C(E) ALL ONES IS AN AC CONTAINING ALL ZEROS SINCE +;THE EQUIVALENCE FUNCTION SHOULD SET ONLY LIKE BITS. FIRST, THE AC +;SHOULD BE CLEARED; THEN, EQV IS EXECUTED WITH E CONTAINING ALL ONES. +;THE AC IS THEN CHECKED FOR ALL ZEROS + +A7700: MOVE [0] ;CLEAR THE AC + EQV [-1] ;*EQV WITH C(AC)=0 AND C(E) ALL ONES SHOULD + ;RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +SN=10000 + ZZ=0 + +A10000: REPEAT ^D18, +<;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN +;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE +;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT +;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS +;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE +;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO +SN=SN+1 + ZZ=ZZ+ZZ+1 + IFE , + MOVE [0] ;PRESET AC TO ALL ZEROS + EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + ;RESULT IN C(AC) NON-ZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +> +SN=10100 + ZZ=0 + +A10100: REPEAT ^D18, +<;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN +;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE +;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT +;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS +;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE +;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO +SN=SN+1 + ZZ=ZZ+ZZ+1 + IFE , + MOVE [0] ;PRESET AC TO ALL ZEROS + EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + ;RESULT IN C(AC) NON-ZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +> +;THIS TEST VERIFIES THAT THE RESULT OF AN AC CONTAINING ALL ZEROS +;EQUIVALENCED WITH AN E CONTAINING ALL ZEROS IS AN AC CONTAINING ALL +;ONES; AND FUTHERMORE, IF THIS AC CONTAINING ALL ONES IS EQUIVALANCED +;WITH AN E CONTAINING ALL ZEROS, THE FINAL RESULT IN THE AC +;SHOULD BE ALL ZEROS. FIRST, THE AC IS INITIALIZED TO ALL ZEROS, THEN, +;THIS AC IS EQUIVALENCED TWICE WITH AN E CONTAINING ALL ZEROS. THE AC IS +;THEN CHECKED FOR ALL ZEROS + +A10200: MOVE [0] ;CLEAR THE AC + EQV [0] ;*EQV SHOULD RESULT IN C(AC)=ALL ONES + EQV [0] ;*EQV SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF AN AC CONTAINING ALL ONES +;EQUIVALENCED WITH AN E CONTAINING ALL ONES IS AN AC CONTAINING ALL ONES; +;AND FUTHERMORE, IF THIS AC CONTAINING ALL ONES ISEQUIVALENCED WITH AN E +;CONTAINING ALL ZEROS, THE FINAL RESULT IN THE AC SHOULD BE ALL ZEROS. +;FIRST, THE AC IS INITIALIZED TO ALL ONES; THEN, THIS AC IS +;EQUIVALENCED WITH AN E OF ALL ONES. NEXT, THIS SAME AC IS EQUIVALENCED +;WITH AN E OF ALL ZEROS. THE AC IS THEN CHECKED FOR ALL ZEROS. + +A10300: MOVE [-1] ;SET C(AC) TO ALL ONES + EQV [-1] ;*EQV SHOULD RESULT IN C(AC) OF ALL ONES + EQV [0] ;*EQV SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +SN=13000 + ZZ=0 + +A13000: REPEAT ^D36, +<;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. +;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN +;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). +;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD +;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, +;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. +;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO +;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + EQV [0] ;RESULTS IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +> +ENDXX: JRST BEGEND ;LOOP PROGRAM + + diff --git a/apps/pdp10/tapes/diags/klad/dakaa/DAKAA.SEQ.txt b/apps/pdp10/tapes/diags/klad/dakaa/DAKAA.SEQ.txt new file mode 100644 index 000000000..a28dc653f --- /dev/null +++ b/apps/pdp10/tapes/diags/klad/dakaa/DAKAA.SEQ.txt @@ -0,0 +1,50131 @@ +MAINDEC-10-DAKAA + SEQ 0001 + + + + + + + IDENTIFICATION + -------------- + + PRODUCT CODE: MAINDEC-10-DAKAA-B-D + + PRODUCT NAME: DECSYSTEM10 PDP-10 KA10 BASIC + INSTRUCTION DIAGNOSTIC (1) + + FUNCTION: BASIC INSTRUCTIONS + + VERSION: 0.2 + + DATE RELEASED: JANUARY 1977 + + MAINTAINED BY: DIAGNOSTIC ENGINEERING GROUP + + AUTHOR: JOHN R. KIRCHOFF + +COPYRIGHT(C) 1976,1977 +DIGITAL EQUIPMENT CORPORATION +MARLBORO, MASS. 01752 + +THIS SOFTWARE IS FURNISHED UNDER A LICENSE FOR USE ONLY +ON A SINGLE COMPUTER SYSTEM AND MAY BE COPIED ONLY WITH +THE INCLUSION OF THE ABOVE COPYRIGHT NOTICE. THIS SOFTWARE, +OR ANY OTHER COPIES THEREOF, MAY NOT BE PROVIDED OR OTHERWISE +MADE AVAILABLE TO ANY OTHER PERSON EXECPT FOR USE ON SUCH SYSTEM +AND TO ONE WHO AGREES TO THESE LICENSE TERMS. TITLE TO AND +OWNERSHIP OF THE SOFTWARE SHALL AT ALL TIMES REMAIN IN DEC. + +THE INFORMATION IN THIS DOCUMENT IS SUBJECT TO CHANGE WITHOUT +NOTICE AND SHOULD NOT BE CONSTRUED AS A COMMITMENT BY DIGITAL +EQUIPMENT CORPORATION. + +DEC ASSUMES NO RESPONSIBILITY FOR THE USE OR RELIABILITY OF ITS +SOFTWARE ON EQUIPMENT WHICH IS NOT SUPPLIED BY DEC. + MAINDEC-10-DAKAA.TXT + PAGE 2 SEQ 0002 + + + TABLE OF CONTENTS + ----------------- + +1.0 ABSTRACT + +2.0 REQUIREMENTS + +2.1 EQUIPMENT + +2.2 STORAGE + +2.3 PRELIMINARY PROGRAMS + +3.0 PROGRAM PROCEDURES + +3.1 LOADING PROCEDURE + +3.2 STARTING PROCEDURE + +3.3 OPERATING PROCEDURE + +4.0 ERRORS + +5.0 ITERATION COUNTER + +6.0 CYCLE TIME + +7.0 OPERATIONAL VARIATIONS + +8.0 MISCELLANEOUS + +9.0 LISTING + MAINDEC-10-DAKAA.TXT + PAGE 3 SEQ 0003 + + +1.0 ABSTRACT + + THIS PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC IS THE FIRST IN A + SERIES OF PDP-10 KA10 PROCESSOR DIAGNOSTICS. IT IS THE MOST + BASIC OF THE PROCESSOR DIAGNOSTICS. THIS DIAGNOSTIC + ASSUMES THE HALT INSTRUCTION AND THE COMPUTER CONSOLE + TO BE OPERATIVE. IT MAKES NO FURTHER ASSUMPTIONS EXCEPT + THAT IT IS LOADED INTO MEMORY CORRECTLY. + + THE DIAGNOSTIC TESTS SOME OF THE MOVE AND SKIP INSTRUCTIONS. + IT ALSO TESTS THE ADDERS AND OTHER LOGIC WITH AND, XOR, EQV, + AND ALL BOOLE INSTRUCTIONS. + +2.0 REQUIREMENTS + +2.1 EQUIPMENT + + A PDP-10 KA10 WITH A MINIMUM OF 32K OF MEMORY + + PAPER TAPE READER + DECTAPE (OPTIONAL) + CONSOLE TELETYPE + +2.2 STORAGE + + THE PROGRAM RUNS WITHIN 32K OF MEMORY. + +2.3 PRELIMINARY PROGRAMS + + CONSOLE FUNCTIONS WORKING PROPERLY + PAPER TAPE OR DECTAPE READ-IN WORKING PROPERLY + MAINDEC-10-DAKAA.TXT + PAGE 4 SEQ 0004 + + +3.0 PROGRAM PROCEDURES + +3.1 LOADING PROCEDURE + + PAPER TAPE - HARDWARE READ-IN (READER DEVICE CODE 104) + DECTAPE - LOAD WITH DIAMON (DECTAPE DEVICE CODE 320) + +3.2 STARTING PROCEDURE + + STAND-ALONE STARTING ADDRESS IS 30000. + + IF THE DIAGNOSTIC FAILS TO START CORRECTLY TRY STARTING AT THE + FIRST TEST INSTEAD OF AT THE BEGINNING OF THE CONTROL SEQUENCE. + (SEE LISTING). + +3.3 OPERATING PROCEDURE + + ONCE STARTED THE PROGRAM WILL CYCLE CONTINUALLY UNTIL STOPPED + OR AN ERROR OCCURS. + +4.0 ERRORS + + ERRORS ARE IN THE FORM OF HALT INSTRUCTIONS. THE LISTING + SHOULD BE CONSULTED TO DETERMINE THE CAUSE OF THE ERROR. A + NO OPERATION (JUMP) INSTRUCTION FOLLOWS EACH HALT. THIS + MAY BE USEFUL IN CONSTRUCTING A SCOPE LOOP TO CYCLE ON THE + FAILING INSTRUCTION. + +5.0 ITERATION COUNTER + + THE ITERATION COUNT OF THE PROGRAM IS DISPLAYED IN THE MEM- + ORY INDICATORS (MI). THIS COUNT IS A DECREMENTING COUNT AND + INITIALLY STARTS AT -1 IN STAND-ALONE OPERATION. + +6.0 CYCLE TIME + + THE CYCLE TIME OF THE PROGRAM IS IN THE MILLISECOND RANGE AND + IS THEREFORE SUITABLE FOR TAKING MARGINS, VIBRATION TESTS, ETC. + + MAINDEC-10-DAKAA.TXT + PAGE 5 SEQ 0005 + + +7.0 OPERATIONAL VARIATIONS + + A. DIAGNOSTIC MONITOR + + THE PROGRAM IS USABLE WITH THE DIAGNOSTIC MONITOR TO PRO- + VIDE RELIABILITY TESTS, ACCEPTANCE TESTS, AND/OR TO PRO- + VIDE A QUICK METHOD OF ISOLATION OF A FAULT TO A PARTICULAR + AREA OF THE PROCESSOR. CERTAIN PROCEDURES ARE USED WHEN + THE PROGRAM IS USED IN THIS MANNER. THEY ARE: + + 1. THE DIAGNOSTIC MONITOR TRANSFERS CONTROL TO THE PRO- + GRAM AND STARTS IT AT LOCATION 30002. + + 2. MONCTL - LOCATION 30043 IS USED AS THE DIAGNOSTIC MON- + ITOR CONTROL FLAG WORD. + + B. USER MODE + + THE PROGRAM WILL OPERATE IN USER MODE AND AS SUCH PROVIDES + ASSURANCE THAT THE PROCESSOR IS PERFORMING ALL FUNCTIONS + CORRECTLY. USER MODE STARTING ADDRESS IS 30000. + + C. SYSTEM EXERCISER + + STARTING ADDRESS IS 30003. NO DATA SWITCHES ARE USED BY + THIS PROGRAM. + +8.0 MISCELLANEOUS + + NONE + +9.0 LISTING + THIS IS A HISTORY OF THE DEVELOPMENT OF MAINDEC-10-DAKAA + SEQ 0006 +************************************************************************ + +PRODUCT CODE: MAINDEC-10-DAKAA + +PRODUCT NAME: BASIC INSTRUCTION DIAGNOSTIC #1 + +DATE RELEASED: JANUARY 1977 + +VERSION: 0.2 + +UPDATE AUTHOR: JOHN R. KIRCHOFF + +CHANGES MADE: + + 1. UPGRADE TO ALLOW COMPATABILITY WITH THE SUBROUTINE PACKAGE. + +************************************************************************ + +ORIGINAL VERSION: 0.1 + +ORIGINAL AUTHOR: RICHARD MALISKA + +ORIGINAL RELEASE: 16-MAR-72 + +************************************************************************ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 1 +DAKAAT MAC 19-JAN-77 13:31 DIAGNOSTIC PARAMETERS SEQ 0007 + + 1 ;DAKAA + 2 + 3 + 4 + 5 000002 DECVER==2 + 6 000000 MCNVER==0 + 7 + 8 XLIST + 9 LIST + 10 LALL + 11 NAME \MCNVER,\DECVER^ + 12 + 13 TITLE DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 + 14 ^ + 15 + 16 ;TEST DESIGNED FOR INITIAL DEBUGGING OF PROCESSOR HARDWARE + 17 ;AND TO DETECT (SOLID) FAILURES IN THE FIELD. + 18 + 19 ;COPYRIGHT 1972,1977 + 20 ;DIGITAL EQUIPMENT CORPORATION + 21 ;MARLBORO, MASS. 01752 + 22 + 23 ;JOHN R. KIRCHOFF + 24 ;DICK MALISKA + 25 + 26 000137 LOC 137 + 27 000137 000000 000002 MCNVER,,DECVER + 28 + 29 NOSYM +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 2 +DAKAAT MAC 19-JAN-77 13:31 DIAGNOSTIC PARAMETERS SEQ 0008 + + 30 SUBTTL DIAGNOSTIC PARAMETERS + 31 + 32 ;PARAMETER DEFINITIONS + 33 000001 EXCASB==1 + 34 000001 USRASB==1 + 35 000001 KA10==1 + 36 + 37 ;FLAG DEFINITIONS + 38 010000 USERF=10000 ;USER MODE FLAG + 39 + 40 + 41 ;MACROS + 42 + 43 ; STOP - USED FOR SCOPE LOOP, IF INSTRUCTION FAILS, CHANGE (JUMPA .+1) + 44 ; TO A (JUMPA .-X) TO CYCLE ON FAILING INSTRUCTION + 45 + 46 DEFINE STOP (A)< + 47 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 48 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 49 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 50 ;IN THE SUBTEST) TO LOOP ON ERROR> + 51 + 52 ;SPECIAL FEATURE PARAMETERS + 53 + 54 030742 SADR1=A00 + 55 030742 SADR2=A00 + 56 030742 SADR3=A00 + 57 030742 SADR4=A00 + 58 254000 030742 SADR5=JRST A00 + 59 254000 030742 SADR6=JRST A00 + 60 254000 030742 SADR7=JRST A00 + 61 254000 030742 SADR8=JRST A00 + 62 254000 030742 SADR9=JRST A00 + 63 254000 030742 SADR10=JRST A00 + 64 254000 030742 SADR11=JRST A00 + 65 + 66 000000 PAREA0=0 + 67 000000 PAREA1=0 + 68 000000 PAREA2=0 + 69 444153 414100 PAREA3=SIXBIT/DAKAA/ + 70 645560 000000 PAREA4=SIXBIT/TMP/ + 71 000000 PAREA5=0 + 72 000000 PAREA6=0 + 73 001000 ITERAT==1000 + 74 777777 777777 PGMEND==-1 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 1 +PARAM KLM 18-JAN-77 11:38 *PARAM* CONSOLE DATA SWITCH ASSIGNMENTS, JAN 18,1977 SEQ 0009 + + 75 SUBTTL *PARAM* CONSOLE DATA SWITCH ASSIGNMENTS, JAN 18,1977 + 76 + 77 DEFINE S,<;*********************************************************************> + 78 + 79 S^;*********************************************************************^ + 80 ;*DATA SWITCHES (READ FROM CONSOLE IN EXEC MODE OR TYPED IN IN USER MODE) + 81 ;*LEFT HALF SWITCHES ARE PRE-ASSIGNED FOR SUBROUTINE PACKAGE USE + 82 ;*AND CONTROL LOOPING, PRINTING (TTY OR OTHER DEVICE) AND MISC. FUNCTIONS + 83 S^;*********************************************************************^ + 84 + 85 400000 ABORT== 400000 ;ABORT PROGRAM ON PASS COMPLETION + 86 200000 RSTART==200000 ;RESTART TEST, PRINT TOTALS + 87 100000 TOTALS==100000 ;PRINT TOTALS, CONTINUE + 88 + 89 040000 NOPNT== 040000 ;INHIBIT ALL PRINT/TYPE OUT (EXCEPT FORCED) + 90 020000 PNTLPT==020000 ;PRINT ALL DATA ON LPT (LOGICAL DEVICE, USER MODE) + 91 010000 DING== 010000 ;RING BELL ON ERROR + 92 + 93 004000 LOOPER==004000 ;ENTER EXERCISE/CHECK LOOP ON ERROR + 94 002000 ERSTOP==002000 ;HALT ON TEST ERROR + 95 001000 PALERS==001000 ;PRINT ALL ERRORS + 96 + 97 000400 RELIAB==000400 ;RELIABILITY MODE + 98 000200 TXTINH==000200 ;INHIBIT ERROR TEXT + 99 000100 INHPAG==000100 ;INHIBIT PAGING + 100 + 101 000040 MODDVC==000040 ;MODIFY DEVICE CODE + 102 000020 INHCSH==000020 ;INHIBIT CACHE + 103 000010 OPRSEL==000010 ;OPERATOR SELECTION + 104 + 105 000004 CHAIN== 000004 ;CHAIN CONTROL SWITCH + 106 + 107 000002 KAHZ50==000002 ;KA10 50 HERTZ POWER + 108 + 109 ;SWITCH 17 RESERVED !!! +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 2 +PARAM KLM 18-JAN-77 11:38 *PARAM* PROGRAM/SUBROUTINE PARAMETERS, JAN 18,1977 SEQ 0010 + + 110 SUBTTL *PARAM* PROGRAM/SUBROUTINE PARAMETERS, JAN 18,1977 + 111 + 112 S^;*********************************************************************^ + 113 ;*SPECIAL SUBPROGRAM LINKAGES + 114 S^;*********************************************************************^ + 115 + 116 027772 FSELNK= 27772 ;FILE SELECT LINK + 117 027773 FRDLNK= 27773 ;FILE READ LINK + 118 027774 LDLNK= 27774 ;LOAD LINKAGE ADDRESS + 119 027775 DDTLNK= 27775 ;DDT LINKAGE ADDRESS + 120 027776 MODLNK= 27776 ;OPERATIONAL MODE CHECK LINKAGE ADDRESS + 121 027777 SUBLNK= 27777 ;SUBROUTINE LINKAGE ADDRESS + 122 + 123 S^;*********************************************************************^ + 124 ;*SPECIAL SUBROUTINE FATAL HALTS + 125 ;*USED TO REPORT ERRORS THAT CAUSE THE SUBROUTINES TO BE UNUSABLE + 126 S^;*********************************************************************^ + 127 + 128 ;ADDRESS TAG REASON + 129 ;--------------------- + 130 + 131 ; 1010 NOEXEC ;PROGRAM NOT CODED FOR EXEC MODE OPERATION + 132 ; 1011 PLERR ;FATAL PUSH LIST POINTER ERROR + 133 ; 1012 PLERR1 ;INITIAL PUSH LIST POINTER ERROR + 134 ; 1013 MUOERR ;MUUO WITH LUUO HANDLER WIPED OUT + 135 ; 1014 DTEBER ;DTE20 INTERRUPT WITHOUT DOORBELL + 136 ; 1015 DTECER ;DTE20 CLOCK INTERRUPT WITHOUT FLAG SET + 137 ; 1016 CPIERR ;CPU INITIALIZATION ERROR + 138 ; 1017 EOPERR ;END OF PROGRAM ERROR + 139 ; 1020 LUOERR ;INTERRUPT WITH LUUO HANDLER WIPED OUT + 140 + 141 S^;*********************************************************************^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 3 +PARAM KLM 18-JAN-77 11:38 *PARAM* PROGRAM/SUBROUTINE PARAMETERS, JAN 18,1977 SEQ 0011 + + 142 S^;*********************************************************************^ + 143 ;OPERATOR DEFINITIONS (NON-UUO'S) + 144 S^;*********************************************************************^ + 145 + 146 260740 000000 OPDEF GO [PUSHJ P,] ;SUBROUTINE CALL + 147 263740 000000 OPDEF RTN [POPJ P,] ;SUBROUTINE RETURN + 148 261740 000000 OPDEF PUT [PUSH P,] ;PUT DATA ON PUSH LIST + 149 262740 000000 OPDEF GET [POP P,] ;GET DATA FROM PUSH LIST + 150 254000 000000 OPDEF PJRST [JRST ] ;JRST TO ROUTINE THAT RTN'S + 151 254200 000000 OPDEF HALT [JRST 4,] ;DEFINITION FOR DDT + 152 254100 000000 OPDEF JRSTF [JRST 2,] ;DEFINITION FOR DDT + 153 254500 000000 OPDEF JEN [JRST 12,] ;DEFINITION FOR DDT + 154 + 155 S^;*********************************************************************^ + 156 ;*SUBROUTINE INITIALIZATION CALL + 157 S^;*********************************************************************^ + 158 + 159 265000 030011 OPDEF PGMINT [JSP 0,SBINIT] ;SUBROUTINE INITIALIZATION + 160 + 161 S^;*********************************************************************^ + 162 ;*HALTING UUO'S (A MORE GRACEFUL HALT THAN SIMPLY USING THE HALT INSTRUCTION). + 163 S^;*********************************************************************^ + 164 + 165 037640 000004 OPDEF FATAL [37B8!15B12!4] ;FATAL PROGRAMMING HALT + 166 037600 000004 OPDEF ERRHLT [37B8!14B12!4] ;PROGRAM ERROR HALT + 167 + 168 S^;*********************************************************************^ + 169 ;*TERMINAL INPUT UUO'S + 170 ;*ALWAYS COME FROM THE CONSOLE TERMINAL IN EXEC MODE OR THE + 171 ;*CONTROLLING TERMINAL (REAL TERMINAL OR PTY) IN USER MODE. + 172 S^;*********************************************************************^ + 173 + 174 037000 000003 OPDEF TTICHR [37B8!0B12!3] ;TTY, INPUT ANY CHARACTER + 175 037040 000003 OPDEF TTIYES [37B8!1B12!3] ;TTY, NORMAL RETURN Y + 176 037100 000003 OPDEF TTINO [37B8!2B12!3] ;TTY, NORMAL RETURN N + 177 037140 000003 OPDEF TTIOCT [37B8!3B12!3] ;TTY, INPUT OCTAL WORD + 178 037200 000003 OPDEF TTIDEC [37B8!4B12!3] ;TTY, INPUT DECIMAL WORD + 179 037240 000003 OPDEF TTICNV [37B8!5B12!3] ;TTY, INPUT CONVERTABLE WORD + 180 037300 000003 OPDEF TTLOOK [37B8!6B12!3] ;TTY, KEYBOARD CHECK + 181 037340 000003 OPDEF TTALTM [37B8!7B12!3] ;TTY, ALT-MODE CHECK + 182 037400 000003 OPDEF TTSIXB [37B8!10B12!3] ;TTY, INPUT SIXBIT WORD + 183 037440 000003 OPDEF TTYINP [37B8!11B12!3] ;TTY, IMAGE MODE INPUT +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 4 +PARAM KLM 18-JAN-77 11:38 *PARAM* PROGRAM/SUBROUTINE PARAMETERS, JAN 18,1977 SEQ 0012 + + 184 ;*TERMINAL OUTPUT UUO'S. + 185 + 186 037000 000000 OPDEF PNTA [37B8!0B12!0] ;PRINT ASCII WORD + 187 037000 000001 OPDEF PNTAF [37B8!0B12!1] ;PRINT ASCII WORD FORCED + 188 037740 000000 OPDEF PNTAL [37B8!17B12!0] ;PRINT ASCIZ LINE + 189 037740 000001 OPDEF PNTALF [37B8!17B12!1] ;PRINT ASCIZ LINE FORCED + 190 037600 000003 OPDEF PSIXL [37B8!14B12!3] ;PRINT SIXBIT'Z LINE + 191 037640 000003 OPDEF PSIXLF [37B8!15B12!3] ;PRINT SIXBIT'Z LINE FORCED + 192 037000 000000 OPDEF PNTMSG [37B8!0B12!0] ;PRINT MESSAGE IMMEDIATE + 193 037040 000000 OPDEF PNTMSF [37B8!1B12!0] ;PRINT MESSAGE IMMEDIATE FORCED + 194 037100 000000 OPDEF PSIXM [37B8!2B12!0] ;PRINT SIXBIT'Z MSG IMMEDIATE + 195 037200 000000 OPDEF PSIXMF [37B8!4B12!0] ;PRINT SIXBIT'Z MSG IMM FORCED + 196 037000 000000 OPDEF PNTCI [37B8!0B12!0] ;PRINT CHARACTER IMMEDIATE + 197 037040 000000 OPDEF PNTCIF [37B8!1B12!0] ;PRINT CHARACTER IMMEDIATE FORCED + 198 037500 000000 OPDEF PNTCHR [37B8!12B12!0] ;PRINT CHARACTER + 199 037500 000001 OPDEF PNTCHF [37B8!12B12!1] ;PRINT CHARACTER FORCED + 200 037040 000000 OPDEF PNT1 [37B8!1B12!0] ;PRINT ONE OCTAL DIGIT + 201 037040 000001 OPDEF PNT1F [37B8!1B12!1] ;PRINT 1 OCTAL DIGIT FORCED + 202 037100 000000 OPDEF PNT2 [37B8!2B12!0] ;PRINT TWO OCTAL DIGITS + 203 037100 000001 OPDEF PNT2F [37B8!2B12!1] ;PRINT 2 OCTAL DIGITS FORCED + 204 037140 000000 OPDEF PNT3 [37B8!3B12!0] ;PRINT THREE OCTAL DIGITS + 205 037140 000001 OPDEF PNT3F [37B8!3B12!1] ;PRINT THREE OCTAL DIGITS FORCED + 206 037200 000000 OPDEF PNT4 [37B8!4B12!0] ;PRINT FOUR OCTAL DIGITS + 207 037200 000001 OPDEF PNT4F [37B8!4B12!1] ;PRINT FOUR OCTAL DIGITS FORCED + 208 037240 000000 OPDEF PNT5 [37B8!5B12!0] ;PRINT FIVE OCTAL DIGITS + 209 037240 000001 OPDEF PNT5F [37B8!5B12!1] ;PRINT FIVE OCTAL DIGITS FORCED + 210 037300 000000 OPDEF PNT6 [37B8!6B12!0] ;PRINT SIX OCTAL DIGITS + 211 037300 000001 OPDEF PNT6F [37B8!6B12!1] ;PRINT SIX OCTAL DIGITS FORCED + 212 037340 000000 OPDEF PNT7 [37B8!7B12!0] ;PRINT 7 OCTAL DIGITS + 213 037340 000001 OPDEF PNT7F [37B8!7B12!1] ;PRINT 7 OCTAL DIGITS FORCED + 214 037440 000000 OPDEF PNT11 [37B8!11B12!0] ;PRINT 11 OCTAL DIGITS + 215 037440 000001 OPDEF PNT11F [37B8!11B12!1] ;PRINT 11 OCTAL DIGITS FORCED. + 216 037400 000000 OPDEF PNTADR [37B8!10B12!0] ;PRINT PHYSICAL ADDRESS + 217 037400 000001 OPDEF PNTADF [37B8!10B12!1] ;PRINT PHYSICAL ADDRESS FORCED + 218 037600 000000 OPDEF PNTOCT [37B8!14B12!0] ;PRINT FULL WORD OCTAL + 219 037600 000001 OPDEF PNTOTF [37B8!14B12!1] ;PRINT FULL WORD OCTAL FORCED + 220 037540 000000 OPDEF PNTHW [37B8!13B12!0] ;PRINT OCTAL HALF WORDS, 6 SP 6 + 221 037540 000001 OPDEF PNTHWF [37B8!13B12!1] ;PRINT OCTAL HALF WORDS, 6 SP 6 FORCED + 222 037700 000003 OPDEF PNTOCS [37B8!16B12!3] ;PRINT OCTAL, SUPPRESS LEADING 0'S + 223 037740 000003 OPDEF PNTOCF [37B8!17B12!3] ;PRINT OCTAL, SUPPRESS LEADING 0'S FORCED + 224 037640 000000 OPDEF PNTDEC [37B8!15B12!0] ;PRINT DECIMAL, SUPRESS LEADING 0'S + 225 037640 000001 OPDEF PNTDCF [37B8!15B12!1] ;PRINT DECIMAL, SUPRESS LEADING 0'S FORCED + 226 037700 000000 OPDEF PNTDS [37B8!16B12!0] ;PRINT DECIMAL, SPACES FOR LD 0'S + 227 037700 000001 OPDEF PNTDSF [37B8!16B12!1] ;PRINT DECIMAL, SPACES FOR LD 0'S FORCED + 228 037200 000002 OPDEF PNTNM [37B8!4B12!2] ;PRINT PROGRAM NAME + 229 037000 000002 OPDEF PNTSIX [37B8!0B12!2] ;PRINT SIXBIT WORD + 230 037040 000002 OPDEF PNTSXF [37B8!1B12!2] ;PRINT SIXBIT WORD FORCED + 231 037240 000002 OPDEF DROPDV [37B8!5B12!2] ;CLOSE LOGICAL FILE, USER MODE + 232 037100 000002 OPDEF PNTCW [37B8!2B12!2] ;PRINT DF10 CONTROL WORD + 233 037140 000002 OPDEF PNTCWF [37B8!3B12!2] ;PRINT DF10 CONTROL WORD FORCED + 234 037000 030242 OPDEF PCRL [37B8!0B12!CRLF] ;PRINT CARRIAGE RETURN/LINE FEED + 235 037040 030242 OPDEF PCRLF [37B8!1B12!CRLF] ;PRINT CARRIAGE RETURN/LINE FEED FORCED + 236 037000 000040 OPDEF PSP [37B8!0B12!40] ;PRINT SPACE + 237 037040 000040 OPDEF PSPF [37B8!1B12!40] ;PRINT SPACE FORCED + 238 037000 030243 OPDEF PCRL2 [37B8!0B12!CRLF2] ;PRINT CARRIAGE RETURN/LINE FEED (TWICE) +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 4-1 +PARAM KLM 18-JAN-77 11:38 *PARAM* PROGRAM/SUBROUTINE PARAMETERS, JAN 18,1977 SEQ 0013 + + 239 037040 030243 OPDEF PCRL2F [37B8!1B12!CRLF2] ;PRINT CARRIAGE RETURN/LINE FEED (TWICE) FORCED + 240 037040 000007 OPDEF PBELL [37B8!1B12!7] ;PRINT TTY BELL + 241 + 242 037040 000026 OPDEF PFORCE [37B8!1B12!26] ;PRINT FORCE, CONTROL O OVERRIDE + 243 + 244 DEFINE PMSG (ARG),< + 245 PSIXM [SIXBIT\ARG'_\]> + 246 + 247 DEFINE PMSGF (ARG),< + 248 PSIXMF [SIXBIT\ARG'_\]> + 249 + 250 ;*SIXBTZ -- MACRO TO GENERATE SIXBIT DATA FOR PRINTING + 251 ;* CONSERVES CORE OVER ASCIZ + 252 + 253 DEFINE SIXBTZ (ARG),< [SIXBIT\ARG'_\]> + 254 + 255 ;*CONSOLE SWITCH INPUT UUO. + 256 ;*READS CONSOLE SWITCHES IF IN EXEC MODE OR ASKS FOR THEM IF + 257 ;* USER MODE. + 258 + 259 037400 000002 OPDEF SWITCH [37B8!10B12!2] ;INPUT CONSOLE SWITCHES + 260 + 261 ;*CLOCK INITIALIZATION UUO - TO SET DESIRED CLOCK OPERATION + 262 ;*EITHER IGNORE CLOCK, ONLY LET IT TICK OR CAUSE INTERRUPT TO OCCUR. + 263 + 264 037540 000004 OPDEF CLOKOP [37B8!13B12!4] ;CLOCK OPERATION UUO - PDP-11 CLOCK + 265 037200 000004 OPDEF MTROP [37B8!4B12!4] ;CLOCK OPERATION UUO - DK20 METER + 266 + 267 ;*KL10 ONLY CACHE OPERATION UUO'S + 268 + 269 037040 000004 OPDEF CINVAL [37B8!1B12!4] ;CACHE INVALIDATE + 270 037100 000004 OPDEF CFLUSH [37B8!2B12!4] ;CACHE FLUSH + 271 037140 000004 OPDEF CWRTBI [37B8!3B12!4] ;CACHE WRITE-BACK & INVALIDATE +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 5 +PARAM KLM 18-JAN-77 11:38 *PARAM* PROGRAM/SUBROUTINE PARAMETERS, JAN 18,1977 SEQ 0014 + + 272 ;*END OF PASS/PROGRAM UUOS + 273 + 274 ;PERFORMS THE END OF PASS FUNCTIONS. INCREMENT PASS COUNT, + 275 ;*DECREMENT ITERATION COUNT, CHECK IF FINISHED WITH THIS PROGRAM ETC. + 276 + 277 037500 000004 OPDEF ENDUUO [37B8!12B12!4] ;UUO TO DISPLAY LIGHTS + 278 037700 000004 OPDEF EOPUUO [37B8!16B12!4] ;END OF PROGRAM UUO + 279 + 280 ;*MEMORY MANAGEMENT UUO'S + 281 ;*UUO'S TO PERFORM VARIOUS MEMORY FUNCTIONS. MAPPING, ZEROING, PAGING, + 282 ;*ADDRESS CONVERSION, ETC... + 283 + 284 037000 000004 OPDEF MAPMEM [37B8!0B12!4] ;MAP MEMORY + 285 037500 000002 OPDEF MEMZRO [37B8!12B12!2] ;ZERO MEMORY + 286 037440 000002 OPDEF MEMSEG [37B8!11B12!2] ;SETUP MEMORY SEGMENT + 287 037540 000002 OPDEF MAPADR [37B8!13B12!2] ;VIRTUAL TO PHYSICAL ADR CONVERT + 288 037640 000002 OPDEF MAPCNK [37B8!15B12!2] ;MAP MEMORY CHUNK + 289 037600 000002 OPDEF MAPSET [37B8!14B12!2] ;SET KI10 EXEC PAGE MAP + 290 037740 000002 OPDEF MAPPNT [37B8!17B12!2] ;PRINT MEMORY MAP + 291 + 292 ;*DEVICE CODE MODIFICATION UUO + 293 ;*ALLOWS THE MODIFICATION OF IOT'S TO ONE DEVICE TO BE CHANGED TO + 294 ;*IOT'S TO A DIFFERENT DEVICE CODE. + 295 + 296 037340 000002 OPDEF MODPCU [37B8!7B12!2] ;MODIFY PERHIPERAL CODE, USER + 297 037300 000002 OPDEF MODPCP [37B8!6B12!2] ;MODIFY PERHIPERAL CODE, PROGRAM + 298 + 299 030000 IFNDEF MODDVL, + 300 030000 IFNDEF MODDVU, + 301 + 302 ;*"DIAMON" FILE SELECTION AND READ UUOS + 303 + 304 037240 000004 OPDEF FSELECT [37B8!5B12!4] ;FILE SELECTION + 305 037300 000004 OPDEF FREAD [37B8!6B12!4] ;FILE READ - ASCII DATA + 306 037340 000004 OPDEF FRD36 [37B8!7B12!4] ;FILE READ - 36 BIT DATA + 307 037400 000004 OPDEF FRD8 [37B8!10B12!4] ;FILE READ - 8 BIT DATA + 308 + 309 ;*KI10 ONLY UUO FOR PRINTING MARGIN VALUES + 310 + 311 037700 000002 OPDEF PNTMGN [37B8!16B12!2] ;PRINT MARGIN VALUE + 312 + 313 XLIST + 314 IFNDEF KLOLD, + 340 + 341 ;*A MACRO TO REPORT AN ERROR AND NOT LOOP + 342 + 343 DEFINE ERROR1 (FORMAT,CORECT,ACTUAL,F,D,ERR)< + 344 SALL + 345 ERUUO FORMAT,[T,,[SIXBIT\F'_\] + 346 CORECT,,ACTUAL + 347 [SIXBIT\D'_\],,ERR] + 348 XALL > + 349 + 350 >;END OF KLOLD CONDITIONAL + 351 + 352 XLIST + 353 LIST +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 1 +FIXED KLM 18-JAN-77 11:39 *FIXED* FIXED CONTROL AND DISPATCH STORAGE, JAN 18,1977 SEQ 0016 + + 354 SUBTTL *FIXED* FIXED CONTROL AND DISPATCH STORAGE, JAN 18,1977 + 355 + 356 030000 LOC 30000 + 357 + 358 S^;*********************************************************************^ + 359 ;*PROGRAM STARTING ADDRESSES + 360 ;*THESE ADDRESSES CALL VARIOUS SPECIAL START ROUTINES AND OR OPTIONS + 361 ;*NORMAL START ADDRESS IS 30000 ALL OTHERS ARE SPECIAL. INVOKED BECAUSE + 362 ;*OF END OF PASS, POWER FAILURE, DDT START, RE-ENTERING(TYPICALLY USER + 363 ;*MODE), OR ANY NUMBER OF SPECIAL FEATURE TESTS. + 364 S^;*********************************************************************^ + 365 + 366 030000 254 00 1 00 027776 BEGIN: JRST @MODLNK ;STAND-ALONE START + 367 030001 254 00 0 00 030712 $START: JRST START ;MODE CHECK STARTING ADDRESS + 368 + 369 030002 254 00 1 00 027774 DIAGMN: JRST @LDLNK ;DIAGNOSTIC MONITOR START + 370 + 371 030003 254 00 1 00 027774 SYSEXR: JRST @LDLNK ;SYSTEM EXERCISER START + 372 + 373 030004 254 00 0 00 030742 SFSTRT: JRST SADR1 ;SPECIAL FEATURE START + 374 + 375 030005 254 00 0 00 030742 PFSTRT: JRST SADR2 ;POWER FAIL RESTART + 376 + 377 030006 254 00 0 00 030742 REENTR: JRST SADR3 ;REENTER START(USUALLY USER MODE ONLY) + 378 + 379 030007 SRTDDT: ;COMMONLY MISTAKEN NAME FOR "DDTSRT" + 380 030007 254 00 1 00 027775 DDTSRT: JRST @DDTLNK ;DDT START + 381 + 382 030010 254 00 0 00 030741 BEGIN1: JRST STARTA ;LOOP START(END OF PASS COMES HERE) + 383 030011 254 00 1 00 027777 SBINIT: JRST @SUBLNK ;PMGINT LINKAGE + 384 030012 000000 000000 RETURN: 0 ;RETURN ADDRESS STORAGE + 385 + 386 030013 254000 030742 START1: SADR7 ;OPTIONAL STARTING ADR/INSTRUCTIONS + 387 030014 254000 030742 START2: SADR8 ; " + 388 030015 254000 030742 START3: SADR9 ; " + 389 030016 254000 030742 START4: SADR10 ; " + 390 030017 254000 030742 START5: SADR11 ; " +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 2 +FIXED KLM 18-JAN-77 11:39 *FIXED* FIXED CONTROL AND DISPATCH STORAGE, JAN 18,1977 SEQ 0017 + + 391 S^;*********************************************************************^ + 392 ;*PROGRAM FIXED PARAMETER AREA + 393 S^;*********************************************************************^ + 394 + 395 030020 444153 414100 PNTNAM: PAREA3 ;SIXBIT PROGRAM NAME + 396 030021 645560 000000 PNTEXT: PAREA4 ;SIXBIT PROGRAM EXTENSION + 397 030022 000000 000000 RANDBS: PAREA1 ;RANDOM BASE NUMBER + 398 030023 000000 000000 SWTEXR: PAREA2 ;SYSTEM EXERCISER SWITCHES + 399 030024 000000 001000 ITRCNT: ITERAT ;PROGRAM ITERATIONS + 400 030025 000000 030725 $PNAME: PGMNAM ;POINTER TO PROGRAMS NAME + 401 030026 000000 000002 $PVER: MCNVER,,DECVER ;MCN & DEC VERSION LEVEL + 402 030027 000000 030000 $MODVL: MODDVL ;DEVICE CODE CHANGE LOWER LIMIT + 403 030030 000000 030000 $MODVU: MODDVU ;DEVICE CODE CHANGE UPPER LIMIT + 404 030031 777777 777777 $EMODE: IFNDEF EXCASB,<0> IFDEF EXCASB,<-1> ;EXEC ALLOWED + 405 030032 777777 777777 $UMODE: IFNDEF USRASB,<0> IFDEF USRASB,<-1> ;USER ALLOWED + 406 030033 000000 000000 $DSKUP: IFNDEF DSKUPD,<0> IFDEF DSKUPD,<-1> ;DISK UPDATE MODE + 407 030034 000000 000000 $MMAP: IFNDEF MEMMAP,<0> IFDEF MEMMAP,<-1> ;ALLOW MEMORY RTNS + 408 030035 000000 000000 PAREA7: PAREA5 ;OPTIONAL PARAMETER + 409 030036 000000 000000 PAREA8: PAREA6 ;OPTIONAL PARAMETER + 410 + 411 S^;*********************************************************************^ + 412 ;*PROGRAM VARIABLE PARAMETER AREA + 413 S^;*********************************************************************^ + 414 + 415 030037 000000 000000 USER: 0 ; 0 = EXEC, -1 = USER MODE FLAG + 416 030040 000000 000000 KAIFLG: 0 ;PROCESSOR TYPE, 0 = KA10, -1 = KI10 + 417 030041 000000 000000 KLFLG: 0 ;PROCESSOR TYPE, 0 = KA/KI, -1 = KL10 + 418 030042 777777 777777 MONFLG: -1 ;DIAG MONITOR SPECIAL USER FLAG + 419 030043 000000 000000 MONCTL: 0 ;DIAG MON/SYS EXR FLAG + 420 030044 000000 000000 MONTEN: 0 ;-1= LOADED BY 10 + 421 030045 000000 000000 CLOCKF: 0 ;CLOCK TICKED FLAG + 422 030046 000000 000000 CONSW: 0 ;CONSOLE SWITCH SETTINGS + 423 030047 000000 000000 PASCNT: 0 ;PROGRAM PASS COUNT + 424 030050 000000 000000 RUNFLG: 0 ;PROGRAM RUN FLAG + 425 030051 000000 000000 TESTPC: 0 ;SUBTEST PC + 426 030052 000000 000000 ERRPC: 0 ;ERROR PC + 427 030053 000000 000000 ERRTLS: 0 ;ERROR TOTALS + 428 030054 000000 000000 TICKS: 0 ;PROGRAM RUNNING TIME + 429 030055 000000 000000 MARGIN: 0 ;KI10 MARGIN WORD VALUE + 430 030056 000000 000000 $ONETM: 0 ;SUBROUTINE INITIALIZATION FLAG +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 3 +FIXED KLM 18-JAN-77 11:39 *FIXED* FIXED CONTROL AND DISPATCH STORAGE, JAN 18,1977 SEQ 0018 + + 431 S^;*********************************************************************^ + 432 ;*SPECIAL PROGRAM DISPATCH ADDRESSES + 433 S^;*********************************************************************^ + 434 + 435 030057 037 12 0 00 000004 BEGEND: ENDUUO ;END OF PASS + 436 030060 254 00 0 00 030010 $BEND1: JRST BEGIN1 ;KEEP RUNNING PROGRAM + 437 030061 037 16 0 00 000004 $BEND2: EOPUUO ;END OF PROGRAM - NO RETURN + 438 030062 254000 030742 CNTLC: SADR5 ;CONTROL C XFER ADDRESS + 439 030063 254000 030742 ALTMGO: SADR6 ;ALTMODE XFER ADDRESS + 440 030064 CPOPJ1: ;SKIP RETURN + 441 030064 350 00 0 17 000000 UUOSKP: AOS (P) ;SKIP RETURN FROM UUO + 442 030065 CPOPJ: ;NON-SKIP REGULAR RETURN + 443 030065 263 17 0 00 000000 UUOEXT: RTN ;UUO RETURN + 444 030066 255 00 0 00 000000 UUORTN: JFCL ;ADDITIONAL USERS UUO ROUTINE + 445 030067 255 00 0 00 000000 $UORTX: JFCL ;ADDITIONAL UUO LINKAGE + 446 030070 255 00 0 00 000000 $UUOER: JFCL ;INITED AS (JRST $UOERX) + 447 030071 255 00 0 00 000000 $ITRHL: JFCL ;ADDITIONAL INTERRUPT LINKAGE + 448 030072 255 00 0 00 000000 $ITRX1: JFCL ; " + 449 030073 255 00 0 00 000000 $USRHL: JFCL ; " + 450 030074 255 00 0 00 000000 $RSRTX: JFCL ;ADDITIONAL POWER FAIL LINKAGE + 451 030075 255 00 0 00 000000 $RSRTY: JFCL ; " + 452 030076 255 00 0 00 000000 RESRT1: JFCL ; INITED AS (JRST RESRTX) + 453 030077 255 00 0 00 000000 RESRT2: JFCL ; " + 454 030100 255 00 0 00 000000 $PARER: JFCL ;ADDITIONAL PARITY ERROR LINKAGE + 455 030101 255 00 0 00 000000 ERMORE: JFCL ;ADDITIONAL ERROR HANDLER LINKAGE + 456 030102 254 04 0 00 030102 HALT . ;IMPROPER TRANSFER HALT + 457 + 458 030103 000000 000000 $PSHER: 0 ;INITED AS (JRST PSHERR) + 459 030104 000000 000000 ITRCH1: 0 ;PC & FLAGS OF CURRENT INTERRUPT + 460 030105 000000 000000 0 ;INITED AS (JRST $ITRC1) + 461 + 462 S^;*********************************************************************^ + 463 ;*PROCESSOR CONTROL STORAGE + 464 S^;*********************************************************************^ + 465 + 466 030106 000000 000000 $ACC0: 0 ;INTERRUPT SAVED AC0 + 467 030107 000000 000000 $SVPI: 0 ;INTERRUPT SAVED PI + 468 030110 000000 000000 $SVAPR: 0 ;INTERRUPT SAVED APR + 469 030111 000000 000000 $SVPAG: 0 ;INTERRUPT SAVED PAG (DATAI) + 470 030112 000000 000000 $SPAG1: 0 ;INTERRUPT SAVED PAG (CONI) + 471 + 472 030113 000000 000000 $SVUUO: 0 ;CURRENT USERS UUO + 473 030114 000000 000000 $SVUPC: 0 ;PC OF CURRENT USERS UUO + 474 + 475 030115 000000 000000 REPTU: 0 ;REPEAT UUO ITERATIONS + 476 030116 000000 000000 SCOPE: 0 ;ERROR HANDLER SCOPE LOOP FLAG + 477 030117 000000 000000 %CORFLG:0 ; " CORRECT FLAG + 478 030120 000000 000000 %COREC: 0 ; " CORRECT DATA + 479 030121 000000 000000 %ACTFL: 0 ; " ACTUAL FLAG + 480 030122 000000 000000 %ACTUL: 0 ; " ACTUAL DATA + 481 030123 000000 000000 %DISCR: 0 ; " DISCREPENCY DATA +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 4 +FIXED KLM 18-JAN-77 11:39 *FIXED* FIXED CONTROL AND DISPATCH STORAGE, JAN 18,1977 SEQ 0019 + + 482 S^;*********************************************************************^ + 483 ;*UUO DISPATCH TABLE + 484 S^;*********************************************************************^ + 485 XLIST + 486 LIST + 487 030124 030070 030070 UUODIS: LUUO1,,$UUOER + 488 030125 030070 030070 LUUO3,,LUUO2 + 489 030126 030070 030070 LUUO5,,LUUO4 + 490 030127 030070 030070 LUUO7,,LUUO6 + 491 030130 030070 030070 LUUO11,,LUUO10 + 492 030131 030070 030070 LUUO13,,LUUO12 + 493 030132 030070 030070 LUUO15,,LUUO14 + 494 030133 030070 030070 LUUO17,,LUUO16 + 495 030134 030070 030070 LUUO21,,LUUO20 + 496 030135 030070 030070 LUUO23,,LUUO22 + 497 030136 030070 030070 LUUO25,,LUUO24 + 498 030137 030070 030070 LUUO27,,LUUO26 + 499 030140 030070 030070 LUUO31,,LUUO30 + 500 030141 030070 030070 LUUO33,,LUUO32 + 501 + 502 S^;*********************************************************************^ + 503 ;*MEMORY MANAGMENT STORAGE + 504 S^;*********************************************************************^ + 505 + 506 030142 000000 000000 DF22F: 0 ;DF10 CONTROL FLAG, 0 = 18, -1 = 22 BIT + 507 030143 000000 000000 MAPNEW: 0 ;MEMORY MAPPING CONTROL FLAG, -1 = 4096K MAPPING + 508 030144 000000 000000 MEMTOT: 0 ;TOTAL MEMORY SIZE IN K (1024.) + 509 030145 000000 000000 MEMLOW: 0 ;LOWEST USABLE MEMORY + 510 030146 MEMSIZ: BLOCK ^D41 ;MEMORY SEGMENT POINTER TABLE + 511 + 512 S^;*********************************************************************^ + 513 ;*PRINT CONTROL STORAGE + 514 S^;*********************************************************************^ + 515 + 516 030217 000000 000000 PNTFLG: 0 ;PRINT FLAG, -1 WHILE IN PRINT ROUTINE + 517 030220 000000 000000 PNTENB: 0 ;PRINT ENABLE + 518 030221 000000 000000 PDISF: 0 ;PRINT DISABLED FLAG + 519 030222 000000 000000 PNTINH: 0 ;INHIBIT PRINT INPUT CHECKS + 520 030223 000000 000000 PNTSPC: 0 ;PRINT SPACE CONTROL + 521 030224 000000 000000 OPTIME: 0 ;TYPE-IN WAIT TIME + 522 030225 000000 000000 $TWCNT: 0 ;TIME WAITED + 523 030226 000000 000000 $DVOFF: 0 ;LOGICAL DEVICE INITED FLAG + 524 030227 000000 000000 TTYFIL: 0 ;TTY EXEC FILLERS FLAG + 525 030230 000000 000000 TTYSPD: 0 ;TTY EXEC BAUD RATE + 526 030231 000000 000000 $TTCHR: 0 ;ACTUAL TYPED IN CHAR + 527 030232 000000 000000 $CHRIN: 0 ;UPPER CASED & PARITY STRIPPED CHAR + 528 030233 000000 000000 $TYPNB: 0 ;TYPED IN NUMBER + 529 030234 000000 000000 $CRLF: 0 ;FREE CR/LF FLAG + 530 030235 000000 000000 $TABF: 0 ;TAB CONVERSION FLAG + 531 030236 000000 000000 $FFF: 0 ;FORM FEED CONVERSION FLAG + 532 030237 000000 000000 $VTF: 0 ;VERTICAL TAB CONVERSION FLAG + 533 030240 000000 000000 USRLFF: 0 ;USER LF FILLERS + 534 030241 000000 000000 USRCRF: 0 ;USER CR FILLERS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 5 +FIXED KLM 18-JAN-77 11:39 *FIXED* FIXED CONTROL AND DISPATCH STORAGE, JAN 18,1977 SEQ 0020 + + 535 S^;*********************************************************************^ + 536 ;*THE FOLLOWING MISCELLANEOUS PRINT CHARACTERS ARE INCLUDED + 537 ;*TO FACILITATE PRINTING AND ARE CALLED AS FOLLOWS: + 538 ;* MOVEI NAME + 539 ;* PNTA ;OR PNTAF + 540 S^;*********************************************************************^ + 541 + 542 030242 CRLF: ASCII/ + 543 030242 015 012 000 000 000 / + 544 030243 CRLF2: ASCII/ + 545 + 546 030243 015 012 015 012 000 / + 547 030244 054 000 000 000 000 COMMA: ASCII/,/ + 548 030245 056 000 000 000 000 PERIOD: ASCII/./ + 549 030246 040 000 000 000 000 SPACE: ASCII/ / + 550 030247 011 000 000 000 000 TAB: ASCII/ / + 551 030250 MINUS: + 552 030250 055 000 000 000 000 HYPEN: ASCII/-/ + 553 030251 053 000 000 000 000 PLUS: ASCII/+/ + 554 030252 052 000 000 000 000 AST: ASCII/*/ + 555 030253 100 000 000 000 000 ATSIN: ASCII/@/ + 556 030254 050 000 000 000 000 LFP: ASCII/(/ + 557 030255 051 000 000 000 000 RTP: ASCII/)/ + 558 030256 007 0000000000 BELL: BYTE (7) 007 + 559 030257 077 000 000 000 000 QUEST: ASCII/?/ + 560 030260 057 000 000 000 000 SLASH: ASCII!/! + 561 030261 044 000 000 000 000 DOLLAR: ASCII/$/ + 562 030262 000000 000012 RADIX: ^D10 ;DECIMAL PRINT RADIX + 563 030263 000000 000040 RADLSP: 40 ;DECIMAL PRINT LEADING CHAR + 564 030264 000000 000012 RADLSC: ^D10 ;DECIMAL PRINT LEADING CHAR COUNT + 565 + 566 S^;*********************************************************************^ + 567 ;*USER MODE OUTPUT FILE INFORMATION + 568 S^;*********************************************************************^ + 569 + 570 030265 $OBUF: BLOCK 3 ;LOGICAL FILE OUTPUT BUFFER HEADER + 571 030270 60 62 51 56 64 00 $OUTNM: SIXBIT /PRINT/ ;FILE NAME + 572 030271 60 56 64 00 00 00 $OUTEX: SIXBIT /PNT/ ;FILE NAME EXTENSION + 573 030272 BLOCK 2 + 574 + 575 S^;*********************************************************************^ + 576 ;*DISK UPDATE MODE FILE INFORMATION + 577 S^;*********************************************************************^ + 578 + 579 030274 $IBUF: BLOCK 3 + 580 030277 60 62 51 56 64 00 $INNM: SIXBIT /PRINT/ + 581 030300 60 56 64 00 00 00 $INEXT: SIXBIT /PNT/ + 582 030301 BLOCK 2 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 6 +FIXED KLM 18-JAN-77 11:39 *FIXED* FIXED CONTROL AND DISPATCH STORAGE, JAN 18,1977 SEQ 0021 + + 583 S^;*********************************************************************^ + 584 ;*PUSHDOWN LIST CONTROL INFORMATION + 585 S^;*********************************************************************^ + 586 + 587 030303 777577 030303 PLIST: PLIST-PLISTE,,PLIST + 588 030304 PLISTS: BLOCK 200 + 589 030504 000000 000000 PLISTE: 0 ;END OF PUSHDOWN LIST + 590 + 591 S^;*********************************************************************^ + 592 ;*POWER LINE CLOCK FREQUENCY FLAG + 593 S^;*********************************************************************^ + 594 + 595 030505 000000 000000 CYCL60: 0 ;0 = 60, -1 = 50 CYCLE + 596 + 597 S^;*********************************************************************^ + 598 ;*KL10 CACHE CONTROL FLAGS + 599 S^;*********************************************************************^ + 600 + 601 030506 000000 000000 CSHFLG: 0 ;ALLOW CACHE IF 0 + 602 030507 000000 000000 CSHMEM: 0 ;CACHE MEMORY SEGMENTS IF 0 + 603 + 604 S^;*********************************************************************^ + 605 ;*NUMBER INPUT DIGIT FLAG + 606 S^;*********************************************************************^ + 607 + 608 030510 000000 000000 TTNBRF: 0 ;-1 IF ANY DIGIT TYPED + 609 + 610 S^;*********************************************************************^ + 611 ;*KL10 & KI10 "INHPAG" SWITCH PAGING PREVENTION + 612 S^;*********************************************************************^ + 613 + 614 030511 000000 000000 PVPAGI: 0 ;IF NON-ZERO, OVERRIDE "INHPAG" SWITCH ACTION + 615 + 616 S^;*********************************************************************^ + 617 ;*ERROR REPORTING ROUTINE ADDITIONAL USERS CONTROL INSTRUCTIONS + 618 S^;*********************************************************************^ + 619 + 620 030512 000000 000000 %ERHI1: 0 ;IF NON-ZERO, XCT'D AT START OF %ERUUO + 621 030513 000000 000000 %ERHI2: 0 ;IF NON-ZERO, XCT'D AT END OF %ERUUO + 622 030514 000000 000000 %ERHI3: 0 ;IF NON-ZERO, XCT'D AFTER "PC" OF %ERUUO + 623 + 624 S^;*********************************************************************^ + 625 ;*SPECIAL USERS UUO INTERCEPT INSTRUCTION + 626 S^;*********************************************************************^ + 627 + 628 030515 000000 000000 $$UUO: 0 ;IF NON-ZERO, XCT'D AT START OF $UORTN +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 7 +FIXED KLM 18-JAN-77 11:39 *FIXED* FIXED CONTROL AND DISPATCH STORAGE, JAN 18,1977 SEQ 0022 + + 629 S^;*********************************************************************^ + 630 ;*KL10 PROCESSOR TYPE FLAG, 0=P0, 1=BBD NEW, 2=BBD OLD + 631 S^;*********************************************************************^ + 632 + 633 030516 000000 000000 KLTYP: 0 + 634 + 635 S^;*********************************************************************^ + 636 ;*SPECIAL USERS MUUO INTERCEPT INSTRUCTION + 637 S^;*********************************************************************^ + 638 + 639 030517 000000 000000 $$MUUO: 0 ;IF NON-ZERO, XCT'D AT START OF MUUOER + 640 + 641 S^;*********************************************************************^ + 642 ;*SPECIAL USERS USER MODE OUTPUT ERROR INTERCEPT INSTUCTION + 643 S^;*********************************************************************^ + 644 + 645 030520 000000 000000 $$OUTER:0 ;IF NON-ZERO, XCT'D AT END OF USER MODE ERROR + 646 + 647 S^;*********************************************************************^ + 648 ;*"SWITCH" CALL USAGE CONTROL + 649 S^;*********************************************************************^ + 650 + 651 030521 000000 000000 $$TOGGLE:0 ;IF NON-ZERO, USE C(CONSW) FOR SWITCHES + 652 + 653 S^;*********************************************************************^ + 654 ;*SPECIAL USERS ALTMODE SWITCH CALL INTERCEPT INSTRUCTIONS + 655 S^;*********************************************************************^ + 656 + 657 030522 000000 000000 $$TAX1: 0 ;IF NON-ZERO, XCT'D AT START OF ALTMODE SWITCH CALL + 658 030523 000000 000000 $$TAX2: 0 ;IF NON-ZERO, XCT'D AT END OF ALTMODE SWITCH CALL + 659 + 660 S^;*********************************************************************^ + 661 ;*SPECIAL FUTURE EXPANSION ROOM + 662 ;*IF ANY FIXED AREA TAGS ARE ADDED, REDUCE THE SIZE OF + 663 ;*THIS BLOCK STATEMENT ACCORDINGLY. THIS MUST BE DONE + 664 ;*SO THAT PREVIOUS FIXED ASSIGNMENTS DO NOT CHANGE. + 665 S^;*********************************************************************^ + 666 + 667 030524 BLOCK 53 ;HOPEFULLY THIS IS ENOUGH FOREVER + 668 + 669 S^;*********************************************************************^ + 670 ;*END OF FIXED STORAGE + 671 S^;*********************************************************************^ + 672 + 673 030577 $ENDFX=&<777700>-1 + 674 030577 LOC $ENDFX + 675 030577 000000 000000 ENDFIX: 0 ;END OF FIXED STORAGE +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 1 +SPCCPU KLM 26-FEB-76 05:50 *SPCCPU* SPECIAL BASIC CPU PROCESSOR CONTROL, 26-FEB-76 SEQ 0023 + + 676 SUBTTL *SPCCPU* SPECIAL BASIC CPU PROCESSOR CONTROL, 26-FEB-76 + 677 + 678 ;NEW DEFINITIONS USED BY THE KL10 SUBROUTINE PACKAGE + 679 + 680 000000 AC0= 0 + 681 030000 DIAGNOS=30000 ;PDP-10 DIAGNOSTIC START ADDRESS + 682 010000 DDT= 10000 ;PDP-10 DDT START ADDRESS + 683 020000 DIAMON= 20000 ;PDP-10 DIAMON LOADER START ADDRESS + 684 020000 DONG11= 1B22 ;11 DOORBELL (FROM THE 10) + 685 + 686 ;DTE20 DEVICE CODES + 687 + 688 000200 DTE== 200 ;DTE0 + 689 000204 DTE0== 204 + 690 000204 DTE1== 204 + 691 000210 DTE2== 210 + 692 000214 DTE3== 214 + 693 + 694 ;KL10 EPT COMMUNICATION AREA + 695 + 696 000440 $STD= 440 ;PDP-10 DIAGNOSTIC START ADDRESS + 697 000441 $DDT= 441 ;PDP-10 DDT START ADDRESS + 698 000442 $STL= 442 ;PDP-10 LOADER START ADDRESS + 699 000443 $STM= 443 ;PDP-10 MONITOR START ADDRESS + 700 + 701 000444 $DTFLG= 444 ;DTE20 OPERATION COMPLETE FLAG + 702 000445 $DTCLK= 445 ;DTE20 CLOCK INTERRUPT FLAG + 703 000446 $DTCI= 446 ;DTE20 CLOCK INTERRUPT INSTRUCTION + 704 000447 $DTT11= 447 ;DTE20 10 TO 11 ARGUMENT + 705 000450 $DTF11= 450 ;DTE20 11 TO 10 ARGUMENT + 706 000451 $DTCMD= 451 ;DTE20 TO 11 COMMAND WORD + 707 000452 $DTSEQ= 452 ;DTE20 OPERATION SEQUENCE NUMBER + 708 000453 $DTOPR= 453 ;DTE20 OPERATIONAL DTE # + 709 000454 $DTCHR= 454 ;DTE20 LAST TYPED CHARACTER + 710 000455 $DTMTD= 455 ;DTE20 MONITOR TTY OUTPUT COMPLETE FLAG + 711 000456 $DTMTI= 456 ;DTE20 MONITOR TTY INPUT FLAG + 712 + 713 000457 $DTSWR= 457 ;DTE20 CONSOLE SWITCH REGISTER +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 2 +SPCCPU KLM 26-FEB-76 05:50 *SPCCPU* SPECIAL BASIC CPU PROCESSOR CONTROL, 26-FEB-76 SEQ 0024 + + 714 ;SPECIAL "FIXED" REASSIGNMENTS + 715 + 716 030600 $$LOC=. ;SAVE CURRENT LOCATION + 717 + 718 030000 LOC 30000 + 719 030000 254 00 0 00 030600 $$BEGIN:JRST $$START ;SETUP SPECIAL START + 720 030001 254 00 0 00 030600 JRST $$START ;"DIAMON" CHAIN START ADDRESS + 721 + 722 000440 LOC 440 + 723 000440 254 00 0 00 030000 $STD: JRST BEGIN ;SETUP FOR "STD" + 724 000443 LOC 443 + 725 000443 254 00 0 00 030636 $STM: JRST $SPEC ;SIMPLE RUN CONTROL + 726 + 727 030057 LOC 30057 + 728 030057 254 00 0 00 030641 $BEGEND:JRST $SPBEND ;SETUP SPECIAL "BEGEND" + 729 + 730 ;SPECIAL MUUO, TRAP & PAGE FAIL SETUP + 731 + 732 000420 LOC 420 + 733 000420 254 04 0 00 000420 $$420: HALT . ;KI10 PAGE FAIL + 734 000421 255 00 0 00 000000 $$421: JFCL ;OVERFLOW + 735 000422 254 04 0 00 000422 $$422: HALT . ;PUSHDOWN OVERFLOW + 736 000423 254 04 0 00 000423 $$423: HALT . ;TRAP 3 + 737 000424 000000 000000 $$424: 0 ;MMUO + 738 000425 000000 000000 $$425: 0 ;MMUO PC + 739 000426 000000 000000 $$426: 0 ;KI10-PAGE FAIL, KL10-PROCESS CONTEXT + 740 000427 254 04 0 00 000427 $$427: HALT . + 741 000430 000000 000427 $$430: 427 ;MMUO NEW PC'S + 742 000431 000000 000427 $$431: 427 + 743 000432 000000 000427 $$432: 427 + 744 000433 000000 000427 $$433: 427 + 745 000434 000000 000427 $$434: 427 + 746 000435 000000 000427 $$435: 427 + 747 000436 000000 000427 $$436: 427 + 748 000437 000000 000427 $$437: 427 + 749 + 750 000500 LOC 500 + 751 000500 000000 000000 $$500: 0 ;KL10 PAGE FAIL WORD + 752 000501 000000 000000 $$501: 0 ;KL10 PAGE FAIL PC + 753 000502 000000 000503 $$502: 503 ;KL10 PAGE FAIL NEW PC + 754 000503 254 04 0 00 000503 $$503: HALT . +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 3 +SPCCPU KLM 26-FEB-76 05:50 *SPCCPU* SPECIAL BASIC CPU PROCESSOR CONTROL, 26-FEB-76 SEQ 0025 + + 755 030600 LOC $$LOC ;RESET CURRENT LOCATION + 756 + 757 ;SPECIAL STARTUP SEQUENCE + 758 + 759 030600 402 00 0 00 030037 $$START:SETZM USER + 760 030601 265 00 0 00 030602 JSP 0,.+1 ;IN USER MODE ? + 761 030602 603 00 0 00 010000 TLNE 0,USERF + 762 030603 476 00 0 00 030037 SETOM USER ;YES, SET CONTROL WORD + 763 030604 336 00 0 00 030042 SKIPN MONFLG ;SPECIAL USER MODE ? + 764 030605 402 00 0 00 030037 SETZM USER ;YES, RUN AS EXEC + 765 030606 332 00 0 00 030037 SKIPE USER + 766 030607 254 00 0 00 030712 JRST START ;USER MODE, DON'T NEED CPU TYPE + 767 + 768 030610 336 00 0 00 030044 $STKIL: SKIPN MONTEN ;LOADED BY "DIAMON" ? + 769 030611 476 00 0 00 030024 SETOM ITRCNT ;NO, RUN FOREVER + 770 030612 402 00 0 00 030516 SETZM KLTYP + 771 030613 402 00 0 00 030041 SETZM KLFLG ;ASSUME KI10 + 772 030614 200 01 0 00 035060 MOVE 1,[1,,1] + 773 030615 251 01 0 00 000001 BLT 1,1 ;HOPE THIS WORKS + 774 030616 316 01 0 00 035060 CAMN 1,[1,,1] ;IF AC NE 1,,1 AFTER BLT, KL10 + 775 030617 254 00 0 00 030712 JRST START ;KI10, NO ADDITIONAL SETUP + 776 + 777 030620 7 000 20 0 00 010040 $STKL: CONO APR,10040 ;SET BBD NOT BIT + 778 030621 7 000 24 0 00 000000 CONI APR,0 + 779 030622 7 000 20 0 00 020040 CONO APR,20040 ;CLEAR BBD NOT BIT + 780 030623 606 00 0 00 000040 TRNN 0,40 ;IF SET, KL10 + 781 030624 350 00 0 00 030516 AOS KLTYP ;IF NOT, BBD + 782 030625 402 00 0 00 000444 SETZM $DTFLG + 783 030626 402 00 0 00 000445 SETZM $DTCLK + 784 030627 200 00 0 00 000453 MOVE $DTOPR ;GET DTE # + 785 030630 436 00 0 00 030670 ORM $$DTE0 ;INSERT IN DTE I/O INSTS + 786 030631 436 00 0 00 030672 ORM $$DTE1 + 787 030632 436 00 0 00 030704 ORM $$DTE2 + 788 030633 436 00 0 00 030706 ORM $$DTE3 + 789 030634 476 00 0 00 030041 SETOM KLFLG ;SET KL10 CONTROL FLAG + 790 030635 254 00 0 00 030712 JRST START + 791 + 792 030636 200 00 0 00 035061 $SPEC: MOVE [JRST STARTA] ;SIMPLE RUN CONTROL + 793 030637 202 00 0 00 030643 MOVEM $SPB1 + 794 030640 254 00 0 00 030712 JRST START +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 4 +SPCCPU KLM 26-FEB-76 05:50 *SPCCPU* SPECIAL BASIC CPU PROCESSOR CONTROL, 26-FEB-76 SEQ 0026 + + 795 ;SPECIAL "BEGEND" ROUTINE + 796 + 797 030641 350 00 0 00 030047 $SPBEND:AOS PASCNT ;INCREMENT PASS COUNT + 798 030642 370 00 0 00 030024 SOS ITRCNT ;DECREMENT ITERATION COUNT + 799 030643 336 00 0 00 030037 $SPB1: SKIPN USER + 800 030644 254 00 0 00 030652 JRST $SPBEX ;EXEC MODE + 801 + 802 030645 332 00 0 00 030024 $SPBUS: SKIPE ITRCNT ;USER MODE, COMPLETED ? + 803 030646 254 00 0 00 030741 JRST STARTA ;NO, KEEP RUNNING + 804 030647 336 00 0 00 030044 SKIPN MONTEN ;DONE, LOADED BY "DIAMON" ? + 805 030650 047 00 0 00 000012 EXIT ;NO, RETURN TO MONITOR + 806 030651 254 00 1 00 030012 JRST @RETURN ;YES, RETURN TO "DIAMON" + 807 + 808 030652 332 00 0 00 030041 $SPBEX: SKIPE KLFLG + 809 030653 254 00 0 00 030660 JRST $SPBKL ;KL10 & EXEC + 810 030654 7 004 14 0 00 030024 DATAO PI,ITRCNT ;KI10 & EXEC, DISPLAY ITER COUNT + 811 030655 332 00 0 00 030024 SKIPE ITRCNT + 812 030656 254 00 0 00 030741 JRST STARTA ;NOT COMPLETED YET + 813 030657 254 00 1 00 030012 JRST @RETURN ;DONE + 814 + 815 030660 336 00 0 00 030024 $SPBKL: SKIPN ITRCNT + 816 030661 254 00 0 00 030676 JRST $SPKLD ;KL10, EXEC & COMPLETED + 817 + 818 030662 335 00 0 00 030043 SKIPGE MONCTL + 819 030663 254 00 0 00 030741 JRST STARTA ;"DIAMON" CONTROL + 820 030664 201 00 0 00 000404 MOVEI 0,404 ;NOTIFY PDP-11 OF END OF PASS + 821 030665 202 00 0 00 000451 MOVEM 0,$DTCMD + 822 030666 402 00 0 00 000444 SETZM $DTFLG + 823 030667 336 00 0 00 030516 SKIPN KLTYP + 824 030670 7 200 20 0 00 020000 $$DTE0: CONO DTE,DONG11 + 825 030671 332 00 0 00 030516 SKIPE KLTYP + 826 030672 7 200 20 0 00 010000 $$DTE1: CONO DTE,10000 + 827 030673 336 00 0 00 000444 SKIPN $DTFLG ;WAIT TILL 11 RESPONDS + 828 030674 254 00 0 00 030673 JRST .-1 + 829 030675 254 00 0 00 030741 JRST STARTA ;KEEP RUNNING + 830 + 831 ;SPECIAL KL10 COMPLETED ROUTINE + 832 + 833 030676 332 00 0 00 030044 $SPKLD: SKIPE MONTEN + 834 030677 254 00 1 00 030012 JRST @RETURN ;LOADED BY "DIAMON" + 835 + 836 030700 201 00 0 00 000403 MOVEI 0,403 ;NOTIFY PDP-11 OF COMPLETION + 837 030701 202 00 0 00 000451 MOVEM 0,$DTCMD + 838 030702 402 00 0 00 000444 SETZM $DTFLG + 839 030703 336 00 0 00 030516 SKIPN KLTYP + 840 030704 7 200 20 0 00 020000 $$DTE2: CONO DTE,DONG11 + 841 030705 332 00 0 00 030516 SKIPE KLTYP + 842 030706 7 200 20 0 00 010000 $$DTE3: CONO DTE,10000 + 843 030707 336 00 0 00 000444 SKIPN $DTFLG ;SHOULD NEVER HAPPEN + 844 030710 254 00 0 00 030707 JRST .-1 ;11 NEVER RETURNS ON END OF PROGRAM + 845 030711 254 04 0 00 030000 HALT BEGIN ;IF IT DOES, HALT. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 1 +DAKAAM MAC 19-JAN-77 13:42 DIAGNOSTIC SECTION SEQ 0027 + + 846 SUBTTL DIAGNOSTIC SECTION + 847 + 848 030712 402 00 0 00 030037 START: SETZM USER# ;CLEAR USER CONTROL WORD + 849 030713 265 00 0 00 030714 JSP 0,.+1 ;GET FLAGS + 850 030714 603 00 0 00 010000 TLNE USERF ;IN USER MODE? + 851 030715 476 00 0 00 030037 SETOM USER ;YES, SET USER CONTROL WORD + 852 030716 336 00 0 00 030042 SKIPN MONFLG ;SPECIAL USER MODE? + 853 030717 402 00 0 00 030037 SETZM USER ;YES, CLEAR USER CONTROL WORD + 854 030720 336 00 0 00 030037 SKIPN USER + 855 030721 254 00 0 00 030741 JRST STARTA + 856 030722 331 00 0 00 030043 SKIPL MONCTL + 857 030723 051 03 0 00 030725 TTCALL 3,PGMNAM ;MENTION OUR NAME + 858 030724 254 00 0 00 030741 JRST STARTA + 859 + 860 030725 PGMNAM: ASCIZ/ + 861 030725 015 012 120 104 120 PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) [DAKAA] + 862 030726 055 061 060 040 113 + 863 030727 101 061 060 040 102 + 864 030730 101 123 111 103 040 + 865 030731 111 116 123 124 122 + 866 030732 125 103 124 111 117 + 867 030733 116 040 104 111 101 + 868 030734 107 116 117 123 124 + 869 030735 111 103 040 050 061 + 870 030736 051 040 133 104 101 + 871 030737 113 101 101 135 015 + 872 030740 012 000 000 000 000 / + 873 + 874 ;BASIC INSTRUCTION TEST (3) + 875 ;THE TEST IS DESIGNED FOR INITIAL DEBUGGING OF + 876 ;PROCESSOR HARDWARE AND TO DETECT (SOLID) FAILURES + 877 ;IN THE FIELD. + 878 + 879 + 880 030741 254 00 0 00 030742 STARTA: JRST .+1 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 2 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0028 + + 881 SUBTTL TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS + 882 + 883 ;NOTE: AN "*" IN THE COMMENT FIELD OF AN INSTRUCTION INDICATES THAT IT IS THE TESTED + 884 ;INSTRUCTION. + 885 ;********** + 886 + 887 ;THIS TEST VERIFIES THAT SKIPA ALWAYS SKIPS THE NEXT INSTRUCTION + 888 + 889 030742 A00=. + 890 030742 334 00 0 00 000000 A12500: SKIPA ;*SKIPA SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + 891 030743 254 04 0 00 000000 HALT ;IF PROGRAM HALTS, SKIPA DID NOT SKIP + 892 + 893 ;IF PROGRAM HANGS UP ON SKIPA INSTRUCTION, CHECK AB PC EN [ABC], AB PC F/F [ABC], + 894 ;AB PC B [ABC OR AB], AB PC (FETCH) EN [ABC], IR SKIPS [IR3], IR SKIPX [IR1], + 895 ;IR 3XX [IR1], IR BITS 00 - 08 [IR1], F CYC ACT EN C [FI], FT6 F/F [F2], + 896 ;PC CLOCK EN [PCC], ET2 J F/F [E], PC CLK (ET2) EN [PCC], ADZ COND P [ADZ] AND + 897 ;ADZ COND R [ADZ], PC CLOCK A [PCC], PC CLOCK B [PCC] + 898 + 899 ;IF PROGRAM HALTED ON HALT INSTRUCTION, CHECK ST1 COND [ST2] + 900 + 901 ;********** + 902 + 903 ;THIS TEST VERIFIES THAT JUMP NEVER JUMPS + 904 + 905 030744 320 00 0 00 030745 A15000: JUMP .+1 ;*JUMP SHOULD NEVER JUMP + 906 030745 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF JUMP IS SUCCESSFUL + 907 030746 254 04 0 00 000000 HALT ;HALT IF JUMP FAILS + 908 + 909 ;IF PROGRAM HANGS UP ON JUMP .+1 INSTRUCTION, THE JUMP INSTRUCTION + 910 ;MAY HAVE FAILED. CHECK ST1 COND [ST2], ST INST FET ST1 EN [ST2], ET2 D F/F [E], + 911 ;IR JUMPS [IR1], IR JUMPX [IR1] + 912 + 913 ;********** + 914 ;THIS TEST VERIFIES THAT JUMP NEVER JUMPS + 915 + 916 030747 320 00 0 00 030751 A15100: JUMP .+2 ;*JUMP SHOULD NEVER JUMP + 917 030750 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF JUMP IS SUCCESSFUL + 918 030751 254 04 0 00 000000 HALT ;HALT IF JUMP FAILS + 919 + 920 ;AB PC EN [ABC] + 921 + 922 ;********** + 923 + 924 ;THIS TEST VERIFIES THAT JUMPA .+1 ALWAYS JUMPS TO THE NEXT INSTRUCTION + 925 + 926 030752 324 00 0 00 030753 A12700: JUMPA .+1 ;*JUMPA .+1 SHOULD NEVER JUMP + 927 030753 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF JUMPA IS SUCCESSFUL + 928 030754 254 04 0 00 000000 HALT ;HALT IF JUMPA .+1 FAILED + 929 + 930 ;********** + 931 + 932 ;THIS TEST VERIFIES THAT JUMPA JUMPS TO THE LOCATION SPECIFIED BY E WHEN E=.+2 + 933 + 934 030755 324 00 0 00 030757 A12600: JUMPA .+2 ;*JUMPA .+2 SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + 935 030756 254 04 0 00 000000 HALT ;PROGRAM HALTS HERE IF JUMPA .+2 FAILS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 2-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0029 + + 936 + 937 ;ST1 COND [ST2] + 938 + 939 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 3 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0030 + + 940 ;THIS TEST VERIFIES THAT SKIPA IS DATA INDEPENDENT. HENCE, IT ALWAYS SKIPS THE NEXT + 941 ;INSTRUCTION. + 942 ;THE MOVE [0] INSTRUCTION IS USED TO TEST THE DATA INDEPENDENCE OF SKIPA. + 943 + 944 030757 200 00 0 00 035062 A100: MOVE [0] ;PRESET LOCATION 0 TO ALL ZEROS + 945 030760 334 00 0 00 000000 SKIPA ;*SKIPA SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + 946 STOP^ + 947 030761 254 04 0 00 030762 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 948 030762 324 00 0 00 030763 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 949 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 950 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 951 + 952 ;********** + 953 ;THIS TEST INSURES THAT SKIPA IS DATA INDEPENDENT. HENCE, IT ALWAYS SKIPS THE NEXT + 954 ;INSTRUCTION. + 955 ;THE MOVE [-1] INSTRUCTION IS USED TO TEST THE DATA INDEPENDENCE OF SKIPA. + 956 + 957 030763 200 00 0 00 035063 A200: MOVE [-1] ;PRESET LOCATION 0 TO ALL ONES + 958 030764 334 00 0 00 000000 SKIPA ;*SKIPA SKOULD ALWAYS SKIP THE NEXT INSTRUCTION + 959 STOP^ + 960 030765 254 04 0 00 030766 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 961 030766 324 00 0 00 030767 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 962 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 963 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 964 + 965 ;COND P [ADZ] + 966 + 967 ;********** + 968 ;THIS TEST VERIFIES THAT SKIPGE SKIPS THE NEXT INSTRUCTION WHEN C(E) + 969 ;IS ALL ZEROS. THE MOVE INSTRUCTION IS USED TO LOAD E WITH ALL ZEROS. + 970 + 971 030767 200 00 0 00 035062 A300: MOVE [0] ;PRESET E TO ALL ZEROS + 972 030770 335 00 0 00 000000 SKIPGE ;*SKIPGE SHOULD SKIP THE NEXT INSTRUCTION + 973 STOP^ + 974 030771 254 04 0 00 030772 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 975 030772 324 00 0 00 030773 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 976 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 977 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 978 + 979 ;COND R [ADZ], AD00 S-A-1, AD AR + EN [ADAP], AD LT AR + EN [ADAP], + 980 ;AD RT AR + EN [ADAP], AD AR + EN C [ADAP], F CYC ACT EN A [F1] + 981 + 982 ;********** + 983 + 984 ;THIS TEST VERIFIES THAT SKIP IS DATA INDEPENDENT. HENCE, SKIP NEVER SKIPS. IT ALWAYS + 985 ;CONTINUES ON TO THE NEXT INSTRUCTION. + 986 + 987 030773 200 00 0 00 035063 A400: MOVE [-1] ;PRESET E TO ALL ONES + 988 030774 330 00 0 00 000000 SKIP ;*SKIP SHOULD NEVER SKIP THE NEXT INSTRUCTION + 989 030775 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + 990 STOP^ + 991 030776 254 04 0 00 030777 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 992 030777 324 00 0 00 031000 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 993 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 994 ;IN THE SUBTEST) TO LOOP ON ERROR^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 3-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0031 + + 995 + 996 ;PC CHANGE [PCC], PC CHANGE A [PCC], SEE MEMORY CONTROL DWGS [MC1-4] + 997 + 998 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 4 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0032 + + 999 ;THIS TEST VERIFIES THAT SKIPGE WILL NOT SKIP THE NEXT INSTRUCTION WHEN C(E) IS + 1000 ;NEGATIVE. THE MOVE INSTRUCTION LOADS E WITH -1, A NEGATIVE NUMBER. + 1001 + 1002 031000 200 00 0 00 035063 A500: MOVE [-1] ;PRESET E WITH ALL ONES, A NEGATIVE NUMBER + 1003 031001 335 00 0 00 000000 SKIPGE ;*SKIPGE SHOULD NOT SKIP THE NEXT INSTRUCTION + 1004 031002 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + 1005 STOP^ + 1006 031003 254 04 0 00 031004 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1007 031004 324 00 0 00 031005 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1008 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1009 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1010 + 1011 ;IF MOVE INSTRUCTION FAILED, CHECK MOVX-: FCE [F1], -IR XXXI [IR1], -IR XXXM [IR1], + 1012 ;IR HWTFWT [IR3], IR FWT [IR2], IR2XX [IR1] + 1013 + 1014 ;IF SKIPGE FAILED, CHECK SKIPX: FCE [F1] + 1015 + 1016 ;********** + 1017 + 1018 ;THIS TEST VERIFIES THAT SKIPN WILL NOT SKIP THE NEXT INSTRUCTION WHEN + 1019 ;C(E) IS ZERO. THE MOVE INSTRUCTION LOADS E WITH ALL ZEROS. + 1020 + 1021 031005 200 00 0 00 035062 A600: MOVE [0] ;PRESET E WITH ALL ZEROS + 1022 031006 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD NOT SKIP THE NEXT INSTRUCTION + 1023 031007 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + 1024 STOP^ + 1025 031010 254 04 0 00 031011 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1026 031011 324 00 0 00 031012 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1027 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1028 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1029 + 1030 ;********** + 1031 + 1032 ;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT SHOULD SKIP + 1033 ;THE NEXT INSTRUCTION WHEN C(E) IS NON-ZERO. THE MOVE INSTRUCTION LOADS + 1034 ;E WITH -1, A NON-ZERO NUMBER. + 1035 + 1036 031012 200 00 0 00 035063 A700: MOVE [-1] ;PRESET E WITH ALL ONES, A NON-ZERO NUMBER + 1037 031013 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1038 STOP^ + 1039 031014 254 04 0 00 031015 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1040 031015 324 00 0 00 031016 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1041 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1042 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1043 + 1044 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 5 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0033 + + 1045 ;THIS TEST VERIFIES THAT SKIPE IS DATA DEPENDENT. HENCE, IT SHOULD NOT + 1046 ;SKIP THE NEXT INSTRUCTION WHEN C(E) IS NON-ZERO. THE MOVE INSTRUCTION LOADS E + 1047 ;WITH -1, A NON-ZERO NUMBER. + 1048 + 1049 031016 200 00 0 00 035063 A1000: MOVE [-1] ;PRESET E WITH ALL ONES, A NON-ZERO NUMBER + 1050 031017 332 00 0 00 000000 SKIPE ;*SKIPE SHOULD NOT SKIP THE NEXT INSTRUCTION + 1051 031020 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + 1052 STOP^ + 1053 031021 254 04 0 00 031022 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1054 031022 324 00 0 00 031023 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1055 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1056 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1057 + 1058 ;********** + 1059 + 1060 ;THIS TEST VERIFIES THAT SKIPE IS DATA DEPENDENT. HENCE, IT SHOULD SKIP + 1061 ;THE NEXT INSTRUCTION WHEN C(E) IS ZERO. THE MOVE INSTRUCTION LOADS ALL + 1062 ;ZEROS INTO E. + 1063 + 1064 031023 200 00 0 00 035062 A1100: MOVE [0] ;PRESET E TO ALL ZEROS + 1065 031024 332 00 0 00 000000 SKIPE ;*SKIPE SHOULD SKIP THE NEXT INSTRUCTION + 1066 STOP^ + 1067 031025 254 04 0 00 031026 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1068 031026 324 00 0 00 031027 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1069 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1070 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1071 + 1072 ;********** + 1073 + 1074 ;THIS TEST VERIFIES THAT SKIP IS DATA INDEPENDENT. HENCE, IT NEVER SKIPS. IT ALWAYS + 1075 ;CONTINUES ON TO THE NEXT INSTRUCTION. + 1076 + 1077 031027 200 00 0 00 035062 A1200: MOVE [0] ;PRESET E TO ALL ZEROS + 1078 031030 330 00 0 00 000000 SKIP ;*SKIP SHOULD NEVER SKIP THE NEXT INSTRUCTION + 1079 031031 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + 1080 STOP^ + 1081 031032 254 04 0 00 031033 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1082 031033 324 00 0 00 031034 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1083 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1084 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1085 + 1086 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 6 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0034 + + 1087 ;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT + 1088 ;INSTRUCTION WHEN C(E) IS NOT NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD + 1089 ;E WITH ALL ZEROS, A NON-NEGATIVE NUMBER. + 1090 + 1091 031034 200 00 0 00 035062 A1300: MOVE [0] ;PRESET E T0 ZERO, A NON-NEGATIVE NUMBER + 1092 031035 331 00 0 00 000000 SKIPL ;*SKIPL SHOULD NOT SKIP THE NEXT INSTRUCTION + 1093 031036 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + 1094 STOP^ + 1095 031037 254 04 0 00 031040 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1096 031040 324 00 0 00 031041 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1097 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1098 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1099 + 1100 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 7 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0035 + + 1101 ;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT + 1102 ;INSTRUCTION WHEN C(E) IS NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD E WITH -1, + 1103 ;A NEGATIVE NUMBER. + 1104 + 1105 031041 200 00 0 00 035063 A1400: MOVE [-1] ;PRESET E TO -1, A NEGATIVE NUMBER + 1106 031042 331 00 0 00 000000 SKIPL ;*SKIPL SHOULD SKIP THE NEXT INSTRUCTION + 1107 STOP^ + 1108 031043 254 04 0 00 031044 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1109 031044 324 00 0 00 031045 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1110 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1111 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1112 + 1113 ;********** + 1114 + 1115 ;THIS TEST VERIFIES THAT SKIPG IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT + 1116 ;INSTRUCTION WHEN C(E) IS GREATER THAN ZERO. THE MOVE INSTRUCTION IS USED TO LOAD + 1117 ;E WITH A NON-ZERO POSITIVE NUMBER. + 1118 + 1119 031045 200 00 0 00 035064 A1500: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A NON-ZERO POSITIVE NUMBER + 1120 031046 337 00 0 00 000000 SKIPG ;*SKIPG SHOULD SKIP THE NEXT INSTRUCTION + 1121 STOP^ + 1122 031047 254 04 0 00 031050 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1123 031050 324 00 0 00 031051 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1124 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1125 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1126 + 1127 ;********** + 1128 + 1129 ;THIS TEST VERIFIES THAT SKIPG IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT + 1130 ;INSTRUCTION WHEN C(E) IS NOT GREATER THAN ZERO. THE MOVE INSTRUCTION IS USED TO + 1131 ;LOAD E WITH -1, A NUMBER NOT GREATER THAN ZERO. + 1132 + 1133 031051 200 00 0 00 035063 A1600: MOVE [-1] ;PRESET E TO -1, A NEGATIVE NUMBER + 1134 031052 337 00 0 00 000000 SKIPG ;*SKIPG SHOULD NOT SKIP THE NEXT INSTRUCTION + 1135 031053 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + 1136 STOP^ + 1137 031054 254 04 0 00 031055 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1138 031055 324 00 0 00 031056 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1139 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1140 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1141 + 1142 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 8 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0036 + + 1143 ;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT + 1144 ;INSTRUCTION WHEN C(E) IS NON-NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD E WITH + 1145 ;377777777777, A NON-NEGATIVE NUMBER. + 1146 + 1147 031056 200 00 0 00 035064 A1700: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A NON-NEGATIVE NUMBER + 1148 031057 331 00 0 00 000000 SKIPL ;*SKIPL SHOULD NOT SKIP THE NEXT INSTRUCTION + 1149 031060 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + 1150 STOP^ + 1151 031061 254 04 0 00 031062 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1152 031062 324 00 0 00 031063 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1153 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1154 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1155 + 1156 ;********** + 1157 + 1158 ;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT + 1159 ;INSTRUCTION WHEN C(E) IS ZERO. THE MOVE INSTRUCTION LOADS E WITH ALL ZEROS. + 1160 + 1161 031063 200 00 0 00 035062 A2000: MOVE [0] ;PRESET E TO ALL ZEROS + 1162 031064 333 00 0 00 000000 SKIPLE ;*SKIPLE SHOULD SKIP THE NEXT INSTRUCTION + 1163 STOP^ + 1164 031065 254 04 0 00 031066 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1165 031066 324 00 0 00 031067 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1166 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1167 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1168 + 1169 ;********** + 1170 + 1171 ;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT + 1172 ;INSTRUCTION WHEN E CONTAINS A NEGATIVE NUMBER. THE MOVE INSTRUCTION IS USED TO LOAD + 1173 ;E WITH 400000000000, A NEGATIVE NUMBER. + 1174 + 1175 031067 200 00 0 00 035065 A2100: MOVE [XWD 400000,0] ;PRESET E TO 400000000000, A NEGATIVE NUMBER. + 1176 031070 333 00 0 00 000000 SKIPLE ;*SKIPLE SHOULD SKIP THE NEXT INSTRUCTION + 1177 STOP^ + 1178 031071 254 04 0 00 031072 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1179 031072 324 00 0 00 031073 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1180 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1181 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1182 + 1183 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 9 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0037 + + 1184 ;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT + 1185 ;INSTRUCTION WHEN E CONTAINS A POSITIVE NUMBER. THE MOVE INSTRUCTION LOADS + 1186 ;E WITH 377777777777, A POSITIVE NUMBER. + 1187 + 1188 031073 200 00 0 00 035064 A2200: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A POSITIVE NUMBER + 1189 031074 333 00 0 00 000000 SKIPLE ;*SKIPLE SHOULD NOT SKIP THE NEXT INSTRUCTION + 1190 031075 334 00 0 00 000000 SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + 1191 STOP^ + 1192 031076 254 04 0 00 031077 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1193 031077 324 00 0 00 031100 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1194 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1195 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1196 + 1197 ;********* +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 10 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0038 + + 1198 ;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT + 1199 ;INSTRUCTION ;WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO + 1200 ;LOAD 777000000000, A NON-ZERO NUMBER INTO E. + 1201 + 1202 031100 200 00 0 00 035066 A2300: MOVE [XWD 777000,0] ;PRESET E WITH 777000000000, A NON-ZERO NUMBER + 1203 031101 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1204 STOP^ + 1205 031102 254 04 0 00 031103 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1206 031103 324 00 0 00 031104 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1207 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1208 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1209 + 1210 ;********** + 1211 + 1212 ;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT + 1213 ;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO + 1214 ;LOAD A NON-ZERO NUMBER, 377000000 INTO E. + 1215 + 1216 031104 200 00 0 00 035067 A2400: MOVE [XWD 377,0] ;PRESET E TO 377000000, A NON ZERO NUMBER + 1217 031105 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1218 STOP^ + 1219 031106 254 04 0 00 031107 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1220 031107 324 00 0 00 031110 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1221 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1222 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1223 + 1224 ;********** + 1225 + 1226 ;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT + 1227 ;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO + 1228 ;LOAD E WITH A NON-ZERO NUMBER, 177000. + 1229 + 1230 031110 200 00 0 00 035070 A2500: MOVE [177000] ;PRESET E WITH 177000, A NON ZERO NUMBER + 1231 031111 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1232 STOP^ + 1233 031112 254 04 0 00 031113 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1234 031113 324 00 0 00 031114 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1235 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1236 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1237 + 1238 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 11 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0039 + + 1239 ;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT + 1240 ;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO + 1241 ;LOAD 777, A NON-ZERO NUMBER INTO E. + 1242 + 1243 031114 200 00 0 00 035071 A2600: MOVE [777] ;PRESET E WITH 777, A NON-ZERO NUMBER + 1244 031115 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1245 STOP^ + 1246 031116 254 04 0 00 031117 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1247 031117 324 00 0 00 031120 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1248 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1249 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1250 + 1251 ;********** + 1252 + 1253 ;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT + 1254 ;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO + 1255 ;LOAD E WITH 600000, A NON-ZERO NUMBER. + 1256 + 1257 031120 200 00 0 00 035072 A2700: MOVE [600000] ;PRESET E WITK 600000, A NON-ZERO NUMBER + 1258 031121 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1259 STOP^ + 1260 031122 254 04 0 00 031123 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1261 031123 324 00 0 00 031124 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1262 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1263 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1264 + 1265 ;********** + 1266 + 1267 ;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT + 1268 ;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO + 1269 ;LOAD E WITH 400000000, A NON-ZERO NUMBER. + 1270 + 1271 031124 200 00 0 00 035073 A3000: MOVE [XWD 400,000] ;PRESET E WITH 400000000, A NON-ZERO NUMBER + 1272 031125 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1273 STOP^ + 1274 031126 254 04 0 00 031127 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1275 031127 324 00 0 00 031130 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1276 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1277 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1278 + 1279 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0040 + + 1280 003100 SN=3100 + 1281 000000 ZZ=0 + 1282 + 1283 A3100: REPEAT ^D36, + 1284 <;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1285 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1286 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1287 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1288 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1289 + 1290 SN=SN+1 + 1291 ZZ=ZZ+ZZ + 1292 IFE ZZ, + 1293 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1294 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1295 STOP + 1296 + 1297 ;********** + 1298 > + 1299 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1300 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1301 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1302 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1303 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1304 + 1305 003101 SN=SN+1 + 1306 000000 ZZ=ZZ+ZZ + 1307 000001 IFE ZZ, + 1308 031130 200 00 0 00 035074 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1309 031131 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1310 STOP^ + 1311 031132 254 04 0 00 031133 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1312 031133 324 00 0 00 031134 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1313 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1314 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1315 + 1316 ;********** + 1317 + 1318 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1319 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1320 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1321 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1322 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1323 + 1324 003102 SN=SN+1 + 1325 000002 ZZ=ZZ+ZZ + 1326 IFE ZZ, + 1327 031134 200 00 0 00 035075 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1328 031135 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1329 STOP^ + 1330 031136 254 04 0 00 031137 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1331 031137 324 00 0 00 031140 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1332 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1333 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1334 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0041 + + 1335 ;********** + 1336 + 1337 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1338 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1339 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1340 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1341 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1342 + 1343 003103 SN=SN+1 + 1344 000004 ZZ=ZZ+ZZ + 1345 IFE ZZ, + 1346 031140 200 00 0 00 035076 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1347 031141 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1348 STOP^ + 1349 031142 254 04 0 00 031143 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1350 031143 324 00 0 00 031144 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1351 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1352 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1353 + 1354 ;********** + 1355 + 1356 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1357 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1358 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1359 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1360 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1361 + 1362 003104 SN=SN+1 + 1363 000010 ZZ=ZZ+ZZ + 1364 IFE ZZ, + 1365 031144 200 00 0 00 035077 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1366 031145 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1367 STOP^ + 1368 031146 254 04 0 00 031147 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1369 031147 324 00 0 00 031150 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1370 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1371 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1372 + 1373 ;********** + 1374 + 1375 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1376 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1377 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1378 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1379 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1380 + 1381 003105 SN=SN+1 + 1382 000020 ZZ=ZZ+ZZ + 1383 IFE ZZ, + 1384 031150 200 00 0 00 035100 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1385 031151 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1386 STOP^ + 1387 031152 254 04 0 00 031153 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1388 031153 324 00 0 00 031154 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1389 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-2 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0042 + + 1390 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1391 + 1392 ;********** + 1393 + 1394 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1395 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1396 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1397 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1398 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1399 + 1400 003106 SN=SN+1 + 1401 000040 ZZ=ZZ+ZZ + 1402 IFE ZZ, + 1403 031154 200 00 0 00 035101 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1404 031155 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1405 STOP^ + 1406 031156 254 04 0 00 031157 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1407 031157 324 00 0 00 031160 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1408 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1409 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1410 + 1411 ;********** + 1412 + 1413 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1414 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1415 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1416 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1417 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1418 + 1419 003107 SN=SN+1 + 1420 000100 ZZ=ZZ+ZZ + 1421 IFE ZZ, + 1422 031160 200 00 0 00 035102 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1423 031161 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1424 STOP^ + 1425 031162 254 04 0 00 031163 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1426 031163 324 00 0 00 031164 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1427 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1428 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1429 + 1430 ;********** + 1431 + 1432 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1433 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1434 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1435 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1436 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1437 + 1438 003110 SN=SN+1 + 1439 000200 ZZ=ZZ+ZZ + 1440 IFE ZZ, + 1441 031164 200 00 0 00 035103 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1442 031165 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1443 STOP^ + 1444 031166 254 04 0 00 031167 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-3 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0043 + + 1445 031167 324 00 0 00 031170 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1446 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1447 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1448 + 1449 ;********** + 1450 + 1451 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1452 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1453 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1454 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1455 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1456 + 1457 003111 SN=SN+1 + 1458 000400 ZZ=ZZ+ZZ + 1459 IFE ZZ, + 1460 031170 200 00 0 00 035104 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1461 031171 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1462 STOP^ + 1463 031172 254 04 0 00 031173 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1464 031173 324 00 0 00 031174 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1465 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1466 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1467 + 1468 ;********** + 1469 + 1470 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1471 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1472 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1473 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1474 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1475 + 1476 003112 SN=SN+1 + 1477 001000 ZZ=ZZ+ZZ + 1478 IFE ZZ, + 1479 031174 200 00 0 00 035105 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1480 031175 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1481 STOP^ + 1482 031176 254 04 0 00 031177 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1483 031177 324 00 0 00 031200 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1484 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1485 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1486 + 1487 ;********** + 1488 + 1489 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1490 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1491 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1492 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1493 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1494 + 1495 003113 SN=SN+1 + 1496 002000 ZZ=ZZ+ZZ + 1497 IFE ZZ, + 1498 031200 200 00 0 00 035106 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1499 031201 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-4 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0044 + + 1500 STOP^ + 1501 031202 254 04 0 00 031203 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1502 031203 324 00 0 00 031204 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1503 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1504 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1505 + 1506 ;********** + 1507 + 1508 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1509 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1510 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1511 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1512 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1513 + 1514 003114 SN=SN+1 + 1515 004000 ZZ=ZZ+ZZ + 1516 IFE ZZ, + 1517 031204 200 00 0 00 035107 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1518 031205 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1519 STOP^ + 1520 031206 254 04 0 00 031207 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1521 031207 324 00 0 00 031210 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1522 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1523 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1524 + 1525 ;********** + 1526 + 1527 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1528 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1529 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1530 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1531 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1532 + 1533 003115 SN=SN+1 + 1534 010000 ZZ=ZZ+ZZ + 1535 IFE ZZ, + 1536 031210 200 00 0 00 035110 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1537 031211 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1538 STOP^ + 1539 031212 254 04 0 00 031213 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1540 031213 324 00 0 00 031214 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1541 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1542 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1543 + 1544 ;********** + 1545 + 1546 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1547 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1548 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1549 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1550 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1551 + 1552 003116 SN=SN+1 + 1553 020000 ZZ=ZZ+ZZ + 1554 IFE ZZ, +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-5 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0045 + + 1555 031214 200 00 0 00 035111 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1556 031215 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1557 STOP^ + 1558 031216 254 04 0 00 031217 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1559 031217 324 00 0 00 031220 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1560 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1561 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1562 + 1563 ;********** + 1564 + 1565 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1566 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1567 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1568 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1569 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1570 + 1571 003117 SN=SN+1 + 1572 040000 ZZ=ZZ+ZZ + 1573 IFE ZZ, + 1574 031220 200 00 0 00 035112 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1575 031221 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1576 STOP^ + 1577 031222 254 04 0 00 031223 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1578 031223 324 00 0 00 031224 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1579 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1580 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1581 + 1582 ;********** + 1583 + 1584 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1585 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1586 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1587 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1588 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1589 + 1590 003120 SN=SN+1 + 1591 100000 ZZ=ZZ+ZZ + 1592 IFE ZZ, + 1593 031224 200 00 0 00 035113 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1594 031225 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1595 STOP^ + 1596 031226 254 04 0 00 031227 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1597 031227 324 00 0 00 031230 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1598 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1599 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1600 + 1601 ;********** + 1602 + 1603 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1604 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1605 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1606 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1607 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1608 + 1609 003121 SN=SN+1 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-6 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0046 + + 1610 200000 ZZ=ZZ+ZZ + 1611 IFE ZZ, + 1612 031230 200 00 0 00 035114 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1613 031231 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1614 STOP^ + 1615 031232 254 04 0 00 031233 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1616 031233 324 00 0 00 031234 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1617 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1618 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1619 + 1620 ;********** + 1621 + 1622 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1623 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1624 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1625 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1626 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1627 + 1628 003122 SN=SN+1 + 1629 400000 ZZ=ZZ+ZZ + 1630 IFE ZZ, + 1631 031234 200 00 0 00 035115 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1632 031235 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1633 STOP^ + 1634 031236 254 04 0 00 031237 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1635 031237 324 00 0 00 031240 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1636 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1637 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1638 + 1639 ;********** + 1640 + 1641 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1642 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1643 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1644 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1645 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1646 + 1647 003123 SN=SN+1 + 1648 000001 000000 ZZ=ZZ+ZZ + 1649 IFE ZZ, + 1650 031240 200 00 0 00 035116 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1651 031241 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1652 STOP^ + 1653 031242 254 04 0 00 031243 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1654 031243 324 00 0 00 031244 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1655 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1656 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1657 + 1658 ;********** + 1659 + 1660 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1661 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1662 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1663 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1664 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-7 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0047 + + 1665 + 1666 003124 SN=SN+1 + 1667 000002 000000 ZZ=ZZ+ZZ + 1668 IFE ZZ, + 1669 031244 200 00 0 00 035117 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1670 031245 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1671 STOP^ + 1672 031246 254 04 0 00 031247 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1673 031247 324 00 0 00 031250 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1674 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1675 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1676 + 1677 ;********** + 1678 + 1679 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1680 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1681 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1682 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1683 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1684 + 1685 003125 SN=SN+1 + 1686 000004 000000 ZZ=ZZ+ZZ + 1687 IFE ZZ, + 1688 031250 200 00 0 00 035120 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1689 031251 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1690 STOP^ + 1691 031252 254 04 0 00 031253 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1692 031253 324 00 0 00 031254 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1693 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1694 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1695 + 1696 ;********** + 1697 + 1698 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1699 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1700 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1701 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1702 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1703 + 1704 003126 SN=SN+1 + 1705 000010 000000 ZZ=ZZ+ZZ + 1706 IFE ZZ, + 1707 031254 200 00 0 00 035121 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1708 031255 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1709 STOP^ + 1710 031256 254 04 0 00 031257 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1711 031257 324 00 0 00 031260 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1712 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1713 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1714 + 1715 ;********** + 1716 + 1717 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1718 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1719 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-8 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0048 + + 1720 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1721 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1722 + 1723 003127 SN=SN+1 + 1724 000020 000000 ZZ=ZZ+ZZ + 1725 IFE ZZ, + 1726 031260 200 00 0 00 035122 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1727 031261 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1728 STOP^ + 1729 031262 254 04 0 00 031263 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1730 031263 324 00 0 00 031264 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1731 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1732 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1733 + 1734 ;********** + 1735 + 1736 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1737 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1738 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1739 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1740 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1741 + 1742 003130 SN=SN+1 + 1743 000040 000000 ZZ=ZZ+ZZ + 1744 IFE ZZ, + 1745 031264 200 00 0 00 035123 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1746 031265 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1747 STOP^ + 1748 031266 254 04 0 00 031267 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1749 031267 324 00 0 00 031270 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1750 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1751 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1752 + 1753 ;********** + 1754 + 1755 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1756 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1757 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1758 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1759 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1760 + 1761 003131 SN=SN+1 + 1762 000100 000000 ZZ=ZZ+ZZ + 1763 IFE ZZ, + 1764 031270 200 00 0 00 035124 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1765 031271 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1766 STOP^ + 1767 031272 254 04 0 00 031273 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1768 031273 324 00 0 00 031274 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1769 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1770 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1771 + 1772 ;********** + 1773 + 1774 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-9 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0049 + + 1775 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1776 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1777 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1778 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1779 + 1780 003132 SN=SN+1 + 1781 000200 000000 ZZ=ZZ+ZZ + 1782 IFE ZZ, + 1783 031274 200 00 0 00 035125 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1784 031275 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1785 STOP^ + 1786 031276 254 04 0 00 031277 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1787 031277 324 00 0 00 031300 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1788 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1789 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1790 + 1791 ;********** + 1792 + 1793 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1794 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1795 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1796 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1797 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1798 + 1799 003133 SN=SN+1 + 1800 000400 000000 ZZ=ZZ+ZZ + 1801 IFE ZZ, + 1802 031300 200 00 0 00 035073 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1803 031301 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1804 STOP^ + 1805 031302 254 04 0 00 031303 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1806 031303 324 00 0 00 031304 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1807 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1808 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1809 + 1810 ;********** + 1811 + 1812 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1813 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1814 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1815 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1816 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1817 + 1818 003134 SN=SN+1 + 1819 001000 000000 ZZ=ZZ+ZZ + 1820 IFE ZZ, + 1821 031304 200 00 0 00 035126 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1822 031305 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1823 STOP^ + 1824 031306 254 04 0 00 031307 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1825 031307 324 00 0 00 031310 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1826 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1827 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1828 + 1829 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-10 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0050 + + 1830 + 1831 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1832 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1833 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1834 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1835 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1836 + 1837 003135 SN=SN+1 + 1838 002000 000000 ZZ=ZZ+ZZ + 1839 IFE ZZ, + 1840 031310 200 00 0 00 035127 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1841 031311 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1842 STOP^ + 1843 031312 254 04 0 00 031313 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1844 031313 324 00 0 00 031314 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1845 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1846 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1847 + 1848 ;********** + 1849 + 1850 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1851 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1852 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1853 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1854 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1855 + 1856 003136 SN=SN+1 + 1857 004000 000000 ZZ=ZZ+ZZ + 1858 IFE ZZ, + 1859 031314 200 00 0 00 035130 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1860 031315 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1861 STOP^ + 1862 031316 254 04 0 00 031317 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1863 031317 324 00 0 00 031320 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1864 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1865 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1866 + 1867 ;********** + 1868 + 1869 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1870 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1871 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1872 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1873 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1874 + 1875 003137 SN=SN+1 + 1876 010000 000000 ZZ=ZZ+ZZ + 1877 IFE ZZ, + 1878 031320 200 00 0 00 035131 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1879 031321 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1880 STOP^ + 1881 031322 254 04 0 00 031323 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1882 031323 324 00 0 00 031324 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1883 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1884 ;IN THE SUBTEST) TO LOOP ON ERROR^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-11 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0051 + + 1885 + 1886 ;********** + 1887 + 1888 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1889 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1890 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1891 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1892 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1893 + 1894 003140 SN=SN+1 + 1895 020000 000000 ZZ=ZZ+ZZ + 1896 IFE ZZ, + 1897 031324 200 00 0 00 035132 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1898 031325 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1899 STOP^ + 1900 031326 254 04 0 00 031327 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1901 031327 324 00 0 00 031330 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1902 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1903 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1904 + 1905 ;********** + 1906 + 1907 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1908 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1909 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1910 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1911 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1912 + 1913 003141 SN=SN+1 + 1914 040000 000000 ZZ=ZZ+ZZ + 1915 IFE ZZ, + 1916 031330 200 00 0 00 035133 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1917 031331 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1918 STOP^ + 1919 031332 254 04 0 00 031333 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1920 031333 324 00 0 00 031334 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1921 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1922 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1923 + 1924 ;********** + 1925 + 1926 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1927 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1928 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1929 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1930 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1931 + 1932 003142 SN=SN+1 + 1933 100000 000000 ZZ=ZZ+ZZ + 1934 IFE ZZ, + 1935 031334 200 00 0 00 035134 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1936 031335 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1937 STOP^ + 1938 031336 254 04 0 00 031337 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1939 031337 324 00 0 00 031340 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 12-12 +DAKAAM MAC 19-JAN-77 13:42 TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS SEQ 0052 + + 1940 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1941 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1942 + 1943 ;********** + 1944 + 1945 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1946 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1947 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1948 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1949 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1950 + 1951 003143 SN=SN+1 + 1952 200000 000000 ZZ=ZZ+ZZ + 1953 IFE ZZ, + 1954 031340 200 00 0 00 035135 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1955 031341 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1956 STOP^ + 1957 031342 254 04 0 00 031343 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1958 031343 324 00 0 00 031344 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1959 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1960 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1961 + 1962 ;********** + 1963 + 1964 ;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST + 1965 ;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE + 1966 ;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO + 1967 ;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). + 1968 ;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + 1969 + 1970 003144 SN=SN+1 + 1971 400000 000000 ZZ=ZZ+ZZ + 1972 IFE ZZ, + 1973 031344 200 00 0 00 035065 MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + 1974 031345 336 00 0 00 000000 SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + 1975 STOP^ + 1976 031346 254 04 0 00 031347 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1977 031347 324 00 0 00 031350 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1978 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1979 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1980 + 1981 ;********** + 1982 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 13 +DAKAAM MAC 19-JAN-77 13:42 TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS SEQ 0053 + + 1983 SUBTTL TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS + 1984 + 1985 ;THIS TEST VERIFIES THAT MOVEI LOADS E INTO THE AC. SKIPG IS USED TO CHECK THAT + 1986 ;THE AC WAS INDEED MODIFIED BY MOVEI. + 1987 + 1988 031350 201 00 0 00 035063 A3200: MOVEI [-1] ;*MOVEI SHOULD LOAD A POSITIVE NUMBER INTO AC0 + 1989 031351 337 00 0 00 000000 SKIPG ;SKIP HALT INSTRUCTION IF MOVEI LOADED AC CORRECTLY + 1990 STOP^ + 1991 031352 254 04 0 00 031353 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 1992 031353 324 00 0 00 031354 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 1993 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 1994 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 1995 + 1996 ;IF PROGRAM HANGS UP ON MOVEI INSTRUCTION, CHECK AD FM +EN [ADFP], AD FM + EN A [ADFP], + 1997 ;AD FM + F/F [ADFP], F CYC ACT EN C [F1] + 1998 + 1999 ;********** + 2000 + 2001 ;THIS TEST VERIFIES THAT SKIPL IS SENSITIVE TO THE DATA IN E AND NOT SENSITIVE TO + 2002 ;THE DATA IN THE AC. HENCE, SKIPL WILL SKIP THE NEXT INSTRUCTION ONLY IF + 2003 ;C(E) IS LESS THAN ZERO. E IS SET TO -1, A NEGATIVE NUMBER; AND + 2004 ;THE AC IS SET TO 0 IN THIS TEST. THEREFORE SKIPL SHOULD SKIP. + 2005 + 2006 031354 200 00 0 00 035062 A3300: MOVE [0] ;SET THE AC TO ALL ZEROS + 2007 031355 331 00 0 00 035063 SKIPL [-1] ;*SKIPL SHOULD SKIP BECAUSE C(E) IS NEGATIVE + 2008 STOP^ + 2009 031356 254 04 0 00 031357 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2010 031357 324 00 0 00 031360 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2011 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2012 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2013 + 2014 ;********** + 2015 + 2016 ;THIS TEST VERIFIES THAT SKIPL IS SENSITIVE TO THE SIGN BIT IN E. SKIPL SHOULD + 2017 ;SKIP THE NEXT INSTRUCTION WHEN AND ONLY WHEN THE SIGN BIT IS SET. IN THIS + 2018 ;TEST, C(AC) IS PRESET TO ALL ZEROS, THEN SKIPL IS EXECUTED WITH C(E)=400000,,0. + 2019 ;A FAILURE UCCURS IF SKIPL DOES NOT SKIP. + 2020 + 2021 031360 200 00 0 00 035062 A13100: MOVE [0] ;SET THE AC TO ALL ZEROS + 2022 031361 331 00 0 00 035065 SKIPL [400000,,0] ;*SKIPL SHOULD SKIP BECAUSE C(E) IS NEGATIVE + 2023 STOP^ + 2024 031362 254 04 0 00 031363 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2025 031363 324 00 0 00 031364 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2026 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2027 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2028 + 2029 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 14 +DAKAAM MAC 19-JAN-77 13:42 TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS SEQ 0054 + + 2030 ;THIS TEST VERIFIES THAT THE CAM INSTRUCTION DOES NOT MODIFY C(AC). THE AC IS + 2031 ;INITIALLY LOADED WITH ALL ZEROS; THEN CAM IS EXECUTED WITH C(E) NON-ZERO. + 2032 ;THE AC IS THEN CHECKED FOR ALL ZEROS. THIS TEST FAILS IF THE AC WAS MODIFIED BY CAM. + 2033 + 2034 031364 200 00 0 00 035062 A3400: MOVE [0] ;PRESET THE AC TO ALL ZEROS + 2035 031365 310 00 0 00 035136 CAM [1234] ;*CAM SHOULD NOT MODIFY THE AC + 2036 031366 200 00 0 00 000000 MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION SERVES AS A NO-OP. + 2037 ;IT IS A PROGRAM FILLER IN CASE CAM SKIPS.) + 2038 031367 332 00 0 00 000000 SKIPE ;SKIP HALT INSTRUCTION IF AC WAS NOT MODIFIED BY CAM + 2039 STOP^ + 2040 031370 254 04 0 00 031371 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2041 031371 324 00 0 00 031372 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2042 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2043 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2044 + 2045 ;IF TEST HANGS ON CAM INSTRUCTION, AB PC EN [ABC], AB PC (FETCH) EN [ABC], + 2046 ;IR CAXX [IR3] + 2047 + 2048 ;IF TEST LOOPS ON CAM INSTRUCTION, MC INSTR FETCH EN [MC1], MC INST FET ET2 EN [MC1], + 2049 ;ET2 J F/F [E], IR CAXX [IR3] + 2050 + 2051 ;IF TEST HALTS AT HALT INSTRUCTION, SAC INH [ST2], IR CAXX [IR3], IR CAMX [IR1] + 2052 + 2053 ;********** + 2054 + 2055 ;THIS TEST VERIFIES THAT THE CAI INSTRUCTION DOES NOT MODIFY C(AC). THE AC IS + 2056 ;INITIALLY LOADED WITH ALL ZEROS; THEN CAI IS EXECUTED WITH E NON-ZERO. THE AC IS + 2057 ;THEN CHECKED FOR ALL ZEROS. THIS TEST FAILS IF THE AC WAS MODIFIED BY CAM. + 2058 + 2059 031372 200 00 0 00 035062 A3500: MOVE [0] ;PRESET THE AC TO ALL ZEROS + 2060 031373 300 00 0 00 035136 CAI [1234] ;*CAI SHOULD NOT MODIFY THE AC + 2061 031374 200 00 0 00 000000 MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A PROGRAM FILLER + 2062 ;IN CASE CAI MODIFIES THE AC. IT SERVES AS A NO-OP) + 2063 031375 332 00 0 00 000000 SKIPE ;SKIP HALT INSTRUCTION IF CAI DID NOT MODIFY THE AC + 2064 STOP^ + 2065 031376 254 04 0 00 031377 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2066 031377 324 00 0 00 031400 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2067 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2068 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2069 + 2070 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 15 +DAKAAM MAC 19-JAN-77 13:42 TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS SEQ 0055 + + 2071 ;THIS TEST VERIFIES THAT CAI DOES NOT MODIFY C(AC0) WHEN AN AC OTHER THAN AC0 IS + 2072 ;SPECIFIED BY THE CAI INSTRUCTION. + 2073 ;THIS IS THE FIRST TEST THAT USES AN AC OTHER THAN AC0. + 2074 ;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, THE CAI INSTRUCTION IS EXECUTED WITH + 2075 ;A NON-ZERO E. AC 0 IS THEN TESTED F0R ALL ZEROS. + 2076 + 2077 031400 200 00 0 00 035062 A3600: MOVE [0] ;LOAD AC0 WITH ALL ZEROS + 2078 031401 300 17 0 00 035136 CAI 17,[1234] ;*CAI SHOULD NOT MODIFY AC0 + 2079 031402 200 00 0 00 000000 MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + 2080 ;CAI SKIPS THE NEXT INSTRUCTION. IT ACTS AS A NO-OP) + 2081 031403 332 00 0 00 000000 SKIPE ;SKIP HALT INSTRUCTION IF C(AC0) WAS NOT MODIFIED + 2082 STOP^ + 2083 031404 254 04 0 00 031405 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2084 031405 324 00 0 00 031406 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2085 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2086 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2087 + 2088 ;********** + 2089 + 2090 ;THIS TEST VERIFIES THAT CAM DOES NOT MODIFY C(AC0) WHEN AN AC OTHER THAN AC0 IS + 2091 ;SPECIFIED BY THE CAM INSTRUCTION. + 2092 ;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, THE CAM INSTRUCTION IS EXECUTED WITH + 2093 ;A NON-ZERO E. AC 0 IS THEN TESTED F0R ALL ZEROS. + 2094 + 2095 031406 200 00 0 00 035062 A3700: MOVE [0] ;PRESET C(AC0) TO ALL ZEROS + 2096 031407 310 17 0 00 035136 CAM 17,[1234] ;*CAM SHOULD NOT MODIFY AC0 + 2097 031410 200 00 0 00 000000 MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + 2098 ;CAM SKIPS THE NEXT INSTRUCTION. IT SERVES AS A NO-OP) + 2099 031411 332 00 0 00 000000 SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS NOT MODIFIED BY CAM + 2100 STOP^ + 2101 031412 254 04 0 00 031413 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2102 031413 324 00 0 00 031414 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2103 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2104 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2105 + 2106 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 16 +DAKAAM MAC 19-JAN-77 13:42 TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS SEQ 0056 + + 2107 ;THIS TEST VERIFIES THAT MOVE WILL PLACE C(E) INTO THE AC. AC, E AND C(E) ARE NON-ZERO + 2108 ;IN THIS TEST. THIS TEST PASSES IF FINAL C(AC) OF THE MOVE INSTRUCTION IS NON-ZERO. + 2109 + 2110 031414 200 01 0 00 035062 A4500: MOVE 1,[0] ;LOAD THE AC WITH ALL ZEROS + 2111 031415 200 01 0 00 035063 MOVE 1,[-1] ;*MOVE SHOULD PLACE ALL ONES IN THE AC + 2112 031416 336 00 0 00 000001 SKIPN 1 ;SKIP HALT INSTRUCTION IF MOVE MODIFIED THE AC + 2113 STOP^ + 2114 031417 254 04 0 00 031420 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2115 031420 324 00 0 00 031421 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2116 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2117 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2118 + 2119 ;********** + 2120 + 2121 ;THIS TEST VERIFIES THAT THE MOVE INSTRUCTION WILL PLACE C(E) INTO THE AC. + 2122 ;AC, E ARE NON-ZERO AND C(E)=0. + 2123 ;THIS TEST FAILS IF THE FINAL C(AC) IS NOT EQUAL TO ZERO, THE ORIGINAL C(E) + 2124 + 2125 031421 200 01 0 00 035062 A4600: MOVE 1,[0] ;*PLACE ALL ZEROS INTO THE AC + 2126 031422 332 00 0 00 000001 SKIPE 1 ;SKIP HALT INSTRUCTION IF MOVE PLACED ALL ZEROS INTO AC + 2127 STOP^ + 2128 031423 254 04 0 00 031424 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2129 031424 324 00 0 00 031425 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2130 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2131 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2132 + 2133 ;IF JUMP .+1 OF PREVIOUS TEST FAILED, CHECK JUMPS: AD FM + EN [ADFP], IR JUMPS [IR1], + 2134 ;IR JUMPX [IR1], AR AD EN [ARMA], AR AD EN A [ARMA], AR LT AD EN A [ARMA], + 2135 ;AR LT ADD EN B [ARMA], AR RT AD EN A [ARMA], + 2136 ;AR RT AD EN B [ARMA], AR A CLK INPUT [ARMA], + 2137 ;ET2 C F/F [E], AR AD ET2 EN [ARMA], AR AD BR (ET2) B [ARMA] + 2138 + 2139 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 17 +DAKAAM MAC 19-JAN-77 13:42 TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS SEQ 0057 + + 2140 ;THIS TEST VERIFIES THAT SOS DOES NOT MODIFY C(AC) WHEN THE AC IS AC0. + 2141 ;FIRST, AC0 IS LOADED WITH ALL ZEROS AND E IS LOADED WITH A POSITIVE NUMBER; + 2142 ;THEN, SOS IS EXECUTED. SOS SHOULD SUBTRACT ONE FROM C((E) + 2143 ;AND EXECUTE THE NEXT INSTRUCTION. IT SHOULD NOT MODIFY AC0. + 2144 ;THE TEST PASSES IF AC0 WAS NOT MODIFIED. + 2145 + 2146 031425 200 00 0 00 035062 A4000: MOVE [0] ;PRESET AC0 TO ALL ZEROS + 2147 031426 200 07 0 00 035136 MOVE 7,[1234] ;PRESET E WITH A POSITIVE NUMBER + 2148 031427 370 00 0 00 000007 SOS 7 ;*SOS SHOULD NOT MODIFY C(AC0) + 2149 031430 200 00 0 00 000000 MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + 2150 ;SOS SKIPS THE NEXT INSTRUCTION. IT SERVES AS A NO-OP) + 2151 031431 332 00 0 00 000000 SKIPE ;SKIP THE HALT INSTRUCTION IF C(AC0) WAS NOT MODIFIED + 2152 STOP^ + 2153 031432 254 04 0 00 031433 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2154 031433 324 00 0 00 031434 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2155 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2156 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2157 + 2158 ;********** + 2159 004100 SN=4100 + 2160 000000 AC=0 + 2161 + 2162 A4100: REPEAT 4, + 2163 <;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL PLACE THE CONTENTS OF E INTO THE AC + 2164 ;WHEN THE AC IS NOT AC0. THIS TEST IS PERFORMED FOR AC1, AC2, AC4 AND AC10. + 2165 ;FIRST, THE AC IS LOADED WITH ALL ZEROS; THEN, THE SKIP INSTRUCTION IS EXECUTED + 2166 ;WITH AC1, AC2, AC4 OR AC10. C(E) IS EQUAL TO -1. THE SKIP INSTRUCTION SHOULD CAUSE + 2167 ;THE AC TO CONTAIN -1. THE AC IS THEN CHECKED FOR + 2168 ;NON-ZERO CONTENTS. THE TEST FAILS IF C(AC)=0. + 2169 ;THIS TEST ALSO VERIFIES THAT 'SKIP AC,[-1]' DOES NOT CLOBBER AC0. + 2170 + 2171 SN=SN+1 + 2172 AC=AC+AC + 2173 IFE AC, + 2174 MOVE 0,[0] ;PRESET AC0 TO ALL ZEROS + 2175 MOVE AC,[0] ;PRESET THE AC TO ALL ZEROS + 2176 SKIP AC,[-1] ;*SKIP SHOULD PLACE -1 INTO THE AC + 2177 SKIPN AC ;SKIP THE HALT INSTRUCTION IF THE AC IS NON-ZERO + 2178 STOP + 2179 SKIPE 0 ;VERIFY THAT AC0 DID NOT GET CLOBBERED BY SKIP + 2180 STOP + 2181 + 2182 ;********** + 2183 > + 2184 ;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL PLACE THE CONTENTS OF E INTO THE AC + 2185 ;WHEN THE AC IS NOT AC0. THIS TEST IS PERFORMED FOR AC1, AC2, AC4 AND AC10. + 2186 ;FIRST, THE AC IS LOADED WITH ALL ZEROS; THEN, THE SKIP INSTRUCTION IS EXECUTED + 2187 ;WITH AC1, AC2, AC4 OR AC10. C(E) IS EQUAL TO -1. THE SKIP INSTRUCTION SHOULD CAUSE + 2188 ;THE AC TO CONTAIN -1. THE AC IS THEN CHECKED FOR + 2189 ;NON-ZERO CONTENTS. THE TEST FAILS IF C(AC)=0. + 2190 ;THIS TEST ALSO VERIFIES THAT 'SKIP AC,[-1]' DOES NOT CLOBBER AC0. + 2191 + 2192 004101 SN=SN+1 + 2193 000000 AC=AC+AC + 2194 000001 IFE AC, +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 17-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS SEQ 0058 + + 2195 031434 200 00 0 00 035062 MOVE 0,[0] ;PRESET AC0 TO ALL ZEROS + 2196 031435 200 01 0 00 035062 MOVE AC,[0] ;PRESET THE AC TO ALL ZEROS + 2197 031436 330 01 0 00 035063 SKIP AC,[-1] ;*SKIP SHOULD PLACE -1 INTO THE AC + 2198 031437 336 00 0 00 000001 SKIPN AC ;SKIP THE HALT INSTRUCTION IF THE AC IS NON-ZERO + 2199 STOP^ + 2200 031440 254 04 0 00 031441 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2201 031441 324 00 0 00 031442 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2202 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2203 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2204 031442 332 00 0 00 000000 SKIPE 0 ;VERIFY THAT AC0 DID NOT GET CLOBBERED BY SKIP + 2205 STOP^ + 2206 031443 254 04 0 00 031444 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2207 031444 324 00 0 00 031445 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2208 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2209 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2210 + 2211 ;********** + 2212 + 2213 ;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL PLACE THE CONTENTS OF E INTO THE AC + 2214 ;WHEN THE AC IS NOT AC0. THIS TEST IS PERFORMED FOR AC1, AC2, AC4 AND AC10. + 2215 ;FIRST, THE AC IS LOADED WITH ALL ZEROS; THEN, THE SKIP INSTRUCTION IS EXECUTED + 2216 ;WITH AC1, AC2, AC4 OR AC10. C(E) IS EQUAL TO -1. THE SKIP INSTRUCTION SHOULD CAUSE + 2217 ;THE AC TO CONTAIN -1. THE AC IS THEN CHECKED FOR + 2218 ;NON-ZERO CONTENTS. THE TEST FAILS IF C(AC)=0. + 2219 ;THIS TEST ALSO VERIFIES THAT 'SKIP AC,[-1]' DOES NOT CLOBBER AC0. + 2220 + 2221 004102 SN=SN+1 + 2222 000002 AC=AC+AC + 2223 IFE AC, + 2224 031445 200 00 0 00 035062 MOVE 0,[0] ;PRESET AC0 TO ALL ZEROS + 2225 031446 200 02 0 00 035062 MOVE AC,[0] ;PRESET THE AC TO ALL ZEROS + 2226 031447 330 02 0 00 035063 SKIP AC,[-1] ;*SKIP SHOULD PLACE -1 INTO THE AC + 2227 031450 336 00 0 00 000002 SKIPN AC ;SKIP THE HALT INSTRUCTION IF THE AC IS NON-ZERO + 2228 STOP^ + 2229 031451 254 04 0 00 031452 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2230 031452 324 00 0 00 031453 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2231 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2232 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2233 031453 332 00 0 00 000000 SKIPE 0 ;VERIFY THAT AC0 DID NOT GET CLOBBERED BY SKIP + 2234 STOP^ + 2235 031454 254 04 0 00 031455 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2236 031455 324 00 0 00 031456 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2237 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2238 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2239 + 2240 ;********** + 2241 + 2242 ;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL PLACE THE CONTENTS OF E INTO THE AC + 2243 ;WHEN THE AC IS NOT AC0. THIS TEST IS PERFORMED FOR AC1, AC2, AC4 AND AC10. + 2244 ;FIRST, THE AC IS LOADED WITH ALL ZEROS; THEN, THE SKIP INSTRUCTION IS EXECUTED + 2245 ;WITH AC1, AC2, AC4 OR AC10. C(E) IS EQUAL TO -1. THE SKIP INSTRUCTION SHOULD CAUSE + 2246 ;THE AC TO CONTAIN -1. THE AC IS THEN CHECKED FOR + 2247 ;NON-ZERO CONTENTS. THE TEST FAILS IF C(AC)=0. + 2248 ;THIS TEST ALSO VERIFIES THAT 'SKIP AC,[-1]' DOES NOT CLOBBER AC0. + 2249 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 17-2 +DAKAAM MAC 19-JAN-77 13:42 TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS SEQ 0059 + + 2250 004103 SN=SN+1 + 2251 000004 AC=AC+AC + 2252 IFE AC, + 2253 031456 200 00 0 00 035062 MOVE 0,[0] ;PRESET AC0 TO ALL ZEROS + 2254 031457 200 04 0 00 035062 MOVE AC,[0] ;PRESET THE AC TO ALL ZEROS + 2255 031460 330 04 0 00 035063 SKIP AC,[-1] ;*SKIP SHOULD PLACE -1 INTO THE AC + 2256 031461 336 00 0 00 000004 SKIPN AC ;SKIP THE HALT INSTRUCTION IF THE AC IS NON-ZERO + 2257 STOP^ + 2258 031462 254 04 0 00 031463 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2259 031463 324 00 0 00 031464 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2260 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2261 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2262 031464 332 00 0 00 000000 SKIPE 0 ;VERIFY THAT AC0 DID NOT GET CLOBBERED BY SKIP + 2263 STOP^ + 2264 031465 254 04 0 00 031466 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2265 031466 324 00 0 00 031467 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2266 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2267 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2268 + 2269 ;********** + 2270 + 2271 ;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL PLACE THE CONTENTS OF E INTO THE AC + 2272 ;WHEN THE AC IS NOT AC0. THIS TEST IS PERFORMED FOR AC1, AC2, AC4 AND AC10. + 2273 ;FIRST, THE AC IS LOADED WITH ALL ZEROS; THEN, THE SKIP INSTRUCTION IS EXECUTED + 2274 ;WITH AC1, AC2, AC4 OR AC10. C(E) IS EQUAL TO -1. THE SKIP INSTRUCTION SHOULD CAUSE + 2275 ;THE AC TO CONTAIN -1. THE AC IS THEN CHECKED FOR + 2276 ;NON-ZERO CONTENTS. THE TEST FAILS IF C(AC)=0. + 2277 ;THIS TEST ALSO VERIFIES THAT 'SKIP AC,[-1]' DOES NOT CLOBBER AC0. + 2278 + 2279 004104 SN=SN+1 + 2280 000010 AC=AC+AC + 2281 IFE AC, + 2282 031467 200 00 0 00 035062 MOVE 0,[0] ;PRESET AC0 TO ALL ZEROS + 2283 031470 200 10 0 00 035062 MOVE AC,[0] ;PRESET THE AC TO ALL ZEROS + 2284 031471 330 10 0 00 035063 SKIP AC,[-1] ;*SKIP SHOULD PLACE -1 INTO THE AC + 2285 031472 336 00 0 00 000010 SKIPN AC ;SKIP THE HALT INSTRUCTION IF THE AC IS NON-ZERO + 2286 STOP^ + 2287 031473 254 04 0 00 031474 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2288 031474 324 00 0 00 031475 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2289 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2290 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2291 031475 332 00 0 00 000000 SKIPE 0 ;VERIFY THAT AC0 DID NOT GET CLOBBERED BY SKIP + 2292 STOP^ + 2293 031476 254 04 0 00 031477 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2294 031477 324 00 0 00 031500 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2295 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2296 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2297 + 2298 ;********** + 2299 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 18 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0060 + + 2300 SUBTTL TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS + 2301 + 2302 ;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL NOT MODIFY THE CONTENTS OF AC0. + 2303 ;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, + 2304 ;THE SKIP INSTRUCTION IS EXECUTED WITH C(E)=-1 AND AC0. AC0 IS THEN TESTED FOR ALL + 2305 ;ZEROS. THE TEST FAILS IF AC0 WAS MODIFIED BY THE SKIP INSTRUCTION. + 2306 + 2307 031500 200 00 0 00 035062 A4200: MOVE [0] ;PRESET AC0 TO ALL ZEROS + 2308 031501 330 00 0 00 035063 SKIP [-1] ;*SKIP SHOULD NOT MODIFY AC0 + 2309 031502 332 00 0 00 000000 SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS NOT MODIFIED BY SKIP + 2310 STOP^ + 2311 031503 254 04 0 00 031504 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2312 031504 324 00 0 00 031505 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2313 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2314 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2315 + 2316 ;AC=0: SAC INH, IR BIT 9, 10, 11 OR 12 S-A-1 + 2317 + 2318 ;********** + 2319 + 2320 ;THIS TEST VERIFIES THAT MOVSS DOES NOT MODIFY C(AC0) WHEN AC0 IS SPECIFIED AS THE AC. + 2321 ;FIRST, E IS LOADED WITH ALL ONES AND THE AC (AC0) + 2322 ;IS LOADED WITH ALL ZEROS; THEN, MOVSS IS EXECUTED. MOVSS SHOULD NOT MODIFY AC0. + 2323 ;AC0 IS THEN CHECKED TO INSURE THAT IT WAS NOT MODIFIED BY MOVSS. + 2324 + 2325 031505 200 07 0 00 035063 A4300: MOVE 7,[-1] ;LOAD E OF THE MOVSS INSTRUCTION WITH ALL ONES + 2326 031506 200 00 0 00 035062 MOVE [0] ;PRESET AC0 TO ALL ZEROS + 2327 031507 207 00 0 00 000007 MOVSS 7 ;*MOVSS SHOULD NOT MODIFY AC0 + 2328 031510 332 00 0 00 000000 SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS INTACT + 2329 STOP^ + 2330 031511 254 04 0 00 031512 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2331 031512 324 00 0 00 031513 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2332 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2333 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2334 + 2335 ;MOVX & AC=0: SAC INH [ST2], IR XXXS [IR1] + 2336 + 2337 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 19 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0061 + + 2338 ;THIS TEST VERIFIES THAT HRRS DOES NOT MODIFY C(AC0) WHEN AC0 IS THE SPECIFIED AC. + 2339 ;FIRST, E IS LOADED WITH ALL ONES AND AC0 IS LOADED WITH ALL ZEROS; + 2340 ;THEN, HRRS IS EXECUTED WITH AC=AC0. AC0 IS THEN CHECKED FOR ALL ZEROS. + 2341 ;IF AC0 WAS MODIFIED BY THE HRRS INSTRUCTION, THE TEST FAIL. + 2342 + 2343 031513 200 07 0 00 035063 A4400: MOVE 7,[-1] ;PRESET E OF THE HRRS INSTRUCTION TO ALL ONES + 2344 031514 200 00 0 00 035062 MOVE [0] ;PRESET AC0 TO ALL ZEROS + 2345 031515 543 00 0 00 000007 HRRS 7 ;*HRRS SHOULD NOT MODIFY THE AC WHEN THE AC IS AC0 + 2346 031516 332 00 0 00 000000 SKIPE ;SKIP THE HALT INSTRUCTION IF AC0 WAS INTACT + 2347 STOP^ + 2348 031517 254 04 0 00 031520 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2349 031520 324 00 0 00 031521 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2350 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2351 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2352 + 2353 ;(HXXXS&AC=0): SAC INH [ST2], IR HWTFWT [IR3], IR XXXS [IR1], IR HWT [IR1], + 2354 ;IR BITS S-A-1 + 2355 + 2356 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 20 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0062 + + 2357 ;THIS TEST VERIFIES THAT THE JUMP INSTRUCTION DOES NOT MODIFY C(AC). + 2358 ;THE AC IS FIRST LOADED WITH ALL ZEROS; + 2359 ;THEN JUMP .+1 IS EXECUTED. THE AC IS THEN CHECKED TO INSURE THAT IT WAS NOT MODIFIED + 2360 ;BY JUMP. THE TEST FAILS IF THE AC WAS MODIFIED + 2361 + 2362 031521 200 00 0 00 035062 A4700: MOVE [0] ; PRESET THE AC TO ALL ZEROS + 2363 031522 320 00 0 00 031523 JUMP .+1 ;*JUMP SHOULD NOT MODIFY THE AC + 2364 031523 332 00 0 00 000000 SKIPE ;SKIP HALT INSTRUCTION IF AC WAS INTACT + 2365 STOP^ + 2366 031524 254 04 0 00 031525 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2367 031525 324 00 0 00 031526 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2368 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2369 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2370 + 2371 ;********** + 2372 + 2373 ;THIS TEST VERIFIES THAT MOVS WILL NOT PICK UP ANY ONES WHEN SWAPPING A WORD + 2374 ;OF ALL ZEROS. FIRST, E IS LOADED WITH A WORD OF ALL ZEROS; + 2375 ;THEN, MOVS IS EXECUTED. THE AC IS THEN CHECKED TO INSURE THAT THE RESULT DOES NOT + 2376 ;CONTAIN ANY ONES. + 2377 + 2378 031526 200 00 0 00 035062 A5000: MOVE [0] ;PRESET E WITH ALL ZEROS + 2379 031527 204 00 0 00 000000 MOVS ;*MOVS SHOULD RESULT WITH AN AC 0F ALL ZEROS + 2380 031530 332 00 0 00 000000 SKIPE ;SKIP HALT INSTRUCTION IF C(AC)=0 + 2381 STOP^ + 2382 031531 254 04 0 00 031532 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2383 031532 324 00 0 00 031533 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2384 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2385 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2386 + 2387 ;********** + 2388 + 2389 ;THIS TEST VERIFIES THAT A MOVS EXECUTED ON AN E CONTAINING A NON-ZERO NUMBER + 2390 ;RESULTS IN A NON-ZERO NUMBER IN THE AC. FIRST A NON-ZERO WORD + 2391 ;IS LOADED INTO E; THEN, MOVS IS EXECUTED. THE AC IS CHECKED FOR A NON-ZERO RESULT. + 2392 ;IF C(AC)=0, THE TEST FAILS. + 2393 + 2394 031533 200 00 0 00 035137 A5100: MOVE [XWD 0,-1] ;LOAD E WITH A NON-ZERO VALUE + 2395 031534 204 00 0 00 000000 MOVS ;*MOVS SHOULD RESULT IN C(AC) NON-ZERO + 2396 031535 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF C(AC) OF MOVS IS NON-ZERO + 2397 STOP^ + 2398 031536 254 04 0 00 031537 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2399 031537 324 00 0 00 031540 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2400 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2401 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2402 + 2403 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 21 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0063 + + 2404 ;THIS TEST VERIFIES THAT A MOVS EXECUTED ON AN E CONTAINING A NON-ZERO NUMBER + 2405 ;RESULTS IN A NON-ZERO NUMBER IN THE AC. FIRST A NON-ZERO WORD + 2406 ;IS LOADED INTO E; THEN, MOVS IS EXECUTED. THE AC IS CHECKED FOR A NON-ZERO + 2407 ;RESULT. IF C(AC)=0, THE TEST FAILS. + 2408 + 2409 031540 200 00 0 00 035140 A5200: MOVE [XWD -1,0] ;LOAD E OF MOVS WITH A NON-ZERO NUMBER + 2410 031541 204 00 0 00 000000 MOVS ;*MOVS SHOULD RESULT WITH C(AC) NON-ZERO + 2411 031542 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF C(AC) OF MOVS IS NON-ZERO + 2412 STOP^ + 2413 031543 254 04 0 00 031544 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2414 031544 324 00 0 00 031545 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2415 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2416 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2417 + 2418 ;********** + 2419 + 2420 ;THIS TEST VERIFIES THAT THE MOVS INSTRUCTION SWAPS BITS 0 AND 188 OF E + 2421 ;AND STORES THEM IN THE AC UNLESS AR BIT 0 IS S-A-0. FIRST, A WORD + 2422 ;WITH BIT 0 SET TO A ONE AND BIT 18 CLEARED IS LOADED INTO E. THEN, MOVS + 2423 ;IS EXECUTED. THE AC IS THEN CHECKED FOR A 0 IN BIT 0, WHICH IS THE + 2424 ;CORRECT RESULT. + 2425 + 2426 031545 200 00 0 00 035140 A5300: MOVE [XWD -1,0] ;LOAD E OF MOVS WITH 777777000000 + 2427 031546 204 00 0 00 000000 MOVS ;*MOVS SHOULD RESULT WITH AC BIT 0 EQUAL TO 0 + 2428 031547 337 00 0 00 000000 SKIPG ;SKIP HALT INSTRUCTION IF BIT 0 OF AC IS 0 + 2429 STOP^ + 2430 031550 254 04 0 00 031551 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2431 031551 324 00 0 00 031552 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2432 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2433 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2434 + 2435 ;MOVSX: AR SWAP [ARMB], AR LT AR RT A [ARMB], AR LT AR RT B [ARMB], + 2436 ;AR RT AR LT A [ARMB], AR RT AR LT B [ARMB], ET2 A F/F [E], IR MOVSX [IR1], + 2437 ;IR MOV(E,S)X [IR1] + 2438 + 2439 ;********** + 2440 + 2441 ;THIS TEST VERIFIES THAT THE MOVS INSTRUCTION SWAPS BITS 0 AND 18 OF E AND + 2442 ;STORES THEM IN THE AC UNLESS AR BIT 0 IS S-A-1. FIRST, A WORD WITH BIT 0 + 2443 ;SET TO A ONE AND BIT 18 CLEARED IS LOADED INTO E. THEN, MOVS IS EXECUTED. + 2444 ;THE AC IS CHECKED FOR A ZERO IN BIT ZERO, WHICH IS THE CORRECT RESULT. + 2445 + 2446 031552 200 00 0 00 035137 A5400: MOVE [XWD 0,-1] ;LOAD E OF MOVS WITH 777777 + 2447 031553 204 00 0 00 000000 MOVS ;*MOVS SHOULD RESULT WITH BIT 0 OF THE AC SET + 2448 031554 331 00 0 00 000000 SKIPL ;PASS IF AC BIT 0 IS SET + 2449 STOP^ + 2450 031555 254 04 0 00 031556 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2451 031556 324 00 0 00 031557 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2452 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2453 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2454 + 2455 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0064 + + 2456 005500 SN=5500 + 2457 000000 ZZ=0 + 2458 + 2459 A5500: REPEAT ^D18, + 2460 <;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2461 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2462 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2463 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2464 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2465 SN=SN+1 + 2466 ZZ=ZZ+ZZ + 2467 IFE ZZ, + 2468 MOVE [0] ;CLEAR AC + 2469 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2470 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2471 STOP + 2472 + 2473 ;********** + 2474 > + 2475 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2476 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2477 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2478 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2479 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2480 005501 SN=SN+1 + 2481 000000 ZZ=ZZ+ZZ + 2482 000001 IFE ZZ, + 2483 031557 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2484 031560 204 00 0 00 035074 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2485 031561 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2486 STOP^ + 2487 031562 254 04 0 00 031563 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2488 031563 324 00 0 00 031564 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2489 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2490 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2491 + 2492 ;********** + 2493 + 2494 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2495 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2496 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2497 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2498 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2499 005502 SN=SN+1 + 2500 000002 ZZ=ZZ+ZZ + 2501 IFE ZZ, + 2502 031564 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2503 031565 204 00 0 00 035075 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2504 031566 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2505 STOP^ + 2506 031567 254 04 0 00 031570 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2507 031570 324 00 0 00 031571 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2508 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2509 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2510 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0065 + + 2511 ;********** + 2512 + 2513 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2514 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2515 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2516 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2517 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2518 005503 SN=SN+1 + 2519 000004 ZZ=ZZ+ZZ + 2520 IFE ZZ, + 2521 031571 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2522 031572 204 00 0 00 035076 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2523 031573 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2524 STOP^ + 2525 031574 254 04 0 00 031575 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2526 031575 324 00 0 00 031576 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2527 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2528 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2529 + 2530 ;********** + 2531 + 2532 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2533 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2534 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2535 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2536 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2537 005504 SN=SN+1 + 2538 000010 ZZ=ZZ+ZZ + 2539 IFE ZZ, + 2540 031576 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2541 031577 204 00 0 00 035077 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2542 031600 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2543 STOP^ + 2544 031601 254 04 0 00 031602 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2545 031602 324 00 0 00 031603 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2546 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2547 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2548 + 2549 ;********** + 2550 + 2551 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2552 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2553 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2554 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2555 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2556 005505 SN=SN+1 + 2557 000020 ZZ=ZZ+ZZ + 2558 IFE ZZ, + 2559 031603 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2560 031604 204 00 0 00 035100 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2561 031605 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2562 STOP^ + 2563 031606 254 04 0 00 031607 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2564 031607 324 00 0 00 031610 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2565 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-2 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0066 + + 2566 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2567 + 2568 ;********** + 2569 + 2570 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2571 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2572 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2573 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2574 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2575 005506 SN=SN+1 + 2576 000040 ZZ=ZZ+ZZ + 2577 IFE ZZ, + 2578 031610 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2579 031611 204 00 0 00 035101 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2580 031612 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2581 STOP^ + 2582 031613 254 04 0 00 031614 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2583 031614 324 00 0 00 031615 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2584 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2585 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2586 + 2587 ;********** + 2588 + 2589 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2590 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2591 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2592 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2593 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2594 005507 SN=SN+1 + 2595 000100 ZZ=ZZ+ZZ + 2596 IFE ZZ, + 2597 031615 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2598 031616 204 00 0 00 035102 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2599 031617 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2600 STOP^ + 2601 031620 254 04 0 00 031621 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2602 031621 324 00 0 00 031622 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2603 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2604 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2605 + 2606 ;********** + 2607 + 2608 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2609 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2610 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2611 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2612 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2613 005510 SN=SN+1 + 2614 000200 ZZ=ZZ+ZZ + 2615 IFE ZZ, + 2616 031622 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2617 031623 204 00 0 00 035103 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2618 031624 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2619 STOP^ + 2620 031625 254 04 0 00 031626 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-3 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0067 + + 2621 031626 324 00 0 00 031627 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2622 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2623 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2624 + 2625 ;********** + 2626 + 2627 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2628 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2629 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2630 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2631 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2632 005511 SN=SN+1 + 2633 000400 ZZ=ZZ+ZZ + 2634 IFE ZZ, + 2635 031627 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2636 031630 204 00 0 00 035104 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2637 031631 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2638 STOP^ + 2639 031632 254 04 0 00 031633 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2640 031633 324 00 0 00 031634 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2641 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2642 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2643 + 2644 ;********** + 2645 + 2646 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2647 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2648 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2649 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2650 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2651 005512 SN=SN+1 + 2652 001000 ZZ=ZZ+ZZ + 2653 IFE ZZ, + 2654 031634 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2655 031635 204 00 0 00 035105 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2656 031636 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2657 STOP^ + 2658 031637 254 04 0 00 031640 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2659 031640 324 00 0 00 031641 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2660 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2661 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2662 + 2663 ;********** + 2664 + 2665 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2666 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2667 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2668 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2669 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2670 005513 SN=SN+1 + 2671 002000 ZZ=ZZ+ZZ + 2672 IFE ZZ, + 2673 031641 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2674 031642 204 00 0 00 035106 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2675 031643 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-4 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0068 + + 2676 STOP^ + 2677 031644 254 04 0 00 031645 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2678 031645 324 00 0 00 031646 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2679 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2680 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2681 + 2682 ;********** + 2683 + 2684 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2685 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2686 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2687 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2688 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2689 005514 SN=SN+1 + 2690 004000 ZZ=ZZ+ZZ + 2691 IFE ZZ, + 2692 031646 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2693 031647 204 00 0 00 035107 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2694 031650 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2695 STOP^ + 2696 031651 254 04 0 00 031652 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2697 031652 324 00 0 00 031653 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2698 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2699 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2700 + 2701 ;********** + 2702 + 2703 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2704 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2705 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2706 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2707 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2708 005515 SN=SN+1 + 2709 010000 ZZ=ZZ+ZZ + 2710 IFE ZZ, + 2711 031653 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2712 031654 204 00 0 00 035110 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2713 031655 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2714 STOP^ + 2715 031656 254 04 0 00 031657 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2716 031657 324 00 0 00 031660 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2717 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2718 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2719 + 2720 ;********** + 2721 + 2722 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2723 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2724 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2725 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2726 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2727 005516 SN=SN+1 + 2728 020000 ZZ=ZZ+ZZ + 2729 IFE ZZ, + 2730 031660 200 00 0 00 035062 MOVE [0] ;CLEAR AC +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-5 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0069 + + 2731 031661 204 00 0 00 035111 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2732 031662 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2733 STOP^ + 2734 031663 254 04 0 00 031664 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2735 031664 324 00 0 00 031665 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2736 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2737 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2738 + 2739 ;********** + 2740 + 2741 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2742 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2743 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2744 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2745 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2746 005517 SN=SN+1 + 2747 040000 ZZ=ZZ+ZZ + 2748 IFE ZZ, + 2749 031665 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2750 031666 204 00 0 00 035112 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2751 031667 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2752 STOP^ + 2753 031670 254 04 0 00 031671 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2754 031671 324 00 0 00 031672 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2755 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2756 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2757 + 2758 ;********** + 2759 + 2760 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2761 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2762 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2763 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2764 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2765 005520 SN=SN+1 + 2766 100000 ZZ=ZZ+ZZ + 2767 IFE ZZ, + 2768 031672 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2769 031673 204 00 0 00 035113 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2770 031674 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2771 STOP^ + 2772 031675 254 04 0 00 031676 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2773 031676 324 00 0 00 031677 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2774 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2775 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2776 + 2777 ;********** + 2778 + 2779 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2780 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2781 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2782 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2783 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2784 005521 SN=SN+1 + 2785 200000 ZZ=ZZ+ZZ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-6 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0070 + + 2786 IFE ZZ, + 2787 031677 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2788 031700 204 00 0 00 035114 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2789 031701 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2790 STOP^ + 2791 031702 254 04 0 00 031703 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2792 031703 324 00 0 00 031704 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2793 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2794 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2795 + 2796 ;********** + 2797 + 2798 ;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, + 2799 ;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. + 2800 ;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES + 2801 ;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT + 2802 ;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. + 2803 005522 SN=SN+1 + 2804 400000 ZZ=ZZ+ZZ + 2805 IFE ZZ, + 2806 031704 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2807 031705 204 00 0 00 035115 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2808 031706 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2809 STOP^ + 2810 031707 254 04 0 00 031710 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2811 031710 324 00 0 00 031711 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2812 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2813 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2814 + 2815 ;********** + 2816 + 2817 005600 SN=5600 + 2818 + 2819 A5600: REPEAT ^D18, + 2820 <;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 2821 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 2822 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 2823 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 2824 ;HALF OF THE AC IS TESTED WITH A 1. + 2825 + 2826 SN=SN+1 + 2827 ZZ=ZZ+ZZ + 2828 MOVE [0] ;CLEAR AC + 2829 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2830 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2831 STOP + 2832 + 2833 ;********** + 2834 > + 2835 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 2836 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 2837 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 2838 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 2839 ;HALF OF THE AC IS TESTED WITH A 1. + 2840 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-7 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0071 + + 2841 005601 SN=SN+1 + 2842 000001 000000 ZZ=ZZ+ZZ + 2843 031711 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2844 031712 204 00 0 00 035116 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2845 031713 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2846 STOP^ + 2847 031714 254 04 0 00 031715 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2848 031715 324 00 0 00 031716 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2849 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2850 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2851 + 2852 ;********** + 2853 + 2854 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 2855 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 2856 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 2857 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 2858 ;HALF OF THE AC IS TESTED WITH A 1. + 2859 + 2860 005602 SN=SN+1 + 2861 000002 000000 ZZ=ZZ+ZZ + 2862 031716 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2863 031717 204 00 0 00 035117 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2864 031720 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2865 STOP^ + 2866 031721 254 04 0 00 031722 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2867 031722 324 00 0 00 031723 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2868 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2869 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2870 + 2871 ;********** + 2872 + 2873 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 2874 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 2875 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 2876 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 2877 ;HALF OF THE AC IS TESTED WITH A 1. + 2878 + 2879 005603 SN=SN+1 + 2880 000004 000000 ZZ=ZZ+ZZ + 2881 031723 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2882 031724 204 00 0 00 035120 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2883 031725 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2884 STOP^ + 2885 031726 254 04 0 00 031727 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2886 031727 324 00 0 00 031730 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2887 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2888 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2889 + 2890 ;********** + 2891 + 2892 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 2893 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 2894 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 2895 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-8 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0072 + + 2896 ;HALF OF THE AC IS TESTED WITH A 1. + 2897 + 2898 005604 SN=SN+1 + 2899 000010 000000 ZZ=ZZ+ZZ + 2900 031730 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2901 031731 204 00 0 00 035121 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2902 031732 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2903 STOP^ + 2904 031733 254 04 0 00 031734 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2905 031734 324 00 0 00 031735 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2906 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2907 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2908 + 2909 ;********** + 2910 + 2911 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 2912 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 2913 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 2914 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 2915 ;HALF OF THE AC IS TESTED WITH A 1. + 2916 + 2917 005605 SN=SN+1 + 2918 000020 000000 ZZ=ZZ+ZZ + 2919 031735 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2920 031736 204 00 0 00 035122 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2921 031737 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2922 STOP^ + 2923 031740 254 04 0 00 031741 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2924 031741 324 00 0 00 031742 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2925 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2926 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2927 + 2928 ;********** + 2929 + 2930 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 2931 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 2932 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 2933 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 2934 ;HALF OF THE AC IS TESTED WITH A 1. + 2935 + 2936 005606 SN=SN+1 + 2937 000040 000000 ZZ=ZZ+ZZ + 2938 031742 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2939 031743 204 00 0 00 035123 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2940 031744 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2941 STOP^ + 2942 031745 254 04 0 00 031746 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2943 031746 324 00 0 00 031747 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2944 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2945 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2946 + 2947 ;********** + 2948 + 2949 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 2950 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-9 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0073 + + 2951 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 2952 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 2953 ;HALF OF THE AC IS TESTED WITH A 1. + 2954 + 2955 005607 SN=SN+1 + 2956 000100 000000 ZZ=ZZ+ZZ + 2957 031747 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2958 031750 204 00 0 00 035124 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2959 031751 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2960 STOP^ + 2961 031752 254 04 0 00 031753 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2962 031753 324 00 0 00 031754 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2963 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2964 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2965 + 2966 ;********** + 2967 + 2968 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 2969 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 2970 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 2971 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 2972 ;HALF OF THE AC IS TESTED WITH A 1. + 2973 + 2974 005610 SN=SN+1 + 2975 000200 000000 ZZ=ZZ+ZZ + 2976 031754 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2977 031755 204 00 0 00 035125 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2978 031756 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2979 STOP^ + 2980 031757 254 04 0 00 031760 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 2981 031760 324 00 0 00 031761 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 2982 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 2983 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 2984 + 2985 ;********** + 2986 + 2987 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 2988 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 2989 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 2990 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 2991 ;HALF OF THE AC IS TESTED WITH A 1. + 2992 + 2993 005611 SN=SN+1 + 2994 000400 000000 ZZ=ZZ+ZZ + 2995 031761 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 2996 031762 204 00 0 00 035073 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 2997 031763 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 2998 STOP^ + 2999 031764 254 04 0 00 031765 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3000 031765 324 00 0 00 031766 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3001 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3002 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3003 + 3004 ;********** + 3005 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-10 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0074 + + 3006 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 3007 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 3008 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 3009 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 3010 ;HALF OF THE AC IS TESTED WITH A 1. + 3011 + 3012 005612 SN=SN+1 + 3013 001000 000000 ZZ=ZZ+ZZ + 3014 031766 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 3015 031767 204 00 0 00 035126 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 3016 031770 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 3017 STOP^ + 3018 031771 254 04 0 00 031772 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3019 031772 324 00 0 00 031773 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3020 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3021 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3022 + 3023 ;********** + 3024 + 3025 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 3026 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 3027 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 3028 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 3029 ;HALF OF THE AC IS TESTED WITH A 1. + 3030 + 3031 005613 SN=SN+1 + 3032 002000 000000 ZZ=ZZ+ZZ + 3033 031773 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 3034 031774 204 00 0 00 035127 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 3035 031775 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 3036 STOP^ + 3037 031776 254 04 0 00 031777 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3038 031777 324 00 0 00 032000 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3039 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3040 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3041 + 3042 ;********** + 3043 + 3044 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 3045 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 3046 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 3047 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 3048 ;HALF OF THE AC IS TESTED WITH A 1. + 3049 + 3050 005614 SN=SN+1 + 3051 004000 000000 ZZ=ZZ+ZZ + 3052 032000 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 3053 032001 204 00 0 00 035130 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 3054 032002 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 3055 STOP^ + 3056 032003 254 04 0 00 032004 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3057 032004 324 00 0 00 032005 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3058 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3059 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3060 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-11 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0075 + + 3061 ;********** + 3062 + 3063 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 3064 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 3065 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 3066 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 3067 ;HALF OF THE AC IS TESTED WITH A 1. + 3068 + 3069 005615 SN=SN+1 + 3070 010000 000000 ZZ=ZZ+ZZ + 3071 032005 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 3072 032006 204 00 0 00 035131 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 3073 032007 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 3074 STOP^ + 3075 032010 254 04 0 00 032011 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3076 032011 324 00 0 00 032012 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3077 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3078 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3079 + 3080 ;********** + 3081 + 3082 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 3083 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 3084 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 3085 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 3086 ;HALF OF THE AC IS TESTED WITH A 1. + 3087 + 3088 005616 SN=SN+1 + 3089 020000 000000 ZZ=ZZ+ZZ + 3090 032012 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 3091 032013 204 00 0 00 035132 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 3092 032014 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 3093 STOP^ + 3094 032015 254 04 0 00 032016 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3095 032016 324 00 0 00 032017 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3096 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3097 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3098 + 3099 ;********** + 3100 + 3101 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 3102 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 3103 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 3104 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 3105 ;HALF OF THE AC IS TESTED WITH A 1. + 3106 + 3107 005617 SN=SN+1 + 3108 040000 000000 ZZ=ZZ+ZZ + 3109 032017 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 3110 032020 204 00 0 00 035133 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 3111 032021 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 3112 STOP^ + 3113 032022 254 04 0 00 032023 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3114 032023 324 00 0 00 032024 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3115 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-12 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0076 + + 3116 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3117 + 3118 ;********** + 3119 + 3120 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 3121 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 3122 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 3123 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 3124 ;HALF OF THE AC IS TESTED WITH A 1. + 3125 + 3126 005620 SN=SN+1 + 3127 100000 000000 ZZ=ZZ+ZZ + 3128 032024 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 3129 032025 204 00 0 00 035134 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 3130 032026 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 3131 STOP^ + 3132 032027 254 04 0 00 032030 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3133 032030 324 00 0 00 032031 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3134 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3135 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3136 + 3137 ;********** + 3138 + 3139 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 3140 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 3141 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 3142 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 3143 ;HALF OF THE AC IS TESTED WITH A 1. + 3144 + 3145 005621 SN=SN+1 + 3146 200000 000000 ZZ=ZZ+ZZ + 3147 032031 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 3148 032032 204 00 0 00 035135 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 3149 032033 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 3150 STOP^ + 3151 032034 254 04 0 00 032035 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3152 032035 324 00 0 00 032036 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3153 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3154 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3155 + 3156 ;********** + 3157 + 3158 ;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL + 3159 ;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS + 3160 ;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS + 3161 ;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT + 3162 ;HALF OF THE AC IS TESTED WITH A 1. + 3163 + 3164 005622 SN=SN+1 + 3165 400000 000000 ZZ=ZZ+ZZ + 3166 032036 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 3167 032037 204 00 0 00 035065 MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + 3168 032040 336 00 0 00 000000 SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + 3169 STOP^ + 3170 032041 254 04 0 00 032042 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 22-13 +DAKAAM MAC 19-JAN-77 13:42 TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS SEQ 0077 + + 3171 032042 324 00 0 00 032043 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3172 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3173 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3174 + 3175 ;********** + 3176 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 23 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0078 + + 3177 SUBTTL TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) + 3178 + 3179 ;THIS TEST VERIFIES THAT SETM LOADS AC FROM E. THE AC IS FIRST SET + 3180 ;TO ZERO; THEN, SETM IS EXECUTED WITH ALL ONES IN E. BIT 0 OF THE AC + 3181 ;IS THEN CHECKED FOR A 1. + 3182 + 3183 032043 200 00 0 00 035062 A5700: MOVE [0] ;PRELOAD AC WITH ALL ZEROS + 3184 032044 414 00 0 00 035063 SETM [-1] ;*SETM SHOULD LOAD AC WITH ALL ONES + 3185 032045 331 00 0 00 000000 SKIPL ;PASS TEST IF AC BIT 0 IS A ONE + 3186 STOP^ + 3187 032046 254 04 0 00 032047 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3188 032047 324 00 0 00 032050 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3189 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3190 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3191 ;FCE [F1], IR BOOLE [IR1], IRXX(0,5,12,17) [IR2] + 3192 + 3193 ;********** + 3194 + 3195 ;THIS TEST VERIFIES THAT SETM LOADS AC FROM E. THE AC IS FIRST SET TO + 3196 ;ALL ONES; THEN, SETM IS EXECUTED WITH ALL ZEROS IN E. THE AC IS CHECKED + 3197 ;FOR ALL ZEROS + 3198 + 3199 032050 200 00 0 00 035063 A6000: MOVE [-1] ;PRESET AC TO ALL ONES + 3200 032051 414 00 0 00 035062 SETM [0] ;*SETM SHOULD LOAD AC WITH ALL ONES + 3201 032052 332 00 0 00 000000 SKIPE ;PASSES TEST IF C(AC) = 0 + 3202 STOP^ + 3203 032053 254 04 0 00 032054 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3204 032054 324 00 0 00 032055 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3205 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3206 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3207 + 3208 ;AD AR + EN [ADAP], AD AR + EN C[ADAP], + 3209 ;F CYC ACT EN A[F1], IR BOOLE AD A +F[IR2], + 3210 ;IR BOOLE [IR1], IRXX(1-3,6,11,15,16) [IR2] + 3211 + 3212 ;********** + 3213 + 3214 ;THIS TEST VERIFIES THAT SETZ LOADS TH AC WITH ALL ZEROS. FIRST, A WORD + 3215 ;OF ALL ZEROS IS LOADED INTO THE AC VIA A MOVE INSTC. THEN, SETZ IS + 3216 ;EXECUTED. THE AC IS CHECKED FOR ALL ZEROS. + 3217 ;IF ANY BITS IN THE AC ARE SET, THE TEST FAILS. + 3218 + 3219 032055 200 00 0 00 035062 A6100: MOVE [0] ;RESET AC TO ALL ZEROS + 3220 032056 400 00 0 00 000000 SETZ ;*SETZ SHOULD CLEAR THE AC + 3221 032057 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC) = 0 + 3222 STOP^ + 3223 032060 254 04 0 00 032061 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3224 032061 324 00 0 00 032062 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3225 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3226 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3227 + 3228 ;AD MASK GEN EN [ADC1], AD MASK GEN F/F'S [ADC1] + 3229 ;AD MASK EN B; F CYC ACT EN B[F1]; IRXX00 [IR1]; + 3230 ;IR BOOLE [IR1] + 3231 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 23-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0079 + + 3232 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0080 + + 3233 ;THIS TEST VERIFIES THAT SETZ LOADS THE AC WITH ALL ZEROS. FIRST, + 3234 ;THE AC IS SET TO ALL ONES; THEN, SETZ IS EXECUTED. THE AC IS THEN + 3235 ;CHECKED FOR ALL ZEROS. IF ANY BITS IN THE AC ARE SET, THE TEST FAILS. + 3236 + 3237 032062 200 00 0 00 035063 A6200: MOVE [-1] ;PRESET AC TO ALL ONES + 3238 032063 400 00 0 00 032063 SETZ . ;*SETZ SHOULD CLEAR THE AC + 3239 032064 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC) = 0 + 3240 STOP^ + 3241 032065 254 04 0 00 032066 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3242 032066 324 00 0 00 032067 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3243 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3244 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3245 + 3246 ;AR AD EN [ARMA], AR AD ET2 EN [ARMA], IR BOOLE [IR] + 3247 + 3248 ;********** + 3249 + 3250 ;THIS TEST VERIFIES THAT THE AND INSTRUCTION OPERATING ON AN AC + 3251 ;CONTAINING ALL ZEROS AND AN E CONTAINING ALL ZEROS RESULTS IN AN AC + 3252 ;OF ALL ZEROS. THE AC AND E ARE THE SAME IN THIS TEST. + 3253 ;AC, E ARE LOADED WITH ZEROS; THEN, AND IS EXECUTED. THE AC IS THEN + 3254 ;CHECKED FOR ALL ZEROS. IF ANY BITS IN THE AC ARE SET BY THE AND INSTRUCTION, + 3255 ;THE TEST FAILS. + 3256 + 3257 + 3258 032067 200 00 0 00 035062 A6300: MOVE [0] ;PRESET AC, E TO ALL ZEROS + 3259 032070 404 00 0 00 000000 AND ;*AND SHOULD RESULT IN C(AC) = 0 + 3260 032071 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC) = 0 + 3261 STOP^ + 3262 032072 254 04 0 00 032073 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3263 032073 324 00 0 00 032074 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3264 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3265 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3266 + 3267 ;********** + 3268 006400 SN=6400 + 3269 000000 ZZ=0 + 3270 + 3271 A6400: REPEAT ^D36, + 3272 <;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3273 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3274 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3275 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3276 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3277 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3278 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3279 + 3280 SN=SN+1 + 3281 ZZ=ZZ+ZZ + 3282 IFE ZZ, + 3283 SETZ ;PRESET E TO ALL ZEROS + 3284 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3285 ;WHERE X VARIES FROM 0 THRU 35 + 3286 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3287 SKIPE 1 ;FAIL IF C(AC) NON-ZERO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0081 + + 3288 STOP + 3289 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3290 STOP + 3291 ;********** + 3292 > + 3293 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3294 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3295 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3296 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3297 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3298 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3299 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3300 + 3301 006401 SN=SN+1 + 3302 000000 ZZ=ZZ+ZZ + 3303 000001 IFE ZZ, + 3304 032074 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3305 032075 200 01 0 00 035074 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3306 ;WHERE X VARIES FROM 0 THRU 35 + 3307 032076 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3308 032077 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3309 STOP^ + 3310 032100 254 04 0 00 032101 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3311 032101 324 00 0 00 032102 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3312 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3313 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3314 032102 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3315 STOP^ + 3316 032103 254 04 0 00 032104 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3317 032104 324 00 0 00 032105 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3318 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3319 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3320 ;********** + 3321 + 3322 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3323 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3324 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3325 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3326 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3327 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3328 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3329 + 3330 006402 SN=SN+1 + 3331 000002 ZZ=ZZ+ZZ + 3332 IFE ZZ, + 3333 032105 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3334 032106 200 01 0 00 035075 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3335 ;WHERE X VARIES FROM 0 THRU 35 + 3336 032107 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3337 032110 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3338 STOP^ + 3339 032111 254 04 0 00 032112 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3340 032112 324 00 0 00 032113 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3341 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3342 ;IN THE SUBTEST) TO LOOP ON ERROR^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-2 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0082 + + 3343 032113 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3344 STOP^ + 3345 032114 254 04 0 00 032115 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3346 032115 324 00 0 00 032116 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3347 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3348 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3349 ;********** + 3350 + 3351 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3352 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3353 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3354 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3355 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3356 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3357 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3358 + 3359 006403 SN=SN+1 + 3360 000004 ZZ=ZZ+ZZ + 3361 IFE ZZ, + 3362 032116 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3363 032117 200 01 0 00 035076 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3364 ;WHERE X VARIES FROM 0 THRU 35 + 3365 032120 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3366 032121 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3367 STOP^ + 3368 032122 254 04 0 00 032123 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3369 032123 324 00 0 00 032124 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3370 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3371 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3372 032124 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3373 STOP^ + 3374 032125 254 04 0 00 032126 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3375 032126 324 00 0 00 032127 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3376 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3377 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3378 ;********** + 3379 + 3380 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3381 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3382 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3383 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3384 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3385 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3386 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3387 + 3388 006404 SN=SN+1 + 3389 000010 ZZ=ZZ+ZZ + 3390 IFE ZZ, + 3391 032127 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3392 032130 200 01 0 00 035077 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3393 ;WHERE X VARIES FROM 0 THRU 35 + 3394 032131 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3395 032132 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3396 STOP^ + 3397 032133 254 04 0 00 032134 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-3 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0083 + + 3398 032134 324 00 0 00 032135 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3399 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3400 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3401 032135 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3402 STOP^ + 3403 032136 254 04 0 00 032137 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3404 032137 324 00 0 00 032140 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3405 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3406 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3407 ;********** + 3408 + 3409 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3410 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3411 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3412 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3413 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3414 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3415 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3416 + 3417 006405 SN=SN+1 + 3418 000020 ZZ=ZZ+ZZ + 3419 IFE ZZ, + 3420 032140 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3421 032141 200 01 0 00 035100 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3422 ;WHERE X VARIES FROM 0 THRU 35 + 3423 032142 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3424 032143 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3425 STOP^ + 3426 032144 254 04 0 00 032145 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3427 032145 324 00 0 00 032146 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3428 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3429 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3430 032146 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3431 STOP^ + 3432 032147 254 04 0 00 032150 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3433 032150 324 00 0 00 032151 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3434 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3435 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3436 ;********** + 3437 + 3438 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3439 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3440 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3441 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3442 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3443 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3444 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3445 + 3446 006406 SN=SN+1 + 3447 000040 ZZ=ZZ+ZZ + 3448 IFE ZZ, + 3449 032151 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3450 032152 200 01 0 00 035101 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3451 ;WHERE X VARIES FROM 0 THRU 35 + 3452 032153 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-4 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0084 + + 3453 032154 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3454 STOP^ + 3455 032155 254 04 0 00 032156 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3456 032156 324 00 0 00 032157 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3457 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3458 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3459 032157 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3460 STOP^ + 3461 032160 254 04 0 00 032161 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3462 032161 324 00 0 00 032162 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3463 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3464 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3465 ;********** + 3466 + 3467 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3468 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3469 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3470 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3471 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3472 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3473 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3474 + 3475 006407 SN=SN+1 + 3476 000100 ZZ=ZZ+ZZ + 3477 IFE ZZ, + 3478 032162 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3479 032163 200 01 0 00 035102 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3480 ;WHERE X VARIES FROM 0 THRU 35 + 3481 032164 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3482 032165 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3483 STOP^ + 3484 032166 254 04 0 00 032167 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3485 032167 324 00 0 00 032170 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3486 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3487 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3488 032170 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3489 STOP^ + 3490 032171 254 04 0 00 032172 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3491 032172 324 00 0 00 032173 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3492 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3493 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3494 ;********** + 3495 + 3496 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3497 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3498 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3499 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3500 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3501 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3502 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3503 + 3504 006410 SN=SN+1 + 3505 000200 ZZ=ZZ+ZZ + 3506 IFE ZZ, + 3507 032173 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-5 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0085 + + 3508 032174 200 01 0 00 035103 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3509 ;WHERE X VARIES FROM 0 THRU 35 + 3510 032175 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3511 032176 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3512 STOP^ + 3513 032177 254 04 0 00 032200 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3514 032200 324 00 0 00 032201 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3515 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3516 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3517 032201 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3518 STOP^ + 3519 032202 254 04 0 00 032203 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3520 032203 324 00 0 00 032204 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3521 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3522 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3523 ;********** + 3524 + 3525 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3526 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3527 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3528 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3529 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3530 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3531 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3532 + 3533 006411 SN=SN+1 + 3534 000400 ZZ=ZZ+ZZ + 3535 IFE ZZ, + 3536 032204 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3537 032205 200 01 0 00 035104 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3538 ;WHERE X VARIES FROM 0 THRU 35 + 3539 032206 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3540 032207 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3541 STOP^ + 3542 032210 254 04 0 00 032211 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3543 032211 324 00 0 00 032212 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3544 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3545 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3546 032212 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3547 STOP^ + 3548 032213 254 04 0 00 032214 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3549 032214 324 00 0 00 032215 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3550 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3551 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3552 ;********** + 3553 + 3554 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3555 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3556 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3557 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3558 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3559 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3560 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3561 + 3562 006412 SN=SN+1 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-6 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0086 + + 3563 001000 ZZ=ZZ+ZZ + 3564 IFE ZZ, + 3565 032215 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3566 032216 200 01 0 00 035105 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3567 ;WHERE X VARIES FROM 0 THRU 35 + 3568 032217 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3569 032220 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3570 STOP^ + 3571 032221 254 04 0 00 032222 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3572 032222 324 00 0 00 032223 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3573 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3574 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3575 032223 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3576 STOP^ + 3577 032224 254 04 0 00 032225 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3578 032225 324 00 0 00 032226 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3579 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3580 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3581 ;********** + 3582 + 3583 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3584 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3585 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3586 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3587 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3588 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3589 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3590 + 3591 006413 SN=SN+1 + 3592 002000 ZZ=ZZ+ZZ + 3593 IFE ZZ, + 3594 032226 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3595 032227 200 01 0 00 035106 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3596 ;WHERE X VARIES FROM 0 THRU 35 + 3597 032230 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3598 032231 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3599 STOP^ + 3600 032232 254 04 0 00 032233 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3601 032233 324 00 0 00 032234 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3602 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3603 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3604 032234 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3605 STOP^ + 3606 032235 254 04 0 00 032236 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3607 032236 324 00 0 00 032237 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3608 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3609 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3610 ;********** + 3611 + 3612 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3613 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3614 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3615 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3616 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3617 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-7 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0087 + + 3618 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3619 + 3620 006414 SN=SN+1 + 3621 004000 ZZ=ZZ+ZZ + 3622 IFE ZZ, + 3623 032237 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3624 032240 200 01 0 00 035107 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3625 ;WHERE X VARIES FROM 0 THRU 35 + 3626 032241 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3627 032242 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3628 STOP^ + 3629 032243 254 04 0 00 032244 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3630 032244 324 00 0 00 032245 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3631 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3632 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3633 032245 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3634 STOP^ + 3635 032246 254 04 0 00 032247 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3636 032247 324 00 0 00 032250 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3637 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3638 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3639 ;********** + 3640 + 3641 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3642 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3643 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3644 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3645 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3646 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3647 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3648 + 3649 006415 SN=SN+1 + 3650 010000 ZZ=ZZ+ZZ + 3651 IFE ZZ, + 3652 032250 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3653 032251 200 01 0 00 035110 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3654 ;WHERE X VARIES FROM 0 THRU 35 + 3655 032252 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3656 032253 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3657 STOP^ + 3658 032254 254 04 0 00 032255 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3659 032255 324 00 0 00 032256 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3660 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3661 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3662 032256 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3663 STOP^ + 3664 032257 254 04 0 00 032260 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3665 032260 324 00 0 00 032261 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3666 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3667 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3668 ;********** + 3669 + 3670 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3671 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3672 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-8 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0088 + + 3673 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3674 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3675 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3676 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3677 + 3678 006416 SN=SN+1 + 3679 020000 ZZ=ZZ+ZZ + 3680 IFE ZZ, + 3681 032261 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3682 032262 200 01 0 00 035111 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3683 ;WHERE X VARIES FROM 0 THRU 35 + 3684 032263 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3685 032264 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3686 STOP^ + 3687 032265 254 04 0 00 032266 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3688 032266 324 00 0 00 032267 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3689 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3690 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3691 032267 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3692 STOP^ + 3693 032270 254 04 0 00 032271 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3694 032271 324 00 0 00 032272 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3695 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3696 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3697 ;********** + 3698 + 3699 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3700 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3701 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3702 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3703 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3704 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3705 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3706 + 3707 006417 SN=SN+1 + 3708 040000 ZZ=ZZ+ZZ + 3709 IFE ZZ, + 3710 032272 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3711 032273 200 01 0 00 035112 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3712 ;WHERE X VARIES FROM 0 THRU 35 + 3713 032274 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3714 032275 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3715 STOP^ + 3716 032276 254 04 0 00 032277 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3717 032277 324 00 0 00 032300 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3718 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3719 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3720 032300 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3721 STOP^ + 3722 032301 254 04 0 00 032302 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3723 032302 324 00 0 00 032303 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3724 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3725 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3726 ;********** + 3727 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-9 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0089 + + 3728 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3729 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3730 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3731 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3732 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3733 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3734 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3735 + 3736 006420 SN=SN+1 + 3737 100000 ZZ=ZZ+ZZ + 3738 IFE ZZ, + 3739 032303 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3740 032304 200 01 0 00 035113 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3741 ;WHERE X VARIES FROM 0 THRU 35 + 3742 032305 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3743 032306 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3744 STOP^ + 3745 032307 254 04 0 00 032310 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3746 032310 324 00 0 00 032311 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3747 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3748 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3749 032311 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3750 STOP^ + 3751 032312 254 04 0 00 032313 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3752 032313 324 00 0 00 032314 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3753 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3754 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3755 ;********** + 3756 + 3757 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3758 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3759 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3760 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3761 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3762 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3763 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3764 + 3765 006421 SN=SN+1 + 3766 200000 ZZ=ZZ+ZZ + 3767 IFE ZZ, + 3768 032314 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3769 032315 200 01 0 00 035114 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3770 ;WHERE X VARIES FROM 0 THRU 35 + 3771 032316 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3772 032317 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3773 STOP^ + 3774 032320 254 04 0 00 032321 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3775 032321 324 00 0 00 032322 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3776 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3777 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3778 032322 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3779 STOP^ + 3780 032323 254 04 0 00 032324 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3781 032324 324 00 0 00 032325 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3782 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-10 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0090 + + 3783 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3784 ;********** + 3785 + 3786 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3787 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3788 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3789 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3790 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3791 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3792 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3793 + 3794 006422 SN=SN+1 + 3795 400000 ZZ=ZZ+ZZ + 3796 IFE ZZ, + 3797 032325 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3798 032326 200 01 0 00 035115 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3799 ;WHERE X VARIES FROM 0 THRU 35 + 3800 032327 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3801 032330 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3802 STOP^ + 3803 032331 254 04 0 00 032332 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3804 032332 324 00 0 00 032333 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3805 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3806 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3807 032333 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3808 STOP^ + 3809 032334 254 04 0 00 032335 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3810 032335 324 00 0 00 032336 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3811 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3812 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3813 ;********** + 3814 + 3815 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3816 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3817 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3818 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3819 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3820 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3821 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3822 + 3823 006423 SN=SN+1 + 3824 000001 000000 ZZ=ZZ+ZZ + 3825 IFE ZZ, + 3826 032336 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3827 032337 200 01 0 00 035116 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3828 ;WHERE X VARIES FROM 0 THRU 35 + 3829 032340 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3830 032341 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3831 STOP^ + 3832 032342 254 04 0 00 032343 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3833 032343 324 00 0 00 032344 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3834 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3835 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3836 032344 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3837 STOP^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-11 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0091 + + 3838 032345 254 04 0 00 032346 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3839 032346 324 00 0 00 032347 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3840 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3841 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3842 ;********** + 3843 + 3844 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3845 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3846 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3847 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3848 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3849 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3850 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3851 + 3852 006424 SN=SN+1 + 3853 000002 000000 ZZ=ZZ+ZZ + 3854 IFE ZZ, + 3855 032347 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3856 032350 200 01 0 00 035117 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3857 ;WHERE X VARIES FROM 0 THRU 35 + 3858 032351 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3859 032352 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3860 STOP^ + 3861 032353 254 04 0 00 032354 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3862 032354 324 00 0 00 032355 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3863 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3864 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3865 032355 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3866 STOP^ + 3867 032356 254 04 0 00 032357 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3868 032357 324 00 0 00 032360 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3869 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3870 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3871 ;********** + 3872 + 3873 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3874 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3875 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3876 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3877 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3878 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3879 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3880 + 3881 006425 SN=SN+1 + 3882 000004 000000 ZZ=ZZ+ZZ + 3883 IFE ZZ, + 3884 032360 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3885 032361 200 01 0 00 035120 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3886 ;WHERE X VARIES FROM 0 THRU 35 + 3887 032362 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3888 032363 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3889 STOP^ + 3890 032364 254 04 0 00 032365 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3891 032365 324 00 0 00 032366 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3892 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-12 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0092 + + 3893 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3894 032366 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3895 STOP^ + 3896 032367 254 04 0 00 032370 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3897 032370 324 00 0 00 032371 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3898 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3899 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3900 ;********** + 3901 + 3902 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3903 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3904 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3905 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3906 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3907 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3908 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3909 + 3910 006426 SN=SN+1 + 3911 000010 000000 ZZ=ZZ+ZZ + 3912 IFE ZZ, + 3913 032371 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3914 032372 200 01 0 00 035121 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3915 ;WHERE X VARIES FROM 0 THRU 35 + 3916 032373 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3917 032374 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3918 STOP^ + 3919 032375 254 04 0 00 032376 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3920 032376 324 00 0 00 032377 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3921 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3922 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3923 032377 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3924 STOP^ + 3925 032400 254 04 0 00 032401 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3926 032401 324 00 0 00 032402 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3927 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3928 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3929 ;********** + 3930 + 3931 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3932 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3933 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3934 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3935 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3936 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3937 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3938 + 3939 006427 SN=SN+1 + 3940 000020 000000 ZZ=ZZ+ZZ + 3941 IFE ZZ, + 3942 032402 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3943 032403 200 01 0 00 035122 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3944 ;WHERE X VARIES FROM 0 THRU 35 + 3945 032404 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3946 032405 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3947 STOP^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-13 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0093 + + 3948 032406 254 04 0 00 032407 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3949 032407 324 00 0 00 032410 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3950 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3951 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3952 032410 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3953 STOP^ + 3954 032411 254 04 0 00 032412 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3955 032412 324 00 0 00 032413 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3956 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3957 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3958 ;********** + 3959 + 3960 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3961 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3962 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3963 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3964 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3965 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3966 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3967 + 3968 006430 SN=SN+1 + 3969 000040 000000 ZZ=ZZ+ZZ + 3970 IFE ZZ, + 3971 032413 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 3972 032414 200 01 0 00 035123 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 3973 ;WHERE X VARIES FROM 0 THRU 35 + 3974 032415 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 3975 032416 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 3976 STOP^ + 3977 032417 254 04 0 00 032420 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3978 032420 324 00 0 00 032421 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3979 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3980 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3981 032421 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 3982 STOP^ + 3983 032422 254 04 0 00 032423 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 3984 032423 324 00 0 00 032424 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 3985 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 3986 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 3987 ;********** + 3988 + 3989 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 3990 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 3991 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 3992 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 3993 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 3994 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 3995 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 3996 + 3997 006431 SN=SN+1 + 3998 000100 000000 ZZ=ZZ+ZZ + 3999 IFE ZZ, + 4000 032424 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4001 032425 200 01 0 00 035124 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4002 ;WHERE X VARIES FROM 0 THRU 35 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-14 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0094 + + 4003 032426 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4004 032427 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4005 STOP^ + 4006 032430 254 04 0 00 032431 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4007 032431 324 00 0 00 032432 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4008 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4009 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4010 032432 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4011 STOP^ + 4012 032433 254 04 0 00 032434 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4013 032434 324 00 0 00 032435 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4014 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4015 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4016 ;********** + 4017 + 4018 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4019 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 4020 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4021 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4022 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 4023 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4024 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4025 + 4026 006432 SN=SN+1 + 4027 000200 000000 ZZ=ZZ+ZZ + 4028 IFE ZZ, + 4029 032435 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4030 032436 200 01 0 00 035125 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4031 ;WHERE X VARIES FROM 0 THRU 35 + 4032 032437 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4033 032440 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4034 STOP^ + 4035 032441 254 04 0 00 032442 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4036 032442 324 00 0 00 032443 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4037 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4038 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4039 032443 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4040 STOP^ + 4041 032444 254 04 0 00 032445 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4042 032445 324 00 0 00 032446 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4043 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4044 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4045 ;********** + 4046 + 4047 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4048 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 4049 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4050 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4051 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 4052 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4053 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4054 + 4055 006433 SN=SN+1 + 4056 000400 000000 ZZ=ZZ+ZZ + 4057 IFE ZZ, +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-15 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0095 + + 4058 032446 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4059 032447 200 01 0 00 035073 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4060 ;WHERE X VARIES FROM 0 THRU 35 + 4061 032450 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4062 032451 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4063 STOP^ + 4064 032452 254 04 0 00 032453 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4065 032453 324 00 0 00 032454 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4066 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4067 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4068 032454 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4069 STOP^ + 4070 032455 254 04 0 00 032456 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4071 032456 324 00 0 00 032457 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4072 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4073 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4074 ;********** + 4075 + 4076 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4077 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 4078 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4079 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4080 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 4081 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4082 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4083 + 4084 006434 SN=SN+1 + 4085 001000 000000 ZZ=ZZ+ZZ + 4086 IFE ZZ, + 4087 032457 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4088 032460 200 01 0 00 035126 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4089 ;WHERE X VARIES FROM 0 THRU 35 + 4090 032461 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4091 032462 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4092 STOP^ + 4093 032463 254 04 0 00 032464 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4094 032464 324 00 0 00 032465 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4095 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4096 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4097 032465 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4098 STOP^ + 4099 032466 254 04 0 00 032467 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4100 032467 324 00 0 00 032470 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4101 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4102 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4103 ;********** + 4104 + 4105 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4106 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 4107 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4108 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4109 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 4110 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4111 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4112 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-16 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0096 + + 4113 006435 SN=SN+1 + 4114 002000 000000 ZZ=ZZ+ZZ + 4115 IFE ZZ, + 4116 032470 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4117 032471 200 01 0 00 035127 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4118 ;WHERE X VARIES FROM 0 THRU 35 + 4119 032472 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4120 032473 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4121 STOP^ + 4122 032474 254 04 0 00 032475 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4123 032475 324 00 0 00 032476 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4124 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4125 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4126 032476 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4127 STOP^ + 4128 032477 254 04 0 00 032500 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4129 032500 324 00 0 00 032501 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4130 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4131 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4132 ;********** + 4133 + 4134 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4135 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 4136 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4137 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4138 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 4139 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4140 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4141 + 4142 006436 SN=SN+1 + 4143 004000 000000 ZZ=ZZ+ZZ + 4144 IFE ZZ, + 4145 032501 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4146 032502 200 01 0 00 035130 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4147 ;WHERE X VARIES FROM 0 THRU 35 + 4148 032503 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4149 032504 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4150 STOP^ + 4151 032505 254 04 0 00 032506 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4152 032506 324 00 0 00 032507 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4153 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4154 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4155 032507 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4156 STOP^ + 4157 032510 254 04 0 00 032511 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4158 032511 324 00 0 00 032512 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4159 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4160 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4161 ;********** + 4162 + 4163 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4164 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 4165 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4166 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4167 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-17 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0097 + + 4168 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4169 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4170 + 4171 006437 SN=SN+1 + 4172 010000 000000 ZZ=ZZ+ZZ + 4173 IFE ZZ, + 4174 032512 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4175 032513 200 01 0 00 035131 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4176 ;WHERE X VARIES FROM 0 THRU 35 + 4177 032514 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4178 032515 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4179 STOP^ + 4180 032516 254 04 0 00 032517 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4181 032517 324 00 0 00 032520 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4182 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4183 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4184 032520 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4185 STOP^ + 4186 032521 254 04 0 00 032522 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4187 032522 324 00 0 00 032523 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4188 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4189 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4190 ;********** + 4191 + 4192 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4193 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 4194 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4195 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4196 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 4197 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4198 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4199 + 4200 006440 SN=SN+1 + 4201 020000 000000 ZZ=ZZ+ZZ + 4202 IFE ZZ, + 4203 032523 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4204 032524 200 01 0 00 035132 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4205 ;WHERE X VARIES FROM 0 THRU 35 + 4206 032525 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4207 032526 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4208 STOP^ + 4209 032527 254 04 0 00 032530 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4210 032530 324 00 0 00 032531 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4211 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4212 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4213 032531 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4214 STOP^ + 4215 032532 254 04 0 00 032533 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4216 032533 324 00 0 00 032534 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4217 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4218 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4219 ;********** + 4220 + 4221 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4222 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-18 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0098 + + 4223 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4224 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4225 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 4226 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4227 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4228 + 4229 006441 SN=SN+1 + 4230 040000 000000 ZZ=ZZ+ZZ + 4231 IFE ZZ, + 4232 032534 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4233 032535 200 01 0 00 035133 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4234 ;WHERE X VARIES FROM 0 THRU 35 + 4235 032536 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4236 032537 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4237 STOP^ + 4238 032540 254 04 0 00 032541 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4239 032541 324 00 0 00 032542 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4240 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4241 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4242 032542 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4243 STOP^ + 4244 032543 254 04 0 00 032544 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4245 032544 324 00 0 00 032545 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4246 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4247 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4248 ;********** + 4249 + 4250 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4251 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 4252 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4253 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4254 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 4255 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4256 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4257 + 4258 006442 SN=SN+1 + 4259 100000 000000 ZZ=ZZ+ZZ + 4260 IFE ZZ, + 4261 032545 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4262 032546 200 01 0 00 035134 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4263 ;WHERE X VARIES FROM 0 THRU 35 + 4264 032547 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4265 032550 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4266 STOP^ + 4267 032551 254 04 0 00 032552 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4268 032552 324 00 0 00 032553 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4269 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4270 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4271 032553 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4272 STOP^ + 4273 032554 254 04 0 00 032555 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4274 032555 324 00 0 00 032556 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4275 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4276 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4277 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-19 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0099 + + 4278 + 4279 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4280 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 4281 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4282 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4283 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 4284 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4285 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4286 + 4287 006443 SN=SN+1 + 4288 200000 000000 ZZ=ZZ+ZZ + 4289 IFE ZZ, + 4290 032556 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4291 032557 200 01 0 00 035135 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4292 ;WHERE X VARIES FROM 0 THRU 35 + 4293 032560 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4294 032561 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4295 STOP^ + 4296 032562 254 04 0 00 032563 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4297 032563 324 00 0 00 032564 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4298 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4299 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4300 032564 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4301 STOP^ + 4302 032565 254 04 0 00 032566 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4303 032566 324 00 0 00 032567 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4304 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4305 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4306 ;********** + 4307 + 4308 ;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION + 4309 ;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH + 4310 ;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X + 4311 ;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. + 4312 ;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, + 4313 ;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS + 4314 ;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + 4315 + 4316 006444 SN=SN+1 + 4317 400000 000000 ZZ=ZZ+ZZ + 4318 IFE ZZ, + 4319 032567 400 00 0 00 000000 SETZ ;PRESET E TO ALL ZEROS + 4320 032570 200 01 0 00 035065 MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + 4321 ;WHERE X VARIES FROM 0 THRU 35 + 4322 032571 404 01 0 00 000000 AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + 4323 032572 332 00 0 00 000001 SKIPE 1 ;FAIL IF C(AC) NON-ZERO + 4324 STOP^ + 4325 032573 254 04 0 00 032574 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4326 032574 324 00 0 00 032575 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4327 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4328 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4329 032575 332 00 0 00 000000 SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + 4330 STOP^ + 4331 032576 254 04 0 00 032577 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4332 032577 324 00 0 00 032600 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 24-20 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0100 + + 4333 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4334 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4335 ;********** + 4336 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0101 + + 4337 006500 SN=6500 + 4338 000000 ZZ=0 + 4339 + 4340 A6500: REPEAT ^D36, + 4341 <;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4342 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4343 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4344 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4345 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4346 SN=SN+1 + 4347 ZZ=ZZ+ZZ + 4348 IFE ZZ, + 4349 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4350 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4351 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4352 ;C(AC) NONZERO + 4353 SKIPN ;PASS TEST IF C(AC) NONZERO + 4354 STOP + 4355 + 4356 ;********** + 4357 > + 4358 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4359 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4360 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4361 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4362 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4363 006501 SN=SN+1 + 4364 000000 ZZ=ZZ+ZZ + 4365 000001 IFE ZZ, + 4366 032600 200 00 0 00 035074 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4367 032601 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4368 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4369 ;C(AC) NONZERO + 4370 032602 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4371 STOP^ + 4372 032603 254 04 0 00 032604 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4373 032604 324 00 0 00 032605 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4374 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4375 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4376 + 4377 ;********** + 4378 + 4379 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4380 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4381 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4382 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4383 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4384 006502 SN=SN+1 + 4385 000002 ZZ=ZZ+ZZ + 4386 IFE ZZ, + 4387 032605 200 00 0 00 035075 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4388 032606 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4389 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4390 ;C(AC) NONZERO + 4391 032607 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0102 + + 4392 STOP^ + 4393 032610 254 04 0 00 032611 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4394 032611 324 00 0 00 032612 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4395 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4396 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4397 + 4398 ;********** + 4399 + 4400 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4401 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4402 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4403 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4404 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4405 006503 SN=SN+1 + 4406 000004 ZZ=ZZ+ZZ + 4407 IFE ZZ, + 4408 032612 200 00 0 00 035076 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4409 032613 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4410 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4411 ;C(AC) NONZERO + 4412 032614 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4413 STOP^ + 4414 032615 254 04 0 00 032616 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4415 032616 324 00 0 00 032617 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4416 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4417 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4418 + 4419 ;********** + 4420 + 4421 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4422 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4423 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4424 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4425 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4426 006504 SN=SN+1 + 4427 000010 ZZ=ZZ+ZZ + 4428 IFE ZZ, + 4429 032617 200 00 0 00 035077 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4430 032620 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4431 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4432 ;C(AC) NONZERO + 4433 032621 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4434 STOP^ + 4435 032622 254 04 0 00 032623 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4436 032623 324 00 0 00 032624 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4437 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4438 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4439 + 4440 ;********** + 4441 + 4442 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4443 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4444 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4445 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4446 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-2 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0103 + + 4447 006505 SN=SN+1 + 4448 000020 ZZ=ZZ+ZZ + 4449 IFE ZZ, + 4450 032624 200 00 0 00 035100 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4451 032625 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4452 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4453 ;C(AC) NONZERO + 4454 032626 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4455 STOP^ + 4456 032627 254 04 0 00 032630 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4457 032630 324 00 0 00 032631 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4458 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4459 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4460 + 4461 ;********** + 4462 + 4463 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4464 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4465 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4466 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4467 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4468 006506 SN=SN+1 + 4469 000040 ZZ=ZZ+ZZ + 4470 IFE ZZ, + 4471 032631 200 00 0 00 035101 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4472 032632 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4473 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4474 ;C(AC) NONZERO + 4475 032633 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4476 STOP^ + 4477 032634 254 04 0 00 032635 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4478 032635 324 00 0 00 032636 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4479 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4480 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4481 + 4482 ;********** + 4483 + 4484 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4485 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4486 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4487 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4488 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4489 006507 SN=SN+1 + 4490 000100 ZZ=ZZ+ZZ + 4491 IFE ZZ, + 4492 032636 200 00 0 00 035102 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4493 032637 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4494 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4495 ;C(AC) NONZERO + 4496 032640 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4497 STOP^ + 4498 032641 254 04 0 00 032642 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4499 032642 324 00 0 00 032643 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4500 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4501 ;IN THE SUBTEST) TO LOOP ON ERROR^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-3 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0104 + + 4502 + 4503 ;********** + 4504 + 4505 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4506 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4507 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4508 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4509 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4510 006510 SN=SN+1 + 4511 000200 ZZ=ZZ+ZZ + 4512 IFE ZZ, + 4513 032643 200 00 0 00 035103 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4514 032644 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4515 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4516 ;C(AC) NONZERO + 4517 032645 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4518 STOP^ + 4519 032646 254 04 0 00 032647 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4520 032647 324 00 0 00 032650 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4521 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4522 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4523 + 4524 ;********** + 4525 + 4526 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4527 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4528 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4529 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4530 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4531 006511 SN=SN+1 + 4532 000400 ZZ=ZZ+ZZ + 4533 IFE ZZ, + 4534 032650 200 00 0 00 035104 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4535 032651 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4536 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4537 ;C(AC) NONZERO + 4538 032652 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4539 STOP^ + 4540 032653 254 04 0 00 032654 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4541 032654 324 00 0 00 032655 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4542 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4543 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4544 + 4545 ;********** + 4546 + 4547 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4548 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4549 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4550 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4551 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4552 006512 SN=SN+1 + 4553 001000 ZZ=ZZ+ZZ + 4554 IFE ZZ, + 4555 032655 200 00 0 00 035105 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4556 032656 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-4 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0105 + + 4557 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4558 ;C(AC) NONZERO + 4559 032657 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4560 STOP^ + 4561 032660 254 04 0 00 032661 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4562 032661 324 00 0 00 032662 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4563 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4564 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4565 + 4566 ;********** + 4567 + 4568 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4569 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4570 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4571 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4572 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4573 006513 SN=SN+1 + 4574 002000 ZZ=ZZ+ZZ + 4575 IFE ZZ, + 4576 032662 200 00 0 00 035106 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4577 032663 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4578 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4579 ;C(AC) NONZERO + 4580 032664 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4581 STOP^ + 4582 032665 254 04 0 00 032666 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4583 032666 324 00 0 00 032667 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4584 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4585 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4586 + 4587 ;********** + 4588 + 4589 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4590 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4591 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4592 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4593 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4594 006514 SN=SN+1 + 4595 004000 ZZ=ZZ+ZZ + 4596 IFE ZZ, + 4597 032667 200 00 0 00 035107 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4598 032670 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4599 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4600 ;C(AC) NONZERO + 4601 032671 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4602 STOP^ + 4603 032672 254 04 0 00 032673 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4604 032673 324 00 0 00 032674 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4605 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4606 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4607 + 4608 ;********** + 4609 + 4610 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4611 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-5 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0106 + + 4612 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4613 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4614 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4615 006515 SN=SN+1 + 4616 010000 ZZ=ZZ+ZZ + 4617 IFE ZZ, + 4618 032674 200 00 0 00 035110 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4619 032675 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4620 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4621 ;C(AC) NONZERO + 4622 032676 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4623 STOP^ + 4624 032677 254 04 0 00 032700 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4625 032700 324 00 0 00 032701 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4626 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4627 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4628 + 4629 ;********** + 4630 + 4631 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4632 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4633 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4634 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4635 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4636 006516 SN=SN+1 + 4637 020000 ZZ=ZZ+ZZ + 4638 IFE ZZ, + 4639 032701 200 00 0 00 035111 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4640 032702 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4641 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4642 ;C(AC) NONZERO + 4643 032703 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4644 STOP^ + 4645 032704 254 04 0 00 032705 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4646 032705 324 00 0 00 032706 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4647 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4648 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4649 + 4650 ;********** + 4651 + 4652 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4653 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4654 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4655 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4656 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4657 006517 SN=SN+1 + 4658 040000 ZZ=ZZ+ZZ + 4659 IFE ZZ, + 4660 032706 200 00 0 00 035112 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4661 032707 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4662 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4663 ;C(AC) NONZERO + 4664 032710 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4665 STOP^ + 4666 032711 254 04 0 00 032712 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-6 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0107 + + 4667 032712 324 00 0 00 032713 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4668 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4669 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4670 + 4671 ;********** + 4672 + 4673 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4674 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4675 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4676 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4677 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4678 006520 SN=SN+1 + 4679 100000 ZZ=ZZ+ZZ + 4680 IFE ZZ, + 4681 032713 200 00 0 00 035113 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4682 032714 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4683 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4684 ;C(AC) NONZERO + 4685 032715 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4686 STOP^ + 4687 032716 254 04 0 00 032717 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4688 032717 324 00 0 00 032720 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4689 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4690 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4691 + 4692 ;********** + 4693 + 4694 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4695 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4696 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4697 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4698 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4699 006521 SN=SN+1 + 4700 200000 ZZ=ZZ+ZZ + 4701 IFE ZZ, + 4702 032720 200 00 0 00 035114 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4703 032721 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4704 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4705 ;C(AC) NONZERO + 4706 032722 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4707 STOP^ + 4708 032723 254 04 0 00 032724 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4709 032724 324 00 0 00 032725 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4710 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4711 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4712 + 4713 ;********** + 4714 + 4715 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4716 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4717 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4718 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4719 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4720 006522 SN=SN+1 + 4721 400000 ZZ=ZZ+ZZ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-7 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0108 + + 4722 IFE ZZ, + 4723 032725 200 00 0 00 035115 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4724 032726 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4725 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4726 ;C(AC) NONZERO + 4727 032727 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4728 STOP^ + 4729 032730 254 04 0 00 032731 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4730 032731 324 00 0 00 032732 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4731 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4732 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4733 + 4734 ;********** + 4735 + 4736 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4737 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4738 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4739 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4740 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4741 006523 SN=SN+1 + 4742 000001 000000 ZZ=ZZ+ZZ + 4743 IFE ZZ, + 4744 032732 200 00 0 00 035116 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4745 032733 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4746 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4747 ;C(AC) NONZERO + 4748 032734 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4749 STOP^ + 4750 032735 254 04 0 00 032736 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4751 032736 324 00 0 00 032737 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4752 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4753 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4754 + 4755 ;********** + 4756 + 4757 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4758 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4759 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4760 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4761 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4762 006524 SN=SN+1 + 4763 000002 000000 ZZ=ZZ+ZZ + 4764 IFE ZZ, + 4765 032737 200 00 0 00 035117 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4766 032740 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4767 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4768 ;C(AC) NONZERO + 4769 032741 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4770 STOP^ + 4771 032742 254 04 0 00 032743 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4772 032743 324 00 0 00 032744 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4773 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4774 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4775 + 4776 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-8 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0109 + + 4777 + 4778 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4779 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4780 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4781 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4782 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4783 006525 SN=SN+1 + 4784 000004 000000 ZZ=ZZ+ZZ + 4785 IFE ZZ, + 4786 032744 200 00 0 00 035120 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4787 032745 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4788 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4789 ;C(AC) NONZERO + 4790 032746 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4791 STOP^ + 4792 032747 254 04 0 00 032750 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4793 032750 324 00 0 00 032751 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4794 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4795 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4796 + 4797 ;********** + 4798 + 4799 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4800 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4801 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4802 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4803 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4804 006526 SN=SN+1 + 4805 000010 000000 ZZ=ZZ+ZZ + 4806 IFE ZZ, + 4807 032751 200 00 0 00 035121 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4808 032752 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4809 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4810 ;C(AC) NONZERO + 4811 032753 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4812 STOP^ + 4813 032754 254 04 0 00 032755 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4814 032755 324 00 0 00 032756 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4815 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4816 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4817 + 4818 ;********** + 4819 + 4820 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4821 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4822 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4823 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4824 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4825 006527 SN=SN+1 + 4826 000020 000000 ZZ=ZZ+ZZ + 4827 IFE ZZ, + 4828 032756 200 00 0 00 035122 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4829 032757 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4830 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4831 ;C(AC) NONZERO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-9 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0110 + + 4832 032760 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4833 STOP^ + 4834 032761 254 04 0 00 032762 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4835 032762 324 00 0 00 032763 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4836 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4837 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4838 + 4839 ;********** + 4840 + 4841 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4842 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4843 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4844 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4845 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4846 006530 SN=SN+1 + 4847 000040 000000 ZZ=ZZ+ZZ + 4848 IFE ZZ, + 4849 032763 200 00 0 00 035123 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4850 032764 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4851 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4852 ;C(AC) NONZERO + 4853 032765 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4854 STOP^ + 4855 032766 254 04 0 00 032767 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4856 032767 324 00 0 00 032770 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4857 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4858 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4859 + 4860 ;********** + 4861 + 4862 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4863 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4864 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4865 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4866 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4867 006531 SN=SN+1 + 4868 000100 000000 ZZ=ZZ+ZZ + 4869 IFE ZZ, + 4870 032770 200 00 0 00 035124 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4871 032771 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4872 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4873 ;C(AC) NONZERO + 4874 032772 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4875 STOP^ + 4876 032773 254 04 0 00 032774 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4877 032774 324 00 0 00 032775 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4878 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4879 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4880 + 4881 ;********** + 4882 + 4883 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4884 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4885 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4886 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-10 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0111 + + 4887 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4888 006532 SN=SN+1 + 4889 000200 000000 ZZ=ZZ+ZZ + 4890 IFE ZZ, + 4891 032775 200 00 0 00 035125 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4892 032776 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4893 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4894 ;C(AC) NONZERO + 4895 032777 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4896 STOP^ + 4897 033000 254 04 0 00 033001 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4898 033001 324 00 0 00 033002 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4899 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4900 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4901 + 4902 ;********** + 4903 + 4904 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4905 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4906 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4907 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4908 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4909 006533 SN=SN+1 + 4910 000400 000000 ZZ=ZZ+ZZ + 4911 IFE ZZ, + 4912 033002 200 00 0 00 035073 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4913 033003 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4914 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4915 ;C(AC) NONZERO + 4916 033004 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4917 STOP^ + 4918 033005 254 04 0 00 033006 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4919 033006 324 00 0 00 033007 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4920 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4921 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4922 + 4923 ;********** + 4924 + 4925 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4926 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4927 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4928 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4929 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4930 006534 SN=SN+1 + 4931 001000 000000 ZZ=ZZ+ZZ + 4932 IFE ZZ, + 4933 033007 200 00 0 00 035126 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4934 033010 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4935 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4936 ;C(AC) NONZERO + 4937 033011 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4938 STOP^ + 4939 033012 254 04 0 00 033013 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4940 033013 324 00 0 00 033014 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4941 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-11 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0112 + + 4942 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4943 + 4944 ;********** + 4945 + 4946 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4947 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4948 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4949 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4950 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4951 006535 SN=SN+1 + 4952 002000 000000 ZZ=ZZ+ZZ + 4953 IFE ZZ, + 4954 033014 200 00 0 00 035127 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4955 033015 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4956 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4957 ;C(AC) NONZERO + 4958 033016 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4959 STOP^ + 4960 033017 254 04 0 00 033020 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4961 033020 324 00 0 00 033021 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4962 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4963 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4964 + 4965 ;********** + 4966 + 4967 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4968 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4969 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4970 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4971 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4972 006536 SN=SN+1 + 4973 004000 000000 ZZ=ZZ+ZZ + 4974 IFE ZZ, + 4975 033021 200 00 0 00 035130 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 4976 033022 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4977 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4978 ;C(AC) NONZERO + 4979 033023 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 4980 STOP^ + 4981 033024 254 04 0 00 033025 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 4982 033025 324 00 0 00 033026 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 4983 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 4984 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 4985 + 4986 ;********** + 4987 + 4988 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 4989 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 4990 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 4991 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 4992 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 4993 006537 SN=SN+1 + 4994 010000 000000 ZZ=ZZ+ZZ + 4995 IFE ZZ, + 4996 033026 200 00 0 00 035131 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-12 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0113 + + 4997 033027 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 4998 ;POSITION WITH ITSELF SHOULD RESULT WITH + 4999 ;C(AC) NONZERO + 5000 033030 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 5001 STOP^ + 5002 033031 254 04 0 00 033032 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5003 033032 324 00 0 00 033033 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5004 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5005 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5006 + 5007 ;********** + 5008 + 5009 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 5010 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 5011 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 5012 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 5013 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 5014 006540 SN=SN+1 + 5015 020000 000000 ZZ=ZZ+ZZ + 5016 IFE ZZ, + 5017 033033 200 00 0 00 035132 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 5018 033034 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 5019 ;POSITION WITH ITSELF SHOULD RESULT WITH + 5020 ;C(AC) NONZERO + 5021 033035 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 5022 STOP^ + 5023 033036 254 04 0 00 033037 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5024 033037 324 00 0 00 033040 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5025 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5026 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5027 + 5028 ;********** + 5029 + 5030 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 5031 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 5032 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 5033 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 5034 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 5035 006541 SN=SN+1 + 5036 040000 000000 ZZ=ZZ+ZZ + 5037 IFE ZZ, + 5038 033040 200 00 0 00 035133 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 5039 033041 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 5040 ;POSITION WITH ITSELF SHOULD RESULT WITH + 5041 ;C(AC) NONZERO + 5042 033042 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 5043 STOP^ + 5044 033043 254 04 0 00 033044 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5045 033044 324 00 0 00 033045 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5046 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5047 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5048 + 5049 ;********** + 5050 + 5051 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-13 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0114 + + 5052 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 5053 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 5054 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 5055 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 5056 006542 SN=SN+1 + 5057 100000 000000 ZZ=ZZ+ZZ + 5058 IFE ZZ, + 5059 033045 200 00 0 00 035134 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 5060 033046 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 5061 ;POSITION WITH ITSELF SHOULD RESULT WITH + 5062 ;C(AC) NONZERO + 5063 033047 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 5064 STOP^ + 5065 033050 254 04 0 00 033051 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5066 033051 324 00 0 00 033052 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5067 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5068 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5069 + 5070 ;********** + 5071 + 5072 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 5073 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 5074 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 5075 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 5076 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 5077 006543 SN=SN+1 + 5078 200000 000000 ZZ=ZZ+ZZ + 5079 IFE ZZ, + 5080 033052 200 00 0 00 035135 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 5081 033053 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 5082 ;POSITION WITH ITSELF SHOULD RESULT WITH + 5083 ;C(AC) NONZERO + 5084 033054 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 5085 STOP^ + 5086 033055 254 04 0 00 033056 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5087 033056 324 00 0 00 033057 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5088 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5089 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5090 + 5091 ;********** + 5092 + 5093 ;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF + 5094 ;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE + 5095 ;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS + 5096 ;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED + 5097 ;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. + 5098 006544 SN=SN+1 + 5099 400000 000000 ZZ=ZZ+ZZ + 5100 IFE ZZ, + 5101 033057 200 00 0 00 035065 MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + 5102 033060 404 00 0 00 000000 AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + 5103 ;POSITION WITH ITSELF SHOULD RESULT WITH + 5104 ;C(AC) NONZERO + 5105 033061 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NONZERO + 5106 STOP^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 25-14 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0115 + + 5107 033062 254 04 0 00 033063 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5108 033063 324 00 0 00 033064 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5109 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5110 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5111 + 5112 ;********** + 5113 + 5114 ;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE + 5115 ;C(AC)=0 AND C(E)=0 AND E OUTSIDE OF THE AC RANGE IS ALL ZEROS IN THE AC. + 5116 ;FIRST, THE AC IS CLEARED; THEN, THE C(AC) IS ANDED WITH C(E) WHEN + 5117 ;E GREATER THAN 20 AND C(E)=0. THE RESULT SHOULD BE C(AC)=0. + 5118 ;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + 5119 + 5120 033064 200 00 0 00 035062 A6600: MOVE [0] ;PRESET AC TO ALL ZEROS + 5121 033065 404 00 0 00 035062 AND [0] ;*AND OF TWO WORDS OF ALL ZEROS + 5122 ;SHOULD RESULT IN C(AC)=0 + 5123 033066 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 5124 STOP^ + 5125 033067 254 04 0 00 033070 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5126 033070 324 00 0 00 033071 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5127 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5128 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5129 + 5130 ;********** + 5131 + 5132 ;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(AC)=0 + 5133 ;AND C(E) IS ALL ONES IS AN AC CONTAINING ALL ZEROS. FIRST, THE AC IS + 5134 ;CLEARED; THEN THE C(AC) IS ANDED WITH AN E CONTAINING ALL ONES, WHERE + 5135 ;E IS BEYOND THE AC RANGE. THE RESULT SHOULD BE C(AC)=0. + 5136 ;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + 5137 + 5138 033071 200 00 0 00 035062 A6700: MOVE [0] ;PRESET AC TO ALL ZEROS + 5139 033072 404 00 0 00 035063 AND [-1] ;*AND OF TWO WORDS WHERE C(AC)=0 SHOULD RESULT IN C(AC)=0 + 5140 033073 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 5141 STOP^ + 5142 033074 254 04 0 00 033075 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5143 033075 324 00 0 00 033076 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5144 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5145 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5146 + 5147 ;AD FM+EN [ADFP],AD FM+EN A [ADFP], + 5148 ;F CYC ACT EN C [F1],IR BOOLE AD FM+F [IR2], + 5149 ;IR BOOLE [IR1],IRXX(1,4,5,11,13,16) [IR2] + 5150 + 5151 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 26 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0116 + + 5152 ;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(E)=0 + 5153 ;AND C(AC) IS ALL ONES IS AN AC CONTAINING ALL ZEROS. FIRST, THE + 5154 ;AC IS SET TO ALL ONES; THEN, THE AC IS ANDED WITH AN E CONTAINING ALL + 5155 ;ZEROS, WHERE E IS BEYOND THE AC RANGE. THE RESULT SHOULD BE C(AC)=0. + 5156 ;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + 5157 + 5158 033076 200 00 0 00 035063 A7000: MOVE [-1] ;PRESET AC TO ALL ONES + 5159 033077 404 00 0 00 035062 AND [0] ;*AND OF TWO WORDS WHERE C(E)=0 SHOULD RESULT IN C(AC)=0 + 5160 033100 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 5161 STOP^ + 5162 033101 254 04 0 00 033102 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5163 033102 324 00 0 00 033103 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5164 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5165 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5166 + 5167 ;********** + 5168 + 5169 ;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(E) IS + 5170 ;ALL ONES AND C(AC) IS ALL ONES IS AN AC CONTAINING ALL ONES. FIRST, THE + 5171 ;AC IS SET TO ALL ONES; THEN, C(AC) IS ANDED WITH AN E CONTAINING ALL + 5172 ;ONES. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5173 ;IF C(AC) IS ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + 5174 + 5175 033103 200 00 0 00 035063 A7100: MOVE [-1] ;PRELOAD AC WITH ALL ONES + 5176 033104 404 00 0 00 035063 AND [-1] ;*AND OF TWO WORDS OF ALL ONES SHOULD + 5177 ;RESULT IN C(AC)=ALL ONES + 5178 033105 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) IS NON ZERO + 5179 STOP^ + 5180 033106 254 04 0 00 033107 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5181 033107 324 00 0 00 033110 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5182 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5183 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5184 + 5185 ;********** + 5186 + 5187 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5188 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHERE CONTENTS IS NON-ZERO. + 5189 ;FIRST, THE AC IS CLEARED, THEN, XOR IS EXECUTED WHERE C(E) IS NON-ZERO. + 5190 ;THE AC IS THEN CHECKED FOR NON-ZERO CONTENTS. + 5191 033110 200 00 0 00 035062 A7200: MOVE [0] ;CLEAR AC + 5192 033111 430 00 0 00 035141 XOR [707070707070] ;*XOR WITH C(AC)=0 AND C(E)=7070707070 SHOULD + 5193 ;RESTULT IN C(AC)=707070707070 + 5194 033112 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5195 STOP^ + 5196 033113 254 04 0 00 033114 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5197 033114 324 00 0 00 033115 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5198 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5199 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5200 + 5201 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0117 + + 5202 007300 SN=7300 + 5203 000000 ZZ=0 + 5204 + 5205 A7300: REPEAT ^D36, + 5206 <;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5207 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5208 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5209 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5210 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5211 SN=SN+1 + 5212 ZZ=ZZ+ZZ + 5213 IFE ZZ, + 5214 MOVE [0] ;CLEAR AC + 5215 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5216 ;RESULT IN C(AC) NONZERO + 5217 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5218 STOP + 5219 + 5220 ;********** + 5221 > + 5222 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5223 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5224 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5225 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5226 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5227 007301 SN=SN+1 + 5228 000000 ZZ=ZZ+ZZ + 5229 000001 IFE ZZ, + 5230 033115 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5231 033116 430 00 0 00 035074 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5232 ;RESULT IN C(AC) NONZERO + 5233 033117 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5234 STOP^ + 5235 033120 254 04 0 00 033121 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5236 033121 324 00 0 00 033122 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5237 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5238 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5239 + 5240 ;********** + 5241 + 5242 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5243 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5244 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5245 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5246 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5247 007302 SN=SN+1 + 5248 000002 ZZ=ZZ+ZZ + 5249 IFE ZZ, + 5250 033122 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5251 033123 430 00 0 00 035075 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5252 ;RESULT IN C(AC) NONZERO + 5253 033124 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5254 STOP^ + 5255 033125 254 04 0 00 033126 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5256 033126 324 00 0 00 033127 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0118 + + 5257 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5258 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5259 + 5260 ;********** + 5261 + 5262 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5263 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5264 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5265 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5266 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5267 007303 SN=SN+1 + 5268 000004 ZZ=ZZ+ZZ + 5269 IFE ZZ, + 5270 033127 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5271 033130 430 00 0 00 035076 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5272 ;RESULT IN C(AC) NONZERO + 5273 033131 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5274 STOP^ + 5275 033132 254 04 0 00 033133 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5276 033133 324 00 0 00 033134 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5277 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5278 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5279 + 5280 ;********** + 5281 + 5282 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5283 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5284 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5285 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5286 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5287 007304 SN=SN+1 + 5288 000010 ZZ=ZZ+ZZ + 5289 IFE ZZ, + 5290 033134 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5291 033135 430 00 0 00 035077 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5292 ;RESULT IN C(AC) NONZERO + 5293 033136 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5294 STOP^ + 5295 033137 254 04 0 00 033140 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5296 033140 324 00 0 00 033141 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5297 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5298 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5299 + 5300 ;********** + 5301 + 5302 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5303 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5304 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5305 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5306 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5307 007305 SN=SN+1 + 5308 000020 ZZ=ZZ+ZZ + 5309 IFE ZZ, + 5310 033141 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5311 033142 430 00 0 00 035100 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-2 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0119 + + 5312 ;RESULT IN C(AC) NONZERO + 5313 033143 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5314 STOP^ + 5315 033144 254 04 0 00 033145 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5316 033145 324 00 0 00 033146 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5317 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5318 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5319 + 5320 ;********** + 5321 + 5322 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5323 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5324 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5325 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5326 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5327 007306 SN=SN+1 + 5328 000040 ZZ=ZZ+ZZ + 5329 IFE ZZ, + 5330 033146 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5331 033147 430 00 0 00 035101 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5332 ;RESULT IN C(AC) NONZERO + 5333 033150 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5334 STOP^ + 5335 033151 254 04 0 00 033152 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5336 033152 324 00 0 00 033153 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5337 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5338 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5339 + 5340 ;********** + 5341 + 5342 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5343 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5344 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5345 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5346 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5347 007307 SN=SN+1 + 5348 000100 ZZ=ZZ+ZZ + 5349 IFE ZZ, + 5350 033153 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5351 033154 430 00 0 00 035102 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5352 ;RESULT IN C(AC) NONZERO + 5353 033155 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5354 STOP^ + 5355 033156 254 04 0 00 033157 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5356 033157 324 00 0 00 033160 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5357 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5358 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5359 + 5360 ;********** + 5361 + 5362 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5363 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5364 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5365 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5366 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-3 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0120 + + 5367 007310 SN=SN+1 + 5368 000200 ZZ=ZZ+ZZ + 5369 IFE ZZ, + 5370 033160 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5371 033161 430 00 0 00 035103 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5372 ;RESULT IN C(AC) NONZERO + 5373 033162 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5374 STOP^ + 5375 033163 254 04 0 00 033164 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5376 033164 324 00 0 00 033165 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5377 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5378 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5379 + 5380 ;********** + 5381 + 5382 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5383 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5384 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5385 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5386 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5387 007311 SN=SN+1 + 5388 000400 ZZ=ZZ+ZZ + 5389 IFE ZZ, + 5390 033165 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5391 033166 430 00 0 00 035104 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5392 ;RESULT IN C(AC) NONZERO + 5393 033167 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5394 STOP^ + 5395 033170 254 04 0 00 033171 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5396 033171 324 00 0 00 033172 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5397 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5398 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5399 + 5400 ;********** + 5401 + 5402 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5403 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5404 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5405 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5406 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5407 007312 SN=SN+1 + 5408 001000 ZZ=ZZ+ZZ + 5409 IFE ZZ, + 5410 033172 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5411 033173 430 00 0 00 035105 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5412 ;RESULT IN C(AC) NONZERO + 5413 033174 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5414 STOP^ + 5415 033175 254 04 0 00 033176 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5416 033176 324 00 0 00 033177 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5417 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5418 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5419 + 5420 ;********** + 5421 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-4 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0121 + + 5422 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5423 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5424 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5425 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5426 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5427 007313 SN=SN+1 + 5428 002000 ZZ=ZZ+ZZ + 5429 IFE ZZ, + 5430 033177 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5431 033200 430 00 0 00 035106 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5432 ;RESULT IN C(AC) NONZERO + 5433 033201 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5434 STOP^ + 5435 033202 254 04 0 00 033203 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5436 033203 324 00 0 00 033204 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5437 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5438 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5439 + 5440 ;********** + 5441 + 5442 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5443 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5444 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5445 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5446 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5447 007314 SN=SN+1 + 5448 004000 ZZ=ZZ+ZZ + 5449 IFE ZZ, + 5450 033204 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5451 033205 430 00 0 00 035107 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5452 ;RESULT IN C(AC) NONZERO + 5453 033206 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5454 STOP^ + 5455 033207 254 04 0 00 033210 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5456 033210 324 00 0 00 033211 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5457 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5458 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5459 + 5460 ;********** + 5461 + 5462 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5463 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5464 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5465 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5466 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5467 007315 SN=SN+1 + 5468 010000 ZZ=ZZ+ZZ + 5469 IFE ZZ, + 5470 033211 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5471 033212 430 00 0 00 035110 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5472 ;RESULT IN C(AC) NONZERO + 5473 033213 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5474 STOP^ + 5475 033214 254 04 0 00 033215 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5476 033215 324 00 0 00 033216 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-5 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0122 + + 5477 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5478 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5479 + 5480 ;********** + 5481 + 5482 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5483 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5484 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5485 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5486 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5487 007316 SN=SN+1 + 5488 020000 ZZ=ZZ+ZZ + 5489 IFE ZZ, + 5490 033216 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5491 033217 430 00 0 00 035111 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5492 ;RESULT IN C(AC) NONZERO + 5493 033220 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5494 STOP^ + 5495 033221 254 04 0 00 033222 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5496 033222 324 00 0 00 033223 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5497 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5498 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5499 + 5500 ;********** + 5501 + 5502 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5503 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5504 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5505 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5506 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5507 007317 SN=SN+1 + 5508 040000 ZZ=ZZ+ZZ + 5509 IFE ZZ, + 5510 033223 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5511 033224 430 00 0 00 035112 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5512 ;RESULT IN C(AC) NONZERO + 5513 033225 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5514 STOP^ + 5515 033226 254 04 0 00 033227 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5516 033227 324 00 0 00 033230 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5517 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5518 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5519 + 5520 ;********** + 5521 + 5522 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5523 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5524 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5525 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5526 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5527 007320 SN=SN+1 + 5528 100000 ZZ=ZZ+ZZ + 5529 IFE ZZ, + 5530 033230 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5531 033231 430 00 0 00 035113 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-6 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0123 + + 5532 ;RESULT IN C(AC) NONZERO + 5533 033232 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5534 STOP^ + 5535 033233 254 04 0 00 033234 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5536 033234 324 00 0 00 033235 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5537 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5538 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5539 + 5540 ;********** + 5541 + 5542 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5543 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5544 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5545 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5546 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5547 007321 SN=SN+1 + 5548 200000 ZZ=ZZ+ZZ + 5549 IFE ZZ, + 5550 033235 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5551 033236 430 00 0 00 035114 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5552 ;RESULT IN C(AC) NONZERO + 5553 033237 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5554 STOP^ + 5555 033240 254 04 0 00 033241 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5556 033241 324 00 0 00 033242 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5557 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5558 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5559 + 5560 ;********** + 5561 + 5562 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5563 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5564 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5565 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5566 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5567 007322 SN=SN+1 + 5568 400000 ZZ=ZZ+ZZ + 5569 IFE ZZ, + 5570 033242 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5571 033243 430 00 0 00 035115 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5572 ;RESULT IN C(AC) NONZERO + 5573 033244 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5574 STOP^ + 5575 033245 254 04 0 00 033246 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5576 033246 324 00 0 00 033247 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5577 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5578 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5579 + 5580 ;********** + 5581 + 5582 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5583 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5584 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5585 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5586 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-7 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0124 + + 5587 007323 SN=SN+1 + 5588 000001 000000 ZZ=ZZ+ZZ + 5589 IFE ZZ, + 5590 033247 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5591 033250 430 00 0 00 035116 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5592 ;RESULT IN C(AC) NONZERO + 5593 033251 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5594 STOP^ + 5595 033252 254 04 0 00 033253 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5596 033253 324 00 0 00 033254 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5597 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5598 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5599 + 5600 ;********** + 5601 + 5602 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5603 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5604 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5605 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5606 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5607 007324 SN=SN+1 + 5608 000002 000000 ZZ=ZZ+ZZ + 5609 IFE ZZ, + 5610 033254 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5611 033255 430 00 0 00 035117 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5612 ;RESULT IN C(AC) NONZERO + 5613 033256 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5614 STOP^ + 5615 033257 254 04 0 00 033260 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5616 033260 324 00 0 00 033261 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5617 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5618 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5619 + 5620 ;********** + 5621 + 5622 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5623 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5624 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5625 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5626 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5627 007325 SN=SN+1 + 5628 000004 000000 ZZ=ZZ+ZZ + 5629 IFE ZZ, + 5630 033261 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5631 033262 430 00 0 00 035120 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5632 ;RESULT IN C(AC) NONZERO + 5633 033263 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5634 STOP^ + 5635 033264 254 04 0 00 033265 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5636 033265 324 00 0 00 033266 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5637 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5638 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5639 + 5640 ;********** + 5641 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-8 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0125 + + 5642 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5643 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5644 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5645 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5646 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5647 007326 SN=SN+1 + 5648 000010 000000 ZZ=ZZ+ZZ + 5649 IFE ZZ, + 5650 033266 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5651 033267 430 00 0 00 035121 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5652 ;RESULT IN C(AC) NONZERO + 5653 033270 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5654 STOP^ + 5655 033271 254 04 0 00 033272 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5656 033272 324 00 0 00 033273 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5657 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5658 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5659 + 5660 ;********** + 5661 + 5662 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5663 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5664 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5665 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5666 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5667 007327 SN=SN+1 + 5668 000020 000000 ZZ=ZZ+ZZ + 5669 IFE ZZ, + 5670 033273 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5671 033274 430 00 0 00 035122 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5672 ;RESULT IN C(AC) NONZERO + 5673 033275 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5674 STOP^ + 5675 033276 254 04 0 00 033277 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5676 033277 324 00 0 00 033300 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5677 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5678 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5679 + 5680 ;********** + 5681 + 5682 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5683 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5684 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5685 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5686 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5687 007330 SN=SN+1 + 5688 000040 000000 ZZ=ZZ+ZZ + 5689 IFE ZZ, + 5690 033300 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5691 033301 430 00 0 00 035123 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5692 ;RESULT IN C(AC) NONZERO + 5693 033302 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5694 STOP^ + 5695 033303 254 04 0 00 033304 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5696 033304 324 00 0 00 033305 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-9 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0126 + + 5697 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5698 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5699 + 5700 ;********** + 5701 + 5702 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5703 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5704 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5705 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5706 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5707 007331 SN=SN+1 + 5708 000100 000000 ZZ=ZZ+ZZ + 5709 IFE ZZ, + 5710 033305 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5711 033306 430 00 0 00 035124 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5712 ;RESULT IN C(AC) NONZERO + 5713 033307 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5714 STOP^ + 5715 033310 254 04 0 00 033311 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5716 033311 324 00 0 00 033312 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5717 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5718 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5719 + 5720 ;********** + 5721 + 5722 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5723 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5724 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5725 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5726 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5727 007332 SN=SN+1 + 5728 000200 000000 ZZ=ZZ+ZZ + 5729 IFE ZZ, + 5730 033312 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5731 033313 430 00 0 00 035125 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5732 ;RESULT IN C(AC) NONZERO + 5733 033314 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5734 STOP^ + 5735 033315 254 04 0 00 033316 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5736 033316 324 00 0 00 033317 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5737 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5738 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5739 + 5740 ;********** + 5741 + 5742 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5743 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5744 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5745 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5746 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5747 007333 SN=SN+1 + 5748 000400 000000 ZZ=ZZ+ZZ + 5749 IFE ZZ, + 5750 033317 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5751 033320 430 00 0 00 035073 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-10 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0127 + + 5752 ;RESULT IN C(AC) NONZERO + 5753 033321 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5754 STOP^ + 5755 033322 254 04 0 00 033323 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5756 033323 324 00 0 00 033324 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5757 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5758 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5759 + 5760 ;********** + 5761 + 5762 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5763 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5764 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5765 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5766 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5767 007334 SN=SN+1 + 5768 001000 000000 ZZ=ZZ+ZZ + 5769 IFE ZZ, + 5770 033324 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5771 033325 430 00 0 00 035126 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5772 ;RESULT IN C(AC) NONZERO + 5773 033326 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5774 STOP^ + 5775 033327 254 04 0 00 033330 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5776 033330 324 00 0 00 033331 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5777 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5778 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5779 + 5780 ;********** + 5781 + 5782 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5783 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5784 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5785 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5786 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5787 007335 SN=SN+1 + 5788 002000 000000 ZZ=ZZ+ZZ + 5789 IFE ZZ, + 5790 033331 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5791 033332 430 00 0 00 035127 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5792 ;RESULT IN C(AC) NONZERO + 5793 033333 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5794 STOP^ + 5795 033334 254 04 0 00 033335 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5796 033335 324 00 0 00 033336 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5797 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5798 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5799 + 5800 ;********** + 5801 + 5802 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5803 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5804 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5805 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5806 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-11 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0128 + + 5807 007336 SN=SN+1 + 5808 004000 000000 ZZ=ZZ+ZZ + 5809 IFE ZZ, + 5810 033336 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5811 033337 430 00 0 00 035130 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5812 ;RESULT IN C(AC) NONZERO + 5813 033340 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5814 STOP^ + 5815 033341 254 04 0 00 033342 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5816 033342 324 00 0 00 033343 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5817 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5818 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5819 + 5820 ;********** + 5821 + 5822 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5823 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5824 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5825 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5826 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5827 007337 SN=SN+1 + 5828 010000 000000 ZZ=ZZ+ZZ + 5829 IFE ZZ, + 5830 033343 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5831 033344 430 00 0 00 035131 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5832 ;RESULT IN C(AC) NONZERO + 5833 033345 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5834 STOP^ + 5835 033346 254 04 0 00 033347 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5836 033347 324 00 0 00 033350 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5837 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5838 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5839 + 5840 ;********** + 5841 + 5842 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5843 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5844 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5845 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5846 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5847 007340 SN=SN+1 + 5848 020000 000000 ZZ=ZZ+ZZ + 5849 IFE ZZ, + 5850 033350 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5851 033351 430 00 0 00 035132 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5852 ;RESULT IN C(AC) NONZERO + 5853 033352 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5854 STOP^ + 5855 033353 254 04 0 00 033354 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5856 033354 324 00 0 00 033355 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5857 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5858 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5859 + 5860 ;********** + 5861 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-12 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0129 + + 5862 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5863 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5864 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5865 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5866 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5867 007341 SN=SN+1 + 5868 040000 000000 ZZ=ZZ+ZZ + 5869 IFE ZZ, + 5870 033355 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5871 033356 430 00 0 00 035133 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5872 ;RESULT IN C(AC) NONZERO + 5873 033357 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5874 STOP^ + 5875 033360 254 04 0 00 033361 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5876 033361 324 00 0 00 033362 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5877 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5878 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5879 + 5880 ;********** + 5881 + 5882 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5883 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5884 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5885 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5886 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5887 007342 SN=SN+1 + 5888 100000 000000 ZZ=ZZ+ZZ + 5889 IFE ZZ, + 5890 033362 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5891 033363 430 00 0 00 035134 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5892 ;RESULT IN C(AC) NONZERO + 5893 033364 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5894 STOP^ + 5895 033365 254 04 0 00 033366 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5896 033366 324 00 0 00 033367 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5897 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5898 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5899 + 5900 ;********** + 5901 + 5902 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5903 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5904 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5905 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5906 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5907 007343 SN=SN+1 + 5908 200000 000000 ZZ=ZZ+ZZ + 5909 IFE ZZ, + 5910 033367 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5911 033370 430 00 0 00 035135 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5912 ;RESULT IN C(AC) NONZERO + 5913 033371 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5914 STOP^ + 5915 033372 254 04 0 00 033373 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5916 033373 324 00 0 00 033374 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-13 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0130 + + 5917 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5918 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5919 + 5920 ;********** + 5921 + 5922 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5923 ;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5924 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED + 5925 ;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. + 5926 ;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5927 007344 SN=SN+1 + 5928 400000 000000 ZZ=ZZ+ZZ + 5929 IFE ZZ, + 5930 033374 200 00 0 00 035062 MOVE [0] ;CLEAR AC + 5931 033375 430 00 0 00 035065 XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + 5932 ;RESULT IN C(AC) NONZERO + 5933 033376 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5934 STOP^ + 5935 033377 254 04 0 00 033400 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5936 033400 324 00 0 00 033401 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5937 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5938 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5939 + 5940 ;********** + 5941 + 5942 007400 SN=7400 + 5943 000000 ZZ=0 + 5944 + 5945 A7400: REPEAT ^D36, + 5946 <;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5947 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5948 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 5949 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 5950 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5951 SN=SN+1 + 5952 ZZ=ZZ+ZZ + 5953 IFE ZZ, + 5954 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 5955 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 5956 ;RESULT IN C(AC) NON-ZERO + 5957 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5958 STOP + 5959 + 5960 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 5961 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 5962 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 5963 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 5964 + 5965 ;********** + 5966 > + 5967 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5968 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5969 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 5970 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 5971 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-14 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0131 + + 5972 007401 SN=SN+1 + 5973 000000 ZZ=ZZ+ZZ + 5974 000001 IFE ZZ, + 5975 033401 200 00 0 00 035074 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 5976 033402 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 5977 ;RESULT IN C(AC) NON-ZERO + 5978 033403 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 5979 STOP^ + 5980 033404 254 04 0 00 033405 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 5981 033405 324 00 0 00 033406 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 5982 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 5983 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 5984 + 5985 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 5986 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 5987 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 5988 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 5989 + 5990 ;********** + 5991 + 5992 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 5993 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 5994 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 5995 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 5996 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 5997 007402 SN=SN+1 + 5998 000002 ZZ=ZZ+ZZ + 5999 IFE ZZ, + 6000 033406 200 00 0 00 035075 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6001 033407 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6002 ;RESULT IN C(AC) NON-ZERO + 6003 033410 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6004 STOP^ + 6005 033411 254 04 0 00 033412 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6006 033412 324 00 0 00 033413 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6007 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6008 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6009 + 6010 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6011 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6012 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6013 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6014 + 6015 ;********** + 6016 + 6017 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6018 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6019 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6020 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6021 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6022 007403 SN=SN+1 + 6023 000004 ZZ=ZZ+ZZ + 6024 IFE ZZ, + 6025 033413 200 00 0 00 035076 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6026 033414 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-15 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0132 + + 6027 ;RESULT IN C(AC) NON-ZERO + 6028 033415 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6029 STOP^ + 6030 033416 254 04 0 00 033417 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6031 033417 324 00 0 00 033420 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6032 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6033 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6034 + 6035 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6036 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6037 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6038 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6039 + 6040 ;********** + 6041 + 6042 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6043 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6044 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6045 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6046 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6047 007404 SN=SN+1 + 6048 000010 ZZ=ZZ+ZZ + 6049 IFE ZZ, + 6050 033420 200 00 0 00 035077 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6051 033421 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6052 ;RESULT IN C(AC) NON-ZERO + 6053 033422 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6054 STOP^ + 6055 033423 254 04 0 00 033424 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6056 033424 324 00 0 00 033425 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6057 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6058 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6059 + 6060 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6061 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6062 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6063 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6064 + 6065 ;********** + 6066 + 6067 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6068 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6069 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6070 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6071 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6072 007405 SN=SN+1 + 6073 000020 ZZ=ZZ+ZZ + 6074 IFE ZZ, + 6075 033425 200 00 0 00 035100 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6076 033426 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6077 ;RESULT IN C(AC) NON-ZERO + 6078 033427 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6079 STOP^ + 6080 033430 254 04 0 00 033431 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6081 033431 324 00 0 00 033432 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-16 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0133 + + 6082 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6083 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6084 + 6085 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6086 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6087 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6088 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6089 + 6090 ;********** + 6091 + 6092 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6093 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6094 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6095 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6096 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6097 007406 SN=SN+1 + 6098 000040 ZZ=ZZ+ZZ + 6099 IFE ZZ, + 6100 033432 200 00 0 00 035101 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6101 033433 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6102 ;RESULT IN C(AC) NON-ZERO + 6103 033434 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6104 STOP^ + 6105 033435 254 04 0 00 033436 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6106 033436 324 00 0 00 033437 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6107 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6108 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6109 + 6110 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6111 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6112 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6113 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6114 + 6115 ;********** + 6116 + 6117 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6118 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6119 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6120 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6121 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6122 007407 SN=SN+1 + 6123 000100 ZZ=ZZ+ZZ + 6124 IFE ZZ, + 6125 033437 200 00 0 00 035102 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6126 033440 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6127 ;RESULT IN C(AC) NON-ZERO + 6128 033441 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6129 STOP^ + 6130 033442 254 04 0 00 033443 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6131 033443 324 00 0 00 033444 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6132 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6133 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6134 + 6135 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6136 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-17 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0134 + + 6137 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6138 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6139 + 6140 ;********** + 6141 + 6142 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6143 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6144 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6145 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6146 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6147 007410 SN=SN+1 + 6148 000200 ZZ=ZZ+ZZ + 6149 IFE ZZ, + 6150 033444 200 00 0 00 035103 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6151 033445 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6152 ;RESULT IN C(AC) NON-ZERO + 6153 033446 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6154 STOP^ + 6155 033447 254 04 0 00 033450 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6156 033450 324 00 0 00 033451 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6157 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6158 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6159 + 6160 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6161 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6162 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6163 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6164 + 6165 ;********** + 6166 + 6167 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6168 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6169 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6170 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6171 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6172 007411 SN=SN+1 + 6173 000400 ZZ=ZZ+ZZ + 6174 IFE ZZ, + 6175 033451 200 00 0 00 035104 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6176 033452 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6177 ;RESULT IN C(AC) NON-ZERO + 6178 033453 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6179 STOP^ + 6180 033454 254 04 0 00 033455 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6181 033455 324 00 0 00 033456 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6182 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6183 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6184 + 6185 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6186 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6187 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6188 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6189 + 6190 ;********** + 6191 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-18 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0135 + + 6192 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6193 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6194 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6195 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6196 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6197 007412 SN=SN+1 + 6198 001000 ZZ=ZZ+ZZ + 6199 IFE ZZ, + 6200 033456 200 00 0 00 035105 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6201 033457 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6202 ;RESULT IN C(AC) NON-ZERO + 6203 033460 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6204 STOP^ + 6205 033461 254 04 0 00 033462 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6206 033462 324 00 0 00 033463 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6207 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6208 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6209 + 6210 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6211 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6212 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6213 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6214 + 6215 ;********** + 6216 + 6217 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6218 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6219 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6220 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6221 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6222 007413 SN=SN+1 + 6223 002000 ZZ=ZZ+ZZ + 6224 IFE ZZ, + 6225 033463 200 00 0 00 035106 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6226 033464 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6227 ;RESULT IN C(AC) NON-ZERO + 6228 033465 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6229 STOP^ + 6230 033466 254 04 0 00 033467 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6231 033467 324 00 0 00 033470 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6232 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6233 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6234 + 6235 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6236 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6237 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6238 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6239 + 6240 ;********** + 6241 + 6242 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6243 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6244 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6245 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6246 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-19 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0136 + + 6247 007414 SN=SN+1 + 6248 004000 ZZ=ZZ+ZZ + 6249 IFE ZZ, + 6250 033470 200 00 0 00 035107 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6251 033471 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6252 ;RESULT IN C(AC) NON-ZERO + 6253 033472 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6254 STOP^ + 6255 033473 254 04 0 00 033474 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6256 033474 324 00 0 00 033475 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6257 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6258 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6259 + 6260 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6261 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6262 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6263 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6264 + 6265 ;********** + 6266 + 6267 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6268 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6269 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6270 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6271 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6272 007415 SN=SN+1 + 6273 010000 ZZ=ZZ+ZZ + 6274 IFE ZZ, + 6275 033475 200 00 0 00 035110 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6276 033476 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6277 ;RESULT IN C(AC) NON-ZERO + 6278 033477 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6279 STOP^ + 6280 033500 254 04 0 00 033501 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6281 033501 324 00 0 00 033502 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6282 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6283 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6284 + 6285 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6286 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6287 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6288 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6289 + 6290 ;********** + 6291 + 6292 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6293 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6294 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6295 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6296 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6297 007416 SN=SN+1 + 6298 020000 ZZ=ZZ+ZZ + 6299 IFE ZZ, + 6300 033502 200 00 0 00 035111 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6301 033503 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-20 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0137 + + 6302 ;RESULT IN C(AC) NON-ZERO + 6303 033504 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6304 STOP^ + 6305 033505 254 04 0 00 033506 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6306 033506 324 00 0 00 033507 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6307 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6308 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6309 + 6310 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6311 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6312 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6313 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6314 + 6315 ;********** + 6316 + 6317 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6318 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6319 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6320 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6321 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6322 007417 SN=SN+1 + 6323 040000 ZZ=ZZ+ZZ + 6324 IFE ZZ, + 6325 033507 200 00 0 00 035112 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6326 033510 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6327 ;RESULT IN C(AC) NON-ZERO + 6328 033511 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6329 STOP^ + 6330 033512 254 04 0 00 033513 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6331 033513 324 00 0 00 033514 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6332 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6333 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6334 + 6335 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6336 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6337 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6338 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6339 + 6340 ;********** + 6341 + 6342 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6343 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6344 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6345 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6346 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6347 007420 SN=SN+1 + 6348 100000 ZZ=ZZ+ZZ + 6349 IFE ZZ, + 6350 033514 200 00 0 00 035113 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6351 033515 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6352 ;RESULT IN C(AC) NON-ZERO + 6353 033516 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6354 STOP^ + 6355 033517 254 04 0 00 033520 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6356 033520 324 00 0 00 033521 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-21 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0138 + + 6357 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6358 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6359 + 6360 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6361 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6362 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6363 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6364 + 6365 ;********** + 6366 + 6367 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6368 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6369 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6370 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6371 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6372 007421 SN=SN+1 + 6373 200000 ZZ=ZZ+ZZ + 6374 IFE ZZ, + 6375 033521 200 00 0 00 035114 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6376 033522 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6377 ;RESULT IN C(AC) NON-ZERO + 6378 033523 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6379 STOP^ + 6380 033524 254 04 0 00 033525 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6381 033525 324 00 0 00 033526 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6382 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6383 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6384 + 6385 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6386 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6387 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6388 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6389 + 6390 ;********** + 6391 + 6392 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6393 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6394 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6395 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6396 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6397 007422 SN=SN+1 + 6398 400000 ZZ=ZZ+ZZ + 6399 IFE ZZ, + 6400 033526 200 00 0 00 035115 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6401 033527 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6402 ;RESULT IN C(AC) NON-ZERO + 6403 033530 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6404 STOP^ + 6405 033531 254 04 0 00 033532 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6406 033532 324 00 0 00 033533 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6407 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6408 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6409 + 6410 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6411 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-22 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0139 + + 6412 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6413 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6414 + 6415 ;********** + 6416 + 6417 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6418 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6419 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6420 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6421 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6422 007423 SN=SN+1 + 6423 000001 000000 ZZ=ZZ+ZZ + 6424 IFE ZZ, + 6425 033533 200 00 0 00 035116 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6426 033534 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6427 ;RESULT IN C(AC) NON-ZERO + 6428 033535 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6429 STOP^ + 6430 033536 254 04 0 00 033537 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6431 033537 324 00 0 00 033540 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6432 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6433 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6434 + 6435 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6436 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6437 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6438 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6439 + 6440 ;********** + 6441 + 6442 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6443 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6444 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6445 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6446 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6447 007424 SN=SN+1 + 6448 000002 000000 ZZ=ZZ+ZZ + 6449 IFE ZZ, + 6450 033540 200 00 0 00 035117 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6451 033541 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6452 ;RESULT IN C(AC) NON-ZERO + 6453 033542 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6454 STOP^ + 6455 033543 254 04 0 00 033544 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6456 033544 324 00 0 00 033545 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6457 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6458 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6459 + 6460 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6461 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6462 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6463 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6464 + 6465 ;********** + 6466 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-23 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0140 + + 6467 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6468 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6469 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6470 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6471 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6472 007425 SN=SN+1 + 6473 000004 000000 ZZ=ZZ+ZZ + 6474 IFE ZZ, + 6475 033545 200 00 0 00 035120 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6476 033546 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6477 ;RESULT IN C(AC) NON-ZERO + 6478 033547 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6479 STOP^ + 6480 033550 254 04 0 00 033551 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6481 033551 324 00 0 00 033552 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6482 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6483 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6484 + 6485 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6486 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6487 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6488 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6489 + 6490 ;********** + 6491 + 6492 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6493 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6494 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6495 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6496 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6497 007426 SN=SN+1 + 6498 000010 000000 ZZ=ZZ+ZZ + 6499 IFE ZZ, + 6500 033552 200 00 0 00 035121 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6501 033553 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6502 ;RESULT IN C(AC) NON-ZERO + 6503 033554 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6504 STOP^ + 6505 033555 254 04 0 00 033556 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6506 033556 324 00 0 00 033557 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6507 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6508 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6509 + 6510 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6511 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6512 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6513 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6514 + 6515 ;********** + 6516 + 6517 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6518 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6519 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6520 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6521 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-24 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0141 + + 6522 007427 SN=SN+1 + 6523 000020 000000 ZZ=ZZ+ZZ + 6524 IFE ZZ, + 6525 033557 200 00 0 00 035122 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6526 033560 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6527 ;RESULT IN C(AC) NON-ZERO + 6528 033561 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6529 STOP^ + 6530 033562 254 04 0 00 033563 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6531 033563 324 00 0 00 033564 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6532 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6533 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6534 + 6535 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6536 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6537 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6538 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6539 + 6540 ;********** + 6541 + 6542 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6543 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6544 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6545 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6546 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6547 007430 SN=SN+1 + 6548 000040 000000 ZZ=ZZ+ZZ + 6549 IFE ZZ, + 6550 033564 200 00 0 00 035123 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6551 033565 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6552 ;RESULT IN C(AC) NON-ZERO + 6553 033566 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6554 STOP^ + 6555 033567 254 04 0 00 033570 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6556 033570 324 00 0 00 033571 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6557 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6558 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6559 + 6560 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6561 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6562 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6563 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6564 + 6565 ;********** + 6566 + 6567 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6568 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6569 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6570 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6571 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6572 007431 SN=SN+1 + 6573 000100 000000 ZZ=ZZ+ZZ + 6574 IFE ZZ, + 6575 033571 200 00 0 00 035124 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6576 033572 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-25 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0142 + + 6577 ;RESULT IN C(AC) NON-ZERO + 6578 033573 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6579 STOP^ + 6580 033574 254 04 0 00 033575 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6581 033575 324 00 0 00 033576 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6582 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6583 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6584 + 6585 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6586 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6587 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6588 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6589 + 6590 ;********** + 6591 + 6592 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6593 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6594 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6595 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6596 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6597 007432 SN=SN+1 + 6598 000200 000000 ZZ=ZZ+ZZ + 6599 IFE ZZ, + 6600 033576 200 00 0 00 035125 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6601 033577 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6602 ;RESULT IN C(AC) NON-ZERO + 6603 033600 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6604 STOP^ + 6605 033601 254 04 0 00 033602 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6606 033602 324 00 0 00 033603 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6607 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6608 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6609 + 6610 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6611 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6612 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6613 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6614 + 6615 ;********** + 6616 + 6617 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6618 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6619 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6620 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6621 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6622 007433 SN=SN+1 + 6623 000400 000000 ZZ=ZZ+ZZ + 6624 IFE ZZ, + 6625 033603 200 00 0 00 035073 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6626 033604 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6627 ;RESULT IN C(AC) NON-ZERO + 6628 033605 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6629 STOP^ + 6630 033606 254 04 0 00 033607 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6631 033607 324 00 0 00 033610 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-26 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0143 + + 6632 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6633 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6634 + 6635 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6636 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6637 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6638 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6639 + 6640 ;********** + 6641 + 6642 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6643 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6644 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6645 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6646 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6647 007434 SN=SN+1 + 6648 001000 000000 ZZ=ZZ+ZZ + 6649 IFE ZZ, + 6650 033610 200 00 0 00 035126 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6651 033611 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6652 ;RESULT IN C(AC) NON-ZERO + 6653 033612 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6654 STOP^ + 6655 033613 254 04 0 00 033614 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6656 033614 324 00 0 00 033615 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6657 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6658 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6659 + 6660 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6661 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6662 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6663 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6664 + 6665 ;********** + 6666 + 6667 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6668 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6669 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6670 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6671 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6672 007435 SN=SN+1 + 6673 002000 000000 ZZ=ZZ+ZZ + 6674 IFE ZZ, + 6675 033615 200 00 0 00 035127 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6676 033616 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6677 ;RESULT IN C(AC) NON-ZERO + 6678 033617 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6679 STOP^ + 6680 033620 254 04 0 00 033621 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6681 033621 324 00 0 00 033622 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6682 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6683 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6684 + 6685 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6686 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-27 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0144 + + 6687 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6688 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6689 + 6690 ;********** + 6691 + 6692 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6693 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6694 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6695 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6696 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6697 007436 SN=SN+1 + 6698 004000 000000 ZZ=ZZ+ZZ + 6699 IFE ZZ, + 6700 033622 200 00 0 00 035130 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6701 033623 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6702 ;RESULT IN C(AC) NON-ZERO + 6703 033624 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6704 STOP^ + 6705 033625 254 04 0 00 033626 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6706 033626 324 00 0 00 033627 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6707 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6708 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6709 + 6710 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6711 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6712 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6713 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6714 + 6715 ;********** + 6716 + 6717 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6718 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6719 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6720 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6721 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6722 007437 SN=SN+1 + 6723 010000 000000 ZZ=ZZ+ZZ + 6724 IFE ZZ, + 6725 033627 200 00 0 00 035131 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6726 033630 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6727 ;RESULT IN C(AC) NON-ZERO + 6728 033631 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6729 STOP^ + 6730 033632 254 04 0 00 033633 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6731 033633 324 00 0 00 033634 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6732 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6733 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6734 + 6735 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6736 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6737 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6738 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6739 + 6740 ;********** + 6741 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-28 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0145 + + 6742 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6743 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6744 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6745 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6746 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6747 007440 SN=SN+1 + 6748 020000 000000 ZZ=ZZ+ZZ + 6749 IFE ZZ, + 6750 033634 200 00 0 00 035132 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6751 033635 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6752 ;RESULT IN C(AC) NON-ZERO + 6753 033636 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6754 STOP^ + 6755 033637 254 04 0 00 033640 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6756 033640 324 00 0 00 033641 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6757 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6758 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6759 + 6760 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6761 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6762 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6763 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6764 + 6765 ;********** + 6766 + 6767 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6768 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6769 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6770 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6771 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6772 007441 SN=SN+1 + 6773 040000 000000 ZZ=ZZ+ZZ + 6774 IFE ZZ, + 6775 033641 200 00 0 00 035133 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6776 033642 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6777 ;RESULT IN C(AC) NON-ZERO + 6778 033643 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6779 STOP^ + 6780 033644 254 04 0 00 033645 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6781 033645 324 00 0 00 033646 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6782 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6783 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6784 + 6785 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6786 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6787 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6788 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6789 + 6790 ;********** + 6791 + 6792 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6793 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6794 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6795 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6796 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-29 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0146 + + 6797 007442 SN=SN+1 + 6798 100000 000000 ZZ=ZZ+ZZ + 6799 IFE ZZ, + 6800 033646 200 00 0 00 035134 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6801 033647 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6802 ;RESULT IN C(AC) NON-ZERO + 6803 033650 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6804 STOP^ + 6805 033651 254 04 0 00 033652 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6806 033652 324 00 0 00 033653 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6807 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6808 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6809 + 6810 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6811 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6812 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6813 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6814 + 6815 ;********** + 6816 + 6817 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6818 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6819 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6820 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6821 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6822 007443 SN=SN+1 + 6823 200000 000000 ZZ=ZZ+ZZ + 6824 IFE ZZ, + 6825 033653 200 00 0 00 035135 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6826 033654 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + 6827 ;RESULT IN C(AC) NON-ZERO + 6828 033655 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6829 STOP^ + 6830 033656 254 04 0 00 033657 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6831 033657 324 00 0 00 033660 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6832 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6833 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6834 + 6835 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6836 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6837 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6838 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6839 + 6840 ;********** + 6841 + 6842 ;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE + 6843 ;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS + 6844 ;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE + 6845 ;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E + 6846 ;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. + 6847 007444 SN=SN+1 + 6848 400000 000000 ZZ=ZZ+ZZ + 6849 IFE ZZ, + 6850 033660 200 00 0 00 035065 MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + 6851 033661 430 00 0 00 035062 XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 27-30 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0147 + + 6852 ;RESULT IN C(AC) NON-ZERO + 6853 033662 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 6854 STOP^ + 6855 033663 254 04 0 00 033664 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6856 033664 324 00 0 00 033665 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6857 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6858 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6859 + 6860 ;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], + 6861 ;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], + 6862 ;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], + 6863 ;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + 6864 + 6865 ;********** + 6866 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0148 + + 6867 007500 SN=7500 + 6868 000000 ZZ=0 + 6869 + 6870 A7500: REPEAT ^D36, + 6871 <;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 6872 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 6873 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 6874 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 6875 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 6876 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 6877 ;INDICATE AN ILLEGAL CARRY GENERATION. + 6878 SN=SN+1 + 6879 ZZ=ZZ+ZZ + 6880 IFE ZZ, + 6881 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 6882 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 6883 SKIPE ;PASS TEST IF C(AC)=0 + 6884 STOP + 6885 + 6886 ;********** + 6887 > + 6888 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 6889 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 6890 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 6891 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 6892 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 6893 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 6894 ;INDICATE AN ILLEGAL CARRY GENERATION. + 6895 007501 SN=SN+1 + 6896 000000 ZZ=ZZ+ZZ + 6897 000001 IFE ZZ, + 6898 033665 200 00 0 00 035074 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 6899 033666 430 00 0 00 035074 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 6900 033667 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 6901 STOP^ + 6902 033670 254 04 0 00 033671 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6903 033671 324 00 0 00 033672 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6904 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6905 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6906 + 6907 ;********** + 6908 + 6909 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 6910 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 6911 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 6912 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 6913 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 6914 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 6915 ;INDICATE AN ILLEGAL CARRY GENERATION. + 6916 007502 SN=SN+1 + 6917 000002 ZZ=ZZ+ZZ + 6918 IFE ZZ, + 6919 033672 200 00 0 00 035075 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 6920 033673 430 00 0 00 035075 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 6921 033674 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0149 + + 6922 STOP^ + 6923 033675 254 04 0 00 033676 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6924 033676 324 00 0 00 033677 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6925 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6926 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6927 + 6928 ;********** + 6929 + 6930 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 6931 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 6932 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 6933 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 6934 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 6935 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 6936 ;INDICATE AN ILLEGAL CARRY GENERATION. + 6937 007503 SN=SN+1 + 6938 000004 ZZ=ZZ+ZZ + 6939 IFE ZZ, + 6940 033677 200 00 0 00 035076 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 6941 033700 430 00 0 00 035076 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 6942 033701 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 6943 STOP^ + 6944 033702 254 04 0 00 033703 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6945 033703 324 00 0 00 033704 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6946 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6947 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6948 + 6949 ;********** + 6950 + 6951 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 6952 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 6953 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 6954 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 6955 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 6956 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 6957 ;INDICATE AN ILLEGAL CARRY GENERATION. + 6958 007504 SN=SN+1 + 6959 000010 ZZ=ZZ+ZZ + 6960 IFE ZZ, + 6961 033704 200 00 0 00 035077 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 6962 033705 430 00 0 00 035077 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 6963 033706 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 6964 STOP^ + 6965 033707 254 04 0 00 033710 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6966 033710 324 00 0 00 033711 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6967 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6968 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6969 + 6970 ;********** + 6971 + 6972 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 6973 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 6974 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 6975 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 6976 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-2 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0150 + + 6977 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 6978 ;INDICATE AN ILLEGAL CARRY GENERATION. + 6979 007505 SN=SN+1 + 6980 000020 ZZ=ZZ+ZZ + 6981 IFE ZZ, + 6982 033711 200 00 0 00 035100 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 6983 033712 430 00 0 00 035100 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 6984 033713 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 6985 STOP^ + 6986 033714 254 04 0 00 033715 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 6987 033715 324 00 0 00 033716 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 6988 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 6989 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 6990 + 6991 ;********** + 6992 + 6993 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 6994 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 6995 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 6996 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 6997 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 6998 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 6999 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7000 007506 SN=SN+1 + 7001 000040 ZZ=ZZ+ZZ + 7002 IFE ZZ, + 7003 033716 200 00 0 00 035101 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7004 033717 430 00 0 00 035101 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7005 033720 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7006 STOP^ + 7007 033721 254 04 0 00 033722 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7008 033722 324 00 0 00 033723 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7009 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7010 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7011 + 7012 ;********** + 7013 + 7014 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7015 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7016 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7017 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7018 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7019 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7020 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7021 007507 SN=SN+1 + 7022 000100 ZZ=ZZ+ZZ + 7023 IFE ZZ, + 7024 033723 200 00 0 00 035102 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7025 033724 430 00 0 00 035102 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7026 033725 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7027 STOP^ + 7028 033726 254 04 0 00 033727 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7029 033727 324 00 0 00 033730 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7030 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7031 ;IN THE SUBTEST) TO LOOP ON ERROR^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-3 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0151 + + 7032 + 7033 ;********** + 7034 + 7035 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7036 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7037 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7038 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7039 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7040 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7041 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7042 007510 SN=SN+1 + 7043 000200 ZZ=ZZ+ZZ + 7044 IFE ZZ, + 7045 033730 200 00 0 00 035103 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7046 033731 430 00 0 00 035103 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7047 033732 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7048 STOP^ + 7049 033733 254 04 0 00 033734 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7050 033734 324 00 0 00 033735 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7051 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7052 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7053 + 7054 ;********** + 7055 + 7056 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7057 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7058 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7059 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7060 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7061 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7062 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7063 007511 SN=SN+1 + 7064 000400 ZZ=ZZ+ZZ + 7065 IFE ZZ, + 7066 033735 200 00 0 00 035104 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7067 033736 430 00 0 00 035104 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7068 033737 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7069 STOP^ + 7070 033740 254 04 0 00 033741 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7071 033741 324 00 0 00 033742 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7072 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7073 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7074 + 7075 ;********** + 7076 + 7077 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7078 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7079 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7080 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7081 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7082 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7083 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7084 007512 SN=SN+1 + 7085 001000 ZZ=ZZ+ZZ + 7086 IFE ZZ, +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-4 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0152 + + 7087 033742 200 00 0 00 035105 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7088 033743 430 00 0 00 035105 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7089 033744 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7090 STOP^ + 7091 033745 254 04 0 00 033746 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7092 033746 324 00 0 00 033747 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7093 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7094 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7095 + 7096 ;********** + 7097 + 7098 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7099 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7100 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7101 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7102 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7103 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7104 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7105 007513 SN=SN+1 + 7106 002000 ZZ=ZZ+ZZ + 7107 IFE ZZ, + 7108 033747 200 00 0 00 035106 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7109 033750 430 00 0 00 035106 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7110 033751 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7111 STOP^ + 7112 033752 254 04 0 00 033753 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7113 033753 324 00 0 00 033754 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7114 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7115 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7116 + 7117 ;********** + 7118 + 7119 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7120 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7121 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7122 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7123 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7124 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7125 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7126 007514 SN=SN+1 + 7127 004000 ZZ=ZZ+ZZ + 7128 IFE ZZ, + 7129 033754 200 00 0 00 035107 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7130 033755 430 00 0 00 035107 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7131 033756 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7132 STOP^ + 7133 033757 254 04 0 00 033760 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7134 033760 324 00 0 00 033761 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7135 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7136 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7137 + 7138 ;********** + 7139 + 7140 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7141 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-5 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0153 + + 7142 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7143 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7144 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7145 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7146 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7147 007515 SN=SN+1 + 7148 010000 ZZ=ZZ+ZZ + 7149 IFE ZZ, + 7150 033761 200 00 0 00 035110 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7151 033762 430 00 0 00 035110 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7152 033763 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7153 STOP^ + 7154 033764 254 04 0 00 033765 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7155 033765 324 00 0 00 033766 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7156 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7157 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7158 + 7159 ;********** + 7160 + 7161 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7162 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7163 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7164 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7165 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7166 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7167 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7168 007516 SN=SN+1 + 7169 020000 ZZ=ZZ+ZZ + 7170 IFE ZZ, + 7171 033766 200 00 0 00 035111 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7172 033767 430 00 0 00 035111 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7173 033770 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7174 STOP^ + 7175 033771 254 04 0 00 033772 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7176 033772 324 00 0 00 033773 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7177 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7178 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7179 + 7180 ;********** + 7181 + 7182 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7183 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7184 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7185 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7186 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7187 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7188 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7189 007517 SN=SN+1 + 7190 040000 ZZ=ZZ+ZZ + 7191 IFE ZZ, + 7192 033773 200 00 0 00 035112 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7193 033774 430 00 0 00 035112 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7194 033775 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7195 STOP^ + 7196 033776 254 04 0 00 033777 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-6 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0154 + + 7197 033777 324 00 0 00 034000 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7198 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7199 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7200 + 7201 ;********** + 7202 + 7203 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7204 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7205 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7206 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7207 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7208 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7209 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7210 007520 SN=SN+1 + 7211 100000 ZZ=ZZ+ZZ + 7212 IFE ZZ, + 7213 034000 200 00 0 00 035113 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7214 034001 430 00 0 00 035113 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7215 034002 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7216 STOP^ + 7217 034003 254 04 0 00 034004 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7218 034004 324 00 0 00 034005 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7219 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7220 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7221 + 7222 ;********** + 7223 + 7224 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7225 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7226 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7227 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7228 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7229 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7230 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7231 007521 SN=SN+1 + 7232 200000 ZZ=ZZ+ZZ + 7233 IFE ZZ, + 7234 034005 200 00 0 00 035114 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7235 034006 430 00 0 00 035114 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7236 034007 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7237 STOP^ + 7238 034010 254 04 0 00 034011 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7239 034011 324 00 0 00 034012 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7240 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7241 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7242 + 7243 ;********** + 7244 + 7245 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7246 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7247 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7248 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7249 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7250 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7251 ;INDICATE AN ILLEGAL CARRY GENERATION. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-7 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0155 + + 7252 007522 SN=SN+1 + 7253 400000 ZZ=ZZ+ZZ + 7254 IFE ZZ, + 7255 034012 200 00 0 00 035115 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7256 034013 430 00 0 00 035115 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7257 034014 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7258 STOP^ + 7259 034015 254 04 0 00 034016 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7260 034016 324 00 0 00 034017 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7261 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7262 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7263 + 7264 ;********** + 7265 + 7266 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7267 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7268 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7269 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7270 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7271 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7272 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7273 007523 SN=SN+1 + 7274 000001 000000 ZZ=ZZ+ZZ + 7275 IFE ZZ, + 7276 034017 200 00 0 00 035116 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7277 034020 430 00 0 00 035116 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7278 034021 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7279 STOP^ + 7280 034022 254 04 0 00 034023 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7281 034023 324 00 0 00 034024 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7282 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7283 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7284 + 7285 ;********** + 7286 + 7287 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7288 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7289 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7290 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7291 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7292 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7293 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7294 007524 SN=SN+1 + 7295 000002 000000 ZZ=ZZ+ZZ + 7296 IFE ZZ, + 7297 034024 200 00 0 00 035117 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7298 034025 430 00 0 00 035117 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7299 034026 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7300 STOP^ + 7301 034027 254 04 0 00 034030 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7302 034030 324 00 0 00 034031 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7303 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7304 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7305 + 7306 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-8 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0156 + + 7307 + 7308 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7309 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7310 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7311 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7312 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7313 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7314 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7315 007525 SN=SN+1 + 7316 000004 000000 ZZ=ZZ+ZZ + 7317 IFE ZZ, + 7318 034031 200 00 0 00 035120 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7319 034032 430 00 0 00 035120 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7320 034033 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7321 STOP^ + 7322 034034 254 04 0 00 034035 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7323 034035 324 00 0 00 034036 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7324 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7325 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7326 + 7327 ;********** + 7328 + 7329 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7330 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7331 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7332 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7333 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7334 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7335 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7336 007526 SN=SN+1 + 7337 000010 000000 ZZ=ZZ+ZZ + 7338 IFE ZZ, + 7339 034036 200 00 0 00 035121 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7340 034037 430 00 0 00 035121 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7341 034040 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7342 STOP^ + 7343 034041 254 04 0 00 034042 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7344 034042 324 00 0 00 034043 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7345 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7346 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7347 + 7348 ;********** + 7349 + 7350 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7351 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7352 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7353 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7354 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7355 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7356 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7357 007527 SN=SN+1 + 7358 000020 000000 ZZ=ZZ+ZZ + 7359 IFE ZZ, + 7360 034043 200 00 0 00 035122 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7361 034044 430 00 0 00 035122 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-9 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0157 + + 7362 034045 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7363 STOP^ + 7364 034046 254 04 0 00 034047 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7365 034047 324 00 0 00 034050 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7366 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7367 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7368 + 7369 ;********** + 7370 + 7371 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7372 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7373 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7374 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7375 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7376 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7377 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7378 007530 SN=SN+1 + 7379 000040 000000 ZZ=ZZ+ZZ + 7380 IFE ZZ, + 7381 034050 200 00 0 00 035123 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7382 034051 430 00 0 00 035123 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7383 034052 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7384 STOP^ + 7385 034053 254 04 0 00 034054 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7386 034054 324 00 0 00 034055 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7387 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7388 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7389 + 7390 ;********** + 7391 + 7392 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7393 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7394 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7395 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7396 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7397 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7398 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7399 007531 SN=SN+1 + 7400 000100 000000 ZZ=ZZ+ZZ + 7401 IFE ZZ, + 7402 034055 200 00 0 00 035124 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7403 034056 430 00 0 00 035124 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7404 034057 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7405 STOP^ + 7406 034060 254 04 0 00 034061 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7407 034061 324 00 0 00 034062 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7408 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7409 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7410 + 7411 ;********** + 7412 + 7413 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7414 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7415 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7416 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-10 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0158 + + 7417 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7418 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7419 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7420 007532 SN=SN+1 + 7421 000200 000000 ZZ=ZZ+ZZ + 7422 IFE ZZ, + 7423 034062 200 00 0 00 035125 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7424 034063 430 00 0 00 035125 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7425 034064 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7426 STOP^ + 7427 034065 254 04 0 00 034066 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7428 034066 324 00 0 00 034067 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7429 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7430 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7431 + 7432 ;********** + 7433 + 7434 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7435 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7436 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7437 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7438 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7439 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7440 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7441 007533 SN=SN+1 + 7442 000400 000000 ZZ=ZZ+ZZ + 7443 IFE ZZ, + 7444 034067 200 00 0 00 035073 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7445 034070 430 00 0 00 035073 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7446 034071 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7447 STOP^ + 7448 034072 254 04 0 00 034073 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7449 034073 324 00 0 00 034074 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7450 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7451 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7452 + 7453 ;********** + 7454 + 7455 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7456 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7457 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7458 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7459 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7460 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7461 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7462 007534 SN=SN+1 + 7463 001000 000000 ZZ=ZZ+ZZ + 7464 IFE ZZ, + 7465 034074 200 00 0 00 035126 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7466 034075 430 00 0 00 035126 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7467 034076 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7468 STOP^ + 7469 034077 254 04 0 00 034100 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7470 034100 324 00 0 00 034101 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7471 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-11 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0159 + + 7472 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7473 + 7474 ;********** + 7475 + 7476 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7477 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7478 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7479 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7480 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7481 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7482 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7483 007535 SN=SN+1 + 7484 002000 000000 ZZ=ZZ+ZZ + 7485 IFE ZZ, + 7486 034101 200 00 0 00 035127 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7487 034102 430 00 0 00 035127 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7488 034103 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7489 STOP^ + 7490 034104 254 04 0 00 034105 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7491 034105 324 00 0 00 034106 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7492 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7493 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7494 + 7495 ;********** + 7496 + 7497 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7498 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7499 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7500 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7501 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7502 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7503 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7504 007536 SN=SN+1 + 7505 004000 000000 ZZ=ZZ+ZZ + 7506 IFE ZZ, + 7507 034106 200 00 0 00 035130 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7508 034107 430 00 0 00 035130 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7509 034110 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7510 STOP^ + 7511 034111 254 04 0 00 034112 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7512 034112 324 00 0 00 034113 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7513 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7514 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7515 + 7516 ;********** + 7517 + 7518 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7519 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7520 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7521 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7522 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7523 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7524 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7525 007537 SN=SN+1 + 7526 010000 000000 ZZ=ZZ+ZZ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-12 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0160 + + 7527 IFE ZZ, + 7528 034113 200 00 0 00 035131 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7529 034114 430 00 0 00 035131 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7530 034115 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7531 STOP^ + 7532 034116 254 04 0 00 034117 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7533 034117 324 00 0 00 034120 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7534 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7535 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7536 + 7537 ;********** + 7538 + 7539 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7540 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7541 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7542 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7543 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7544 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7545 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7546 007540 SN=SN+1 + 7547 020000 000000 ZZ=ZZ+ZZ + 7548 IFE ZZ, + 7549 034120 200 00 0 00 035132 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7550 034121 430 00 0 00 035132 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7551 034122 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7552 STOP^ + 7553 034123 254 04 0 00 034124 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7554 034124 324 00 0 00 034125 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7555 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7556 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7557 + 7558 ;********** + 7559 + 7560 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7561 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7562 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7563 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7564 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7565 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7566 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7567 007541 SN=SN+1 + 7568 040000 000000 ZZ=ZZ+ZZ + 7569 IFE ZZ, + 7570 034125 200 00 0 00 035133 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7571 034126 430 00 0 00 035133 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7572 034127 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7573 STOP^ + 7574 034130 254 04 0 00 034131 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7575 034131 324 00 0 00 034132 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7576 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7577 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7578 + 7579 ;********** + 7580 + 7581 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-13 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0161 + + 7582 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7583 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7584 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7585 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7586 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7587 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7588 007542 SN=SN+1 + 7589 100000 000000 ZZ=ZZ+ZZ + 7590 IFE ZZ, + 7591 034132 200 00 0 00 035134 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7592 034133 430 00 0 00 035134 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7593 034134 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7594 STOP^ + 7595 034135 254 04 0 00 034136 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7596 034136 324 00 0 00 034137 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7597 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7598 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7599 + 7600 ;********** + 7601 + 7602 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7603 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7604 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7605 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7606 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7607 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7608 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7609 007543 SN=SN+1 + 7610 200000 000000 ZZ=ZZ+ZZ + 7611 IFE ZZ, + 7612 034137 200 00 0 00 035135 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7613 034140 430 00 0 00 035135 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7614 034141 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7615 STOP^ + 7616 034142 254 04 0 00 034143 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7617 034143 324 00 0 00 034144 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7618 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7619 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7620 + 7621 ;********** + 7622 + 7623 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION + 7624 ;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, + 7625 ;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT + 7626 ;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME + 7627 ;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST + 7628 ;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY + 7629 ;INDICATE AN ILLEGAL CARRY GENERATION. + 7630 007544 SN=SN+1 + 7631 400000 000000 ZZ=ZZ+ZZ + 7632 IFE ZZ, + 7633 034144 200 00 0 00 035065 MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + 7634 034145 430 00 0 00 035065 XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + 7635 034146 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7636 STOP^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 28-14 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0162 + + 7637 034147 254 04 0 00 034150 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7638 034150 324 00 0 00 034151 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7639 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7640 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7641 + 7642 ;********** + 7643 + 7644 ;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION WHERE + 7645 ;C(AC)=C(E)=0 IS AN AC CONTAINING ALL ZEROS; FIRST THE AC IS CLEARED; THEN + 7646 ;XOR IS EXECUTED WITH C(E)=0. THE RESULT IN THE AC IS CHECKED FOR ALL ZEROS + 7647 + 7648 034151 200 00 0 00 035062 A7600: MOVE [0] ;INITIALIZE AC TO ALL ZEROS + 7649 034152 430 00 0 00 035062 XOR [0] ;*XOR WITH C(AC)=C(E)=0 SHOULD RESULT INC(AC)=0 + 7650 034153 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7651 STOP^ + 7652 034154 254 04 0 00 034155 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7653 034155 324 00 0 00 034156 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7654 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7655 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7656 + 7657 ;********** + 7658 + 7659 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7660 ;C(AC)=0 AND C(E) ALL ONES IS AN AC CONTAINING ALL ZEROS SINCE + 7661 ;THE EQUIVALENCE FUNCTION SHOULD SET ONLY LIKE BITS. FIRST, THE AC + 7662 ;SHOULD BE CLEARED; THEN, EQV IS EXECUTED WITH E CONTAINING ALL ONES. + 7663 ;THE AC IS THEN CHECKED FOR ALL ZEROS + 7664 + 7665 034156 200 00 0 00 035062 A7700: MOVE [0] ;CLEAR THE AC + 7666 034157 444 00 0 00 035063 EQV [-1] ;*EQV WITH C(AC)=0 AND C(E) ALL ONES SHOULD + 7667 ;RESULT IN C(AC)=0 + 7668 034160 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 7669 STOP^ + 7670 034161 254 04 0 00 034162 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7671 034162 324 00 0 00 034163 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7672 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7673 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7674 + 7675 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0163 + + 7676 010000 SN=10000 + 7677 000000 ZZ=0 + 7678 + 7679 A10000: REPEAT ^D18, + 7680 <;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7681 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7682 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7683 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7684 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7685 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7686 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7687 SN=SN+1 + 7688 ZZ=ZZ+ZZ+1 + 7689 IFE , + 7690 MOVE [0] ;PRESET AC TO ALL ZEROS + 7691 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7692 ;RESULT IN C(AC) NON-ZERO + 7693 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7694 STOP + 7695 + 7696 ;********** + 7697 > + 7698 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7699 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7700 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7701 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7702 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7703 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7704 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7705 010001 SN=SN+1 + 7706 000001 ZZ=ZZ+ZZ+1 + 7707 777777 777776 IFE , + 7708 034163 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7709 034164 444 00 0 00 035142 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7710 ;RESULT IN C(AC) NON-ZERO + 7711 034165 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7712 STOP^ + 7713 034166 254 04 0 00 034167 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7714 034167 324 00 0 00 034170 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7715 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7716 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7717 + 7718 ;********** + 7719 + 7720 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7721 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7722 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7723 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7724 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7725 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7726 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7727 010002 SN=SN+1 + 7728 777777 777775 ZZ=ZZ+ZZ+1 + 7729 IFE , + 7730 034170 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0164 + + 7731 034171 444 00 0 00 035143 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7732 ;RESULT IN C(AC) NON-ZERO + 7733 034172 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7734 STOP^ + 7735 034173 254 04 0 00 034174 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7736 034174 324 00 0 00 034175 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7737 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7738 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7739 + 7740 ;********** + 7741 + 7742 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7743 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7744 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7745 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7746 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7747 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7748 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7749 010003 SN=SN+1 + 7750 777777 777773 ZZ=ZZ+ZZ+1 + 7751 IFE , + 7752 034175 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7753 034176 444 00 0 00 035144 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7754 ;RESULT IN C(AC) NON-ZERO + 7755 034177 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7756 STOP^ + 7757 034200 254 04 0 00 034201 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7758 034201 324 00 0 00 034202 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7759 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7760 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7761 + 7762 ;********** + 7763 + 7764 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7765 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7766 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7767 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7768 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7769 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7770 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7771 010004 SN=SN+1 + 7772 777777 777767 ZZ=ZZ+ZZ+1 + 7773 IFE , + 7774 034202 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7775 034203 444 00 0 00 035145 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7776 ;RESULT IN C(AC) NON-ZERO + 7777 034204 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7778 STOP^ + 7779 034205 254 04 0 00 034206 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7780 034206 324 00 0 00 034207 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7781 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7782 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7783 + 7784 ;********** + 7785 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-2 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0165 + + 7786 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7787 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7788 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7789 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7790 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7791 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7792 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7793 010005 SN=SN+1 + 7794 777777 777757 ZZ=ZZ+ZZ+1 + 7795 IFE , + 7796 034207 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7797 034210 444 00 0 00 035146 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7798 ;RESULT IN C(AC) NON-ZERO + 7799 034211 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7800 STOP^ + 7801 034212 254 04 0 00 034213 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7802 034213 324 00 0 00 034214 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7803 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7804 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7805 + 7806 ;********** + 7807 + 7808 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7809 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7810 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7811 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7812 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7813 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7814 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7815 010006 SN=SN+1 + 7816 777777 777737 ZZ=ZZ+ZZ+1 + 7817 IFE , + 7818 034214 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7819 034215 444 00 0 00 035147 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7820 ;RESULT IN C(AC) NON-ZERO + 7821 034216 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7822 STOP^ + 7823 034217 254 04 0 00 034220 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7824 034220 324 00 0 00 034221 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7825 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7826 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7827 + 7828 ;********** + 7829 + 7830 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7831 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7832 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7833 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7834 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7835 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7836 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7837 010007 SN=SN+1 + 7838 777777 777677 ZZ=ZZ+ZZ+1 + 7839 IFE , + 7840 034221 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-3 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0166 + + 7841 034222 444 00 0 00 035150 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7842 ;RESULT IN C(AC) NON-ZERO + 7843 034223 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7844 STOP^ + 7845 034224 254 04 0 00 034225 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7846 034225 324 00 0 00 034226 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7847 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7848 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7849 + 7850 ;********** + 7851 + 7852 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7853 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7854 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7855 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7856 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7857 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7858 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7859 010010 SN=SN+1 + 7860 777777 777577 ZZ=ZZ+ZZ+1 + 7861 IFE , + 7862 034226 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7863 034227 444 00 0 00 035151 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7864 ;RESULT IN C(AC) NON-ZERO + 7865 034230 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7866 STOP^ + 7867 034231 254 04 0 00 034232 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7868 034232 324 00 0 00 034233 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7869 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7870 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7871 + 7872 ;********** + 7873 + 7874 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7875 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7876 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7877 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7878 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7879 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7880 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7881 010011 SN=SN+1 + 7882 777777 777377 ZZ=ZZ+ZZ+1 + 7883 IFE , + 7884 034233 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7885 034234 444 00 0 00 035152 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7886 ;RESULT IN C(AC) NON-ZERO + 7887 034235 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7888 STOP^ + 7889 034236 254 04 0 00 034237 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7890 034237 324 00 0 00 034240 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7891 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7892 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7893 + 7894 ;********** + 7895 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-4 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0167 + + 7896 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7897 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7898 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7899 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7900 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7901 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7902 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7903 010012 SN=SN+1 + 7904 777777 776777 ZZ=ZZ+ZZ+1 + 7905 IFE , + 7906 034240 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7907 034241 444 00 0 00 035153 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7908 ;RESULT IN C(AC) NON-ZERO + 7909 034242 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7910 STOP^ + 7911 034243 254 04 0 00 034244 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7912 034244 324 00 0 00 034245 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7913 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7914 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7915 + 7916 ;********** + 7917 + 7918 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7919 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7920 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7921 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7922 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7923 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7924 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7925 010013 SN=SN+1 + 7926 777777 775777 ZZ=ZZ+ZZ+1 + 7927 IFE , + 7928 034245 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7929 034246 444 00 0 00 035154 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7930 ;RESULT IN C(AC) NON-ZERO + 7931 034247 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7932 STOP^ + 7933 034250 254 04 0 00 034251 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7934 034251 324 00 0 00 034252 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7935 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7936 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7937 + 7938 ;********** + 7939 + 7940 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7941 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7942 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7943 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7944 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7945 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7946 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7947 010014 SN=SN+1 + 7948 777777 773777 ZZ=ZZ+ZZ+1 + 7949 IFE , + 7950 034252 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-5 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0168 + + 7951 034253 444 00 0 00 035155 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7952 ;RESULT IN C(AC) NON-ZERO + 7953 034254 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7954 STOP^ + 7955 034255 254 04 0 00 034256 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7956 034256 324 00 0 00 034257 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7957 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7958 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7959 + 7960 ;********** + 7961 + 7962 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7963 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7964 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7965 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7966 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7967 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7968 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7969 010015 SN=SN+1 + 7970 777777 767777 ZZ=ZZ+ZZ+1 + 7971 IFE , + 7972 034257 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7973 034260 444 00 0 00 035156 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7974 ;RESULT IN C(AC) NON-ZERO + 7975 034261 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7976 STOP^ + 7977 034262 254 04 0 00 034263 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 7978 034263 324 00 0 00 034264 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 7979 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 7980 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 7981 + 7982 ;********** + 7983 + 7984 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 7985 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 7986 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 7987 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 7988 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 7989 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 7990 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 7991 010016 SN=SN+1 + 7992 777777 757777 ZZ=ZZ+ZZ+1 + 7993 IFE , + 7994 034264 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 7995 034265 444 00 0 00 035157 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 7996 ;RESULT IN C(AC) NON-ZERO + 7997 034266 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 7998 STOP^ + 7999 034267 254 04 0 00 034270 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8000 034270 324 00 0 00 034271 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8001 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8002 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8003 + 8004 ;********** + 8005 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-6 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0169 + + 8006 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8007 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8008 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8009 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8010 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8011 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8012 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8013 010017 SN=SN+1 + 8014 777777 737777 ZZ=ZZ+ZZ+1 + 8015 IFE , + 8016 034271 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8017 034272 444 00 0 00 035160 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8018 ;RESULT IN C(AC) NON-ZERO + 8019 034273 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8020 STOP^ + 8021 034274 254 04 0 00 034275 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8022 034275 324 00 0 00 034276 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8023 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8024 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8025 + 8026 ;********** + 8027 + 8028 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8029 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8030 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8031 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8032 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8033 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8034 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8035 010020 SN=SN+1 + 8036 777777 677777 ZZ=ZZ+ZZ+1 + 8037 IFE , + 8038 034276 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8039 034277 444 00 0 00 035161 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8040 ;RESULT IN C(AC) NON-ZERO + 8041 034300 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8042 STOP^ + 8043 034301 254 04 0 00 034302 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8044 034302 324 00 0 00 034303 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8045 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8046 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8047 + 8048 ;********** + 8049 + 8050 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8051 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8052 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8053 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8054 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8055 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8056 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8057 010021 SN=SN+1 + 8058 777777 577777 ZZ=ZZ+ZZ+1 + 8059 IFE , + 8060 034303 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-7 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0170 + + 8061 034304 444 00 0 00 035162 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8062 ;RESULT IN C(AC) NON-ZERO + 8063 034305 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8064 STOP^ + 8065 034306 254 04 0 00 034307 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8066 034307 324 00 0 00 034310 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8067 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8068 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8069 + 8070 ;********** + 8071 + 8072 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8073 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8074 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8075 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8076 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8077 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8078 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8079 010022 SN=SN+1 + 8080 777777 377777 ZZ=ZZ+ZZ+1 + 8081 IFE , + 8082 034310 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8083 034311 444 00 0 00 035163 EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8084 ;RESULT IN C(AC) NON-ZERO + 8085 034312 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8086 STOP^ + 8087 034313 254 04 0 00 034314 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8088 034314 324 00 0 00 034315 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8089 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8090 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8091 + 8092 ;********** + 8093 + 8094 010100 SN=10100 + 8095 000000 ZZ=0 + 8096 + 8097 A10100: REPEAT ^D18, + 8098 <;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8099 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8100 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8101 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8102 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8103 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8104 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8105 SN=SN+1 + 8106 ZZ=ZZ+ZZ+1 + 8107 IFE , + 8108 MOVE [0] ;PRESET AC TO ALL ZEROS + 8109 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8110 ;RESULT IN C(AC) NON-ZERO + 8111 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8112 STOP + 8113 + 8114 ;********** + 8115 > +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-8 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0171 + + 8116 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8117 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8118 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8119 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8120 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8121 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8122 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8123 010101 SN=SN+1 + 8124 000001 ZZ=ZZ+ZZ+1 + 8125 777777 777776 IFE , + 8126 034315 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8127 034316 444 00 0 00 035164 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8128 ;RESULT IN C(AC) NON-ZERO + 8129 034317 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8130 STOP^ + 8131 034320 254 04 0 00 034321 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8132 034321 324 00 0 00 034322 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8133 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8134 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8135 + 8136 ;********** + 8137 + 8138 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8139 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8140 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8141 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8142 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8143 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8144 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8145 010102 SN=SN+1 + 8146 777777 777775 ZZ=ZZ+ZZ+1 + 8147 IFE , + 8148 034322 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8149 034323 444 00 0 00 035165 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8150 ;RESULT IN C(AC) NON-ZERO + 8151 034324 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8152 STOP^ + 8153 034325 254 04 0 00 034326 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8154 034326 324 00 0 00 034327 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8155 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8156 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8157 + 8158 ;********** + 8159 + 8160 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8161 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8162 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8163 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8164 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8165 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8166 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8167 010103 SN=SN+1 + 8168 777777 777773 ZZ=ZZ+ZZ+1 + 8169 IFE , + 8170 034327 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-9 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0172 + + 8171 034330 444 00 0 00 035166 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8172 ;RESULT IN C(AC) NON-ZERO + 8173 034331 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8174 STOP^ + 8175 034332 254 04 0 00 034333 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8176 034333 324 00 0 00 034334 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8177 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8178 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8179 + 8180 ;********** + 8181 + 8182 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8183 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8184 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8185 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8186 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8187 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8188 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8189 010104 SN=SN+1 + 8190 777777 777767 ZZ=ZZ+ZZ+1 + 8191 IFE , + 8192 034334 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8193 034335 444 00 0 00 035167 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8194 ;RESULT IN C(AC) NON-ZERO + 8195 034336 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8196 STOP^ + 8197 034337 254 04 0 00 034340 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8198 034340 324 00 0 00 034341 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8199 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8200 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8201 + 8202 ;********** + 8203 + 8204 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8205 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8206 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8207 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8208 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8209 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8210 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8211 010105 SN=SN+1 + 8212 777777 777757 ZZ=ZZ+ZZ+1 + 8213 IFE , + 8214 034341 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8215 034342 444 00 0 00 035170 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8216 ;RESULT IN C(AC) NON-ZERO + 8217 034343 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8218 STOP^ + 8219 034344 254 04 0 00 034345 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8220 034345 324 00 0 00 034346 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8221 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8222 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8223 + 8224 ;********** + 8225 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-10 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0173 + + 8226 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8227 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8228 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8229 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8230 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8231 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8232 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8233 010106 SN=SN+1 + 8234 777777 777737 ZZ=ZZ+ZZ+1 + 8235 IFE , + 8236 034346 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8237 034347 444 00 0 00 035171 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8238 ;RESULT IN C(AC) NON-ZERO + 8239 034350 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8240 STOP^ + 8241 034351 254 04 0 00 034352 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8242 034352 324 00 0 00 034353 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8243 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8244 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8245 + 8246 ;********** + 8247 + 8248 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8249 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8250 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8251 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8252 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8253 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8254 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8255 010107 SN=SN+1 + 8256 777777 777677 ZZ=ZZ+ZZ+1 + 8257 IFE , + 8258 034353 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8259 034354 444 00 0 00 035172 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8260 ;RESULT IN C(AC) NON-ZERO + 8261 034355 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8262 STOP^ + 8263 034356 254 04 0 00 034357 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8264 034357 324 00 0 00 034360 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8265 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8266 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8267 + 8268 ;********** + 8269 + 8270 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8271 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8272 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8273 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8274 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8275 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8276 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8277 010110 SN=SN+1 + 8278 777777 777577 ZZ=ZZ+ZZ+1 + 8279 IFE , + 8280 034360 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-11 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0174 + + 8281 034361 444 00 0 00 035173 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8282 ;RESULT IN C(AC) NON-ZERO + 8283 034362 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8284 STOP^ + 8285 034363 254 04 0 00 034364 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8286 034364 324 00 0 00 034365 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8287 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8288 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8289 + 8290 ;********** + 8291 + 8292 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8293 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8294 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8295 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8296 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8297 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8298 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8299 010111 SN=SN+1 + 8300 777777 777377 ZZ=ZZ+ZZ+1 + 8301 IFE , + 8302 034365 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8303 034366 444 00 0 00 035174 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8304 ;RESULT IN C(AC) NON-ZERO + 8305 034367 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8306 STOP^ + 8307 034370 254 04 0 00 034371 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8308 034371 324 00 0 00 034372 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8309 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8310 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8311 + 8312 ;********** + 8313 + 8314 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8315 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8316 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8317 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8318 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8319 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8320 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8321 010112 SN=SN+1 + 8322 777777 776777 ZZ=ZZ+ZZ+1 + 8323 IFE , + 8324 034372 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8325 034373 444 00 0 00 035175 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8326 ;RESULT IN C(AC) NON-ZERO + 8327 034374 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8328 STOP^ + 8329 034375 254 04 0 00 034376 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8330 034376 324 00 0 00 034377 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8331 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8332 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8333 + 8334 ;********** + 8335 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-12 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0175 + + 8336 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8337 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8338 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8339 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8340 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8341 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8342 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8343 010113 SN=SN+1 + 8344 777777 775777 ZZ=ZZ+ZZ+1 + 8345 IFE , + 8346 034377 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8347 034400 444 00 0 00 035176 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8348 ;RESULT IN C(AC) NON-ZERO + 8349 034401 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8350 STOP^ + 8351 034402 254 04 0 00 034403 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8352 034403 324 00 0 00 034404 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8353 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8354 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8355 + 8356 ;********** + 8357 + 8358 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8359 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8360 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8361 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8362 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8363 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8364 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8365 010114 SN=SN+1 + 8366 777777 773777 ZZ=ZZ+ZZ+1 + 8367 IFE , + 8368 034404 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8369 034405 444 00 0 00 035177 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8370 ;RESULT IN C(AC) NON-ZERO + 8371 034406 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8372 STOP^ + 8373 034407 254 04 0 00 034410 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8374 034410 324 00 0 00 034411 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8375 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8376 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8377 + 8378 ;********** + 8379 + 8380 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8381 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8382 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8383 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8384 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8385 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8386 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8387 010115 SN=SN+1 + 8388 777777 767777 ZZ=ZZ+ZZ+1 + 8389 IFE , + 8390 034411 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-13 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0176 + + 8391 034412 444 00 0 00 035200 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8392 ;RESULT IN C(AC) NON-ZERO + 8393 034413 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8394 STOP^ + 8395 034414 254 04 0 00 034415 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8396 034415 324 00 0 00 034416 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8397 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8398 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8399 + 8400 ;********** + 8401 + 8402 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8403 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8404 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8405 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8406 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8407 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8408 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8409 010116 SN=SN+1 + 8410 777777 757777 ZZ=ZZ+ZZ+1 + 8411 IFE , + 8412 034416 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8413 034417 444 00 0 00 035201 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8414 ;RESULT IN C(AC) NON-ZERO + 8415 034420 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8416 STOP^ + 8417 034421 254 04 0 00 034422 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8418 034422 324 00 0 00 034423 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8419 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8420 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8421 + 8422 ;********** + 8423 + 8424 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8425 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8426 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8427 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8428 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8429 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8430 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8431 010117 SN=SN+1 + 8432 777777 737777 ZZ=ZZ+ZZ+1 + 8433 IFE , + 8434 034423 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8435 034424 444 00 0 00 035202 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8436 ;RESULT IN C(AC) NON-ZERO + 8437 034425 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8438 STOP^ + 8439 034426 254 04 0 00 034427 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8440 034427 324 00 0 00 034430 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8441 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8442 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8443 + 8444 ;********** + 8445 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-14 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0177 + + 8446 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8447 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8448 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8449 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8450 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8451 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8452 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8453 010120 SN=SN+1 + 8454 777777 677777 ZZ=ZZ+ZZ+1 + 8455 IFE , + 8456 034430 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8457 034431 444 00 0 00 035203 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8458 ;RESULT IN C(AC) NON-ZERO + 8459 034432 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8460 STOP^ + 8461 034433 254 04 0 00 034434 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8462 034434 324 00 0 00 034435 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8463 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8464 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8465 + 8466 ;********** + 8467 + 8468 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8469 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8470 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8471 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8472 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8473 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8474 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8475 010121 SN=SN+1 + 8476 777777 577777 ZZ=ZZ+ZZ+1 + 8477 IFE , + 8478 034435 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS + 8479 034436 444 00 0 00 035204 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8480 ;RESULT IN C(AC) NON-ZERO + 8481 034437 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8482 STOP^ + 8483 034440 254 04 0 00 034441 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8484 034441 324 00 0 00 034442 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8485 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8486 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8487 + 8488 ;********** + 8489 + 8490 ;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE + 8491 ;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN + 8492 ;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE + 8493 ;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT + 8494 ;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS + 8495 ;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE + 8496 ;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO + 8497 010122 SN=SN+1 + 8498 777777 377777 ZZ=ZZ+ZZ+1 + 8499 IFE , + 8500 034442 200 00 0 00 035062 MOVE [0] ;PRESET AC TO ALL ZEROS +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 29-15 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0178 + + 8501 034443 444 00 0 00 035064 EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + 8502 ;RESULT IN C(AC) NON-ZERO + 8503 034444 336 00 0 00 000000 SKIPN ;PASS TEST IF C(AC) NON-ZERO + 8504 STOP^ + 8505 034445 254 04 0 00 034446 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8506 034446 324 00 0 00 034447 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8507 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8508 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8509 + 8510 ;********** + 8511 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 30 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0179 + + 8512 ;THIS TEST VERIFIES THAT THE RESULT OF AN AC CONTAINING ALL ZEROS + 8513 ;EQUIVALENCED WITH AN E CONTAINING ALL ZEROS IS AN AC CONTAINING ALL + 8514 ;ONES; AND FUTHERMORE, IF THIS AC CONTAINING ALL ONES IS EQUIVALANCED + 8515 ;WITH AN E CONTAINING ALL ZEROS, THE FINAL RESULT IN THE AC + 8516 ;SHOULD BE ALL ZEROS. FIRST, THE AC IS INITIALIZED TO ALL ZEROS, THEN, + 8517 ;THIS AC IS EQUIVALENCED TWICE WITH AN E CONTAINING ALL ZEROS. THE AC IS + 8518 ;THEN CHECKED FOR ALL ZEROS + 8519 + 8520 034447 200 00 0 00 035062 A10200: MOVE [0] ;CLEAR THE AC + 8521 034450 444 00 0 00 035062 EQV [0] ;*EQV SHOULD RESULT IN C(AC)=ALL ONES + 8522 034451 444 00 0 00 035062 EQV [0] ;*EQV SHOULD RESULT IN C(AC)=0 + 8523 034452 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8524 STOP^ + 8525 034453 254 04 0 00 034454 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8526 034454 324 00 0 00 034455 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8527 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8528 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8529 + 8530 ;********** + 8531 + 8532 ;THIS TEST VERIFIES THAT THE RESULT OF AN AC CONTAINING ALL ONES + 8533 ;EQUIVALENCED WITH AN E CONTAINING ALL ONES IS AN AC CONTAINING ALL ONES; + 8534 ;AND FUTHERMORE, IF THIS AC CONTAINING ALL ONES ISEQUIVALENCED WITH AN E + 8535 ;CONTAINING ALL ZEROS, THE FINAL RESULT IN THE AC SHOULD BE ALL ZEROS. + 8536 ;FIRST, THE AC IS INITIALIZED TO ALL ONES; THEN, THIS AC IS + 8537 ;EQUIVALENCED WITH AN E OF ALL ONES. NEXT, THIS SAME AC IS EQUIVALENCED + 8538 ;WITH AN E OF ALL ZEROS. THE AC IS THEN CHECKED FOR ALL ZEROS. + 8539 + 8540 034455 200 00 0 00 035063 A10300: MOVE [-1] ;SET C(AC) TO ALL ONES + 8541 034456 444 00 0 00 035063 EQV [-1] ;*EQV SHOULD RESULT IN C(AC) OF ALL ONES + 8542 034457 444 00 0 00 035062 EQV [0] ;*EQV SHOULD RESULT IN C(AC)=0 + 8543 034460 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8544 STOP^ + 8545 034461 254 04 0 00 034462 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8546 034462 324 00 0 00 034463 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8547 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8548 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8549 + 8550 ;********** +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0180 + + 8551 013000 SN=13000 + 8552 000000 ZZ=0 + 8553 + 8554 A13000: REPEAT ^D36, + 8555 <;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8556 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8557 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8558 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8559 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8560 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8561 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8562 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8563 SN=SN+1 + 8564 ZZ=ZZ+ZZ + 8565 IFE ZZ, + 8566 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8567 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8568 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8569 EQV [0] ;RESULTS IN C(AC)=0 + 8570 SKIPE ;PASS TEST IF C(AC)=0 + 8571 STOP + 8572 + 8573 ;********** + 8574 > + 8575 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8576 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8577 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8578 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8579 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8580 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8581 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8582 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8583 013001 SN=SN+1 + 8584 000000 ZZ=ZZ+ZZ + 8585 000001 IFE ZZ, + 8586 034463 200 00 0 00 035074 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8587 034464 404 00 0 00 035074 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8588 034465 444 00 0 00 035074 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8589 034466 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8590 034467 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8591 STOP^ + 8592 034470 254 04 0 00 034471 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8593 034471 324 00 0 00 034472 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8594 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8595 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8596 + 8597 ;********** + 8598 + 8599 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8600 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8601 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8602 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8603 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8604 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8605 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-1 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0181 + + 8606 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8607 013002 SN=SN+1 + 8608 000002 ZZ=ZZ+ZZ + 8609 IFE ZZ, + 8610 034472 200 00 0 00 035075 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8611 034473 404 00 0 00 035075 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8612 034474 444 00 0 00 035075 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8613 034475 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8614 034476 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8615 STOP^ + 8616 034477 254 04 0 00 034500 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8617 034500 324 00 0 00 034501 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8618 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8619 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8620 + 8621 ;********** + 8622 + 8623 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8624 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8625 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8626 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8627 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8628 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8629 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8630 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8631 013003 SN=SN+1 + 8632 000004 ZZ=ZZ+ZZ + 8633 IFE ZZ, + 8634 034501 200 00 0 00 035076 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8635 034502 404 00 0 00 035076 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8636 034503 444 00 0 00 035076 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8637 034504 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8638 034505 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8639 STOP^ + 8640 034506 254 04 0 00 034507 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8641 034507 324 00 0 00 034510 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8642 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8643 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8644 + 8645 ;********** + 8646 + 8647 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8648 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8649 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8650 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8651 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8652 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8653 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8654 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8655 013004 SN=SN+1 + 8656 000010 ZZ=ZZ+ZZ + 8657 IFE ZZ, + 8658 034510 200 00 0 00 035077 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8659 034511 404 00 0 00 035077 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8660 034512 444 00 0 00 035077 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-2 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0182 + + 8661 034513 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8662 034514 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8663 STOP^ + 8664 034515 254 04 0 00 034516 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8665 034516 324 00 0 00 034517 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8666 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8667 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8668 + 8669 ;********** + 8670 + 8671 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8672 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8673 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8674 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8675 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8676 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8677 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8678 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8679 013005 SN=SN+1 + 8680 000020 ZZ=ZZ+ZZ + 8681 IFE ZZ, + 8682 034517 200 00 0 00 035100 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8683 034520 404 00 0 00 035100 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8684 034521 444 00 0 00 035100 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8685 034522 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8686 034523 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8687 STOP^ + 8688 034524 254 04 0 00 034525 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8689 034525 324 00 0 00 034526 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8690 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8691 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8692 + 8693 ;********** + 8694 + 8695 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8696 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8697 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8698 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8699 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8700 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8701 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8702 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8703 013006 SN=SN+1 + 8704 000040 ZZ=ZZ+ZZ + 8705 IFE ZZ, + 8706 034526 200 00 0 00 035101 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8707 034527 404 00 0 00 035101 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8708 034530 444 00 0 00 035101 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8709 034531 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8710 034532 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8711 STOP^ + 8712 034533 254 04 0 00 034534 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8713 034534 324 00 0 00 034535 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8714 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8715 ;IN THE SUBTEST) TO LOOP ON ERROR^ +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-3 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0183 + + 8716 + 8717 ;********** + 8718 + 8719 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8720 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8721 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8722 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8723 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8724 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8725 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8726 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8727 013007 SN=SN+1 + 8728 000100 ZZ=ZZ+ZZ + 8729 IFE ZZ, + 8730 034535 200 00 0 00 035102 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8731 034536 404 00 0 00 035102 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8732 034537 444 00 0 00 035102 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8733 034540 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8734 034541 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8735 STOP^ + 8736 034542 254 04 0 00 034543 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8737 034543 324 00 0 00 034544 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8738 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8739 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8740 + 8741 ;********** + 8742 + 8743 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8744 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8745 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8746 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8747 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8748 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8749 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8750 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8751 013010 SN=SN+1 + 8752 000200 ZZ=ZZ+ZZ + 8753 IFE ZZ, + 8754 034544 200 00 0 00 035103 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8755 034545 404 00 0 00 035103 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8756 034546 444 00 0 00 035103 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8757 034547 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8758 034550 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8759 STOP^ + 8760 034551 254 04 0 00 034552 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8761 034552 324 00 0 00 034553 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8762 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8763 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8764 + 8765 ;********** + 8766 + 8767 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8768 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8769 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8770 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-4 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0184 + + 8771 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8772 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8773 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8774 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8775 013011 SN=SN+1 + 8776 000400 ZZ=ZZ+ZZ + 8777 IFE ZZ, + 8778 034553 200 00 0 00 035104 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8779 034554 404 00 0 00 035104 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8780 034555 444 00 0 00 035104 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8781 034556 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8782 034557 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8783 STOP^ + 8784 034560 254 04 0 00 034561 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8785 034561 324 00 0 00 034562 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8786 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8787 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8788 + 8789 ;********** + 8790 + 8791 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8792 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8793 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8794 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8795 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8796 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8797 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8798 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8799 013012 SN=SN+1 + 8800 001000 ZZ=ZZ+ZZ + 8801 IFE ZZ, + 8802 034562 200 00 0 00 035105 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8803 034563 404 00 0 00 035105 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8804 034564 444 00 0 00 035105 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8805 034565 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8806 034566 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8807 STOP^ + 8808 034567 254 04 0 00 034570 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8809 034570 324 00 0 00 034571 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8810 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8811 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8812 + 8813 ;********** + 8814 + 8815 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8816 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8817 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8818 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8819 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8820 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8821 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8822 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8823 013013 SN=SN+1 + 8824 002000 ZZ=ZZ+ZZ + 8825 IFE ZZ, +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-5 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0185 + + 8826 034571 200 00 0 00 035106 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8827 034572 404 00 0 00 035106 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8828 034573 444 00 0 00 035106 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8829 034574 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8830 034575 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8831 STOP^ + 8832 034576 254 04 0 00 034577 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8833 034577 324 00 0 00 034600 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8834 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8835 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8836 + 8837 ;********** + 8838 + 8839 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8840 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8841 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8842 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8843 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8844 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8845 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8846 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8847 013014 SN=SN+1 + 8848 004000 ZZ=ZZ+ZZ + 8849 IFE ZZ, + 8850 034600 200 00 0 00 035107 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8851 034601 404 00 0 00 035107 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8852 034602 444 00 0 00 035107 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8853 034603 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8854 034604 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8855 STOP^ + 8856 034605 254 04 0 00 034606 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8857 034606 324 00 0 00 034607 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8858 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8859 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8860 + 8861 ;********** + 8862 + 8863 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8864 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8865 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8866 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8867 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8868 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8869 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8870 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8871 013015 SN=SN+1 + 8872 010000 ZZ=ZZ+ZZ + 8873 IFE ZZ, + 8874 034607 200 00 0 00 035110 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8875 034610 404 00 0 00 035110 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8876 034611 444 00 0 00 035110 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8877 034612 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8878 034613 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8879 STOP^ + 8880 034614 254 04 0 00 034615 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-6 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0186 + + 8881 034615 324 00 0 00 034616 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8882 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8883 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8884 + 8885 ;********** + 8886 + 8887 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8888 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8889 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8890 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8891 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8892 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8893 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8894 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8895 013016 SN=SN+1 + 8896 020000 ZZ=ZZ+ZZ + 8897 IFE ZZ, + 8898 034616 200 00 0 00 035111 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8899 034617 404 00 0 00 035111 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8900 034620 444 00 0 00 035111 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8901 034621 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8902 034622 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8903 STOP^ + 8904 034623 254 04 0 00 034624 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8905 034624 324 00 0 00 034625 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8906 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8907 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8908 + 8909 ;********** + 8910 + 8911 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8912 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8913 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8914 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8915 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8916 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8917 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8918 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8919 013017 SN=SN+1 + 8920 040000 ZZ=ZZ+ZZ + 8921 IFE ZZ, + 8922 034625 200 00 0 00 035112 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8923 034626 404 00 0 00 035112 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8924 034627 444 00 0 00 035112 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8925 034630 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8926 034631 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8927 STOP^ + 8928 034632 254 04 0 00 034633 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8929 034633 324 00 0 00 034634 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8930 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8931 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8932 + 8933 ;********** + 8934 + 8935 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-7 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0187 + + 8936 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8937 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8938 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8939 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8940 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8941 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8942 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8943 013020 SN=SN+1 + 8944 100000 ZZ=ZZ+ZZ + 8945 IFE ZZ, + 8946 034634 200 00 0 00 035113 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8947 034635 404 00 0 00 035113 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8948 034636 444 00 0 00 035113 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8949 034637 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8950 034640 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8951 STOP^ + 8952 034641 254 04 0 00 034642 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8953 034642 324 00 0 00 034643 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8954 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8955 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8956 + 8957 ;********** + 8958 + 8959 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8960 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8961 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8962 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8963 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8964 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8965 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8966 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 8967 013021 SN=SN+1 + 8968 200000 ZZ=ZZ+ZZ + 8969 IFE ZZ, + 8970 034643 200 00 0 00 035114 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8971 034644 404 00 0 00 035114 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8972 034645 444 00 0 00 035114 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8973 034646 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8974 034647 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8975 STOP^ + 8976 034650 254 04 0 00 034651 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 8977 034651 324 00 0 00 034652 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 8978 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 8979 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 8980 + 8981 ;********** + 8982 + 8983 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 8984 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 8985 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 8986 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 8987 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 8988 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 8989 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 8990 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-8 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0188 + + 8991 013022 SN=SN+1 + 8992 400000 ZZ=ZZ+ZZ + 8993 IFE ZZ, + 8994 034652 200 00 0 00 035115 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 8995 034653 404 00 0 00 035115 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 8996 034654 444 00 0 00 035115 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 8997 034655 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 8998 034656 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 8999 STOP^ + 9000 034657 254 04 0 00 034660 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9001 034660 324 00 0 00 034661 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9002 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9003 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9004 + 9005 ;********** + 9006 + 9007 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9008 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9009 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9010 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9011 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9012 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9013 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9014 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9015 013023 SN=SN+1 + 9016 000001 000000 ZZ=ZZ+ZZ + 9017 IFE ZZ, + 9018 034661 200 00 0 00 035116 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9019 034662 404 00 0 00 035116 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9020 034663 444 00 0 00 035116 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9021 034664 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9022 034665 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9023 STOP^ + 9024 034666 254 04 0 00 034667 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9025 034667 324 00 0 00 034670 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9026 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9027 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9028 + 9029 ;********** + 9030 + 9031 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9032 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9033 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9034 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9035 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9036 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9037 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9038 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9039 013024 SN=SN+1 + 9040 000002 000000 ZZ=ZZ+ZZ + 9041 IFE ZZ, + 9042 034670 200 00 0 00 035117 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9043 034671 404 00 0 00 035117 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9044 034672 444 00 0 00 035117 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9045 034673 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-9 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0189 + + 9046 034674 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9047 STOP^ + 9048 034675 254 04 0 00 034676 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9049 034676 324 00 0 00 034677 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9050 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9051 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9052 + 9053 ;********** + 9054 + 9055 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9056 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9057 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9058 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9059 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9060 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9061 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9062 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9063 013025 SN=SN+1 + 9064 000004 000000 ZZ=ZZ+ZZ + 9065 IFE ZZ, + 9066 034677 200 00 0 00 035120 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9067 034700 404 00 0 00 035120 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9068 034701 444 00 0 00 035120 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9069 034702 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9070 034703 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9071 STOP^ + 9072 034704 254 04 0 00 034705 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9073 034705 324 00 0 00 034706 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9074 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9075 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9076 + 9077 ;********** + 9078 + 9079 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9080 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9081 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9082 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9083 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9084 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9085 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9086 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9087 013026 SN=SN+1 + 9088 000010 000000 ZZ=ZZ+ZZ + 9089 IFE ZZ, + 9090 034706 200 00 0 00 035121 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9091 034707 404 00 0 00 035121 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9092 034710 444 00 0 00 035121 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9093 034711 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9094 034712 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9095 STOP^ + 9096 034713 254 04 0 00 034714 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9097 034714 324 00 0 00 034715 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9098 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9099 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9100 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-10 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0190 + + 9101 ;********** + 9102 + 9103 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9104 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9105 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9106 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9107 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9108 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9109 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9110 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9111 013027 SN=SN+1 + 9112 000020 000000 ZZ=ZZ+ZZ + 9113 IFE ZZ, + 9114 034715 200 00 0 00 035122 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9115 034716 404 00 0 00 035122 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9116 034717 444 00 0 00 035122 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9117 034720 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9118 034721 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9119 STOP^ + 9120 034722 254 04 0 00 034723 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9121 034723 324 00 0 00 034724 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9122 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9123 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9124 + 9125 ;********** + 9126 + 9127 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9128 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9129 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9130 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9131 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9132 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9133 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9134 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9135 013030 SN=SN+1 + 9136 000040 000000 ZZ=ZZ+ZZ + 9137 IFE ZZ, + 9138 034724 200 00 0 00 035123 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9139 034725 404 00 0 00 035123 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9140 034726 444 00 0 00 035123 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9141 034727 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9142 034730 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9143 STOP^ + 9144 034731 254 04 0 00 034732 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9145 034732 324 00 0 00 034733 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9146 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9147 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9148 + 9149 ;********** + 9150 + 9151 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9152 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9153 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9154 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9155 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-11 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0191 + + 9156 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9157 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9158 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9159 013031 SN=SN+1 + 9160 000100 000000 ZZ=ZZ+ZZ + 9161 IFE ZZ, + 9162 034733 200 00 0 00 035124 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9163 034734 404 00 0 00 035124 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9164 034735 444 00 0 00 035124 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9165 034736 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9166 034737 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9167 STOP^ + 9168 034740 254 04 0 00 034741 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9169 034741 324 00 0 00 034742 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9170 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9171 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9172 + 9173 ;********** + 9174 + 9175 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9176 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9177 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9178 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9179 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9180 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9181 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9182 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9183 013032 SN=SN+1 + 9184 000200 000000 ZZ=ZZ+ZZ + 9185 IFE ZZ, + 9186 034742 200 00 0 00 035125 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9187 034743 404 00 0 00 035125 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9188 034744 444 00 0 00 035125 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9189 034745 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9190 034746 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9191 STOP^ + 9192 034747 254 04 0 00 034750 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9193 034750 324 00 0 00 034751 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9194 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9195 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9196 + 9197 ;********** + 9198 + 9199 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9200 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9201 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9202 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9203 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9204 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9205 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9206 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9207 013033 SN=SN+1 + 9208 000400 000000 ZZ=ZZ+ZZ + 9209 IFE ZZ, + 9210 034751 200 00 0 00 035073 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-12 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0192 + + 9211 034752 404 00 0 00 035073 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9212 034753 444 00 0 00 035073 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9213 034754 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9214 034755 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9215 STOP^ + 9216 034756 254 04 0 00 034757 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9217 034757 324 00 0 00 034760 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9218 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9219 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9220 + 9221 ;********** + 9222 + 9223 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9224 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9225 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9226 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9227 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9228 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9229 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9230 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9231 013034 SN=SN+1 + 9232 001000 000000 ZZ=ZZ+ZZ + 9233 IFE ZZ, + 9234 034760 200 00 0 00 035126 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9235 034761 404 00 0 00 035126 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9236 034762 444 00 0 00 035126 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9237 034763 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9238 034764 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9239 STOP^ + 9240 034765 254 04 0 00 034766 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9241 034766 324 00 0 00 034767 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9242 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9243 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9244 + 9245 ;********** + 9246 + 9247 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9248 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9249 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9250 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9251 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9252 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9253 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9254 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9255 013035 SN=SN+1 + 9256 002000 000000 ZZ=ZZ+ZZ + 9257 IFE ZZ, + 9258 034767 200 00 0 00 035127 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9259 034770 404 00 0 00 035127 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9260 034771 444 00 0 00 035127 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9261 034772 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9262 034773 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9263 STOP^ + 9264 034774 254 04 0 00 034775 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9265 034775 324 00 0 00 034776 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-13 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0193 + + 9266 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9267 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9268 + 9269 ;********** + 9270 + 9271 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9272 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9273 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9274 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9275 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9276 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9277 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9278 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9279 013036 SN=SN+1 + 9280 004000 000000 ZZ=ZZ+ZZ + 9281 IFE ZZ, + 9282 034776 200 00 0 00 035130 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9283 034777 404 00 0 00 035130 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9284 035000 444 00 0 00 035130 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9285 035001 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9286 035002 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9287 STOP^ + 9288 035003 254 04 0 00 035004 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9289 035004 324 00 0 00 035005 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9290 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9291 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9292 + 9293 ;********** + 9294 + 9295 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9296 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9297 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9298 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9299 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9300 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9301 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9302 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9303 013037 SN=SN+1 + 9304 010000 000000 ZZ=ZZ+ZZ + 9305 IFE ZZ, + 9306 035005 200 00 0 00 035131 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9307 035006 404 00 0 00 035131 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9308 035007 444 00 0 00 035131 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9309 035010 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9310 035011 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9311 STOP^ + 9312 035012 254 04 0 00 035013 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9313 035013 324 00 0 00 035014 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9314 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9315 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9316 + 9317 ;********** + 9318 + 9319 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9320 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-14 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0194 + + 9321 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9322 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9323 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9324 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9325 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9326 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9327 013040 SN=SN+1 + 9328 020000 000000 ZZ=ZZ+ZZ + 9329 IFE ZZ, + 9330 035014 200 00 0 00 035132 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9331 035015 404 00 0 00 035132 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9332 035016 444 00 0 00 035132 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9333 035017 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9334 035020 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9335 STOP^ + 9336 035021 254 04 0 00 035022 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9337 035022 324 00 0 00 035023 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9338 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9339 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9340 + 9341 ;********** + 9342 + 9343 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9344 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9345 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9346 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9347 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9348 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9349 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9350 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9351 013041 SN=SN+1 + 9352 040000 000000 ZZ=ZZ+ZZ + 9353 IFE ZZ, + 9354 035023 200 00 0 00 035133 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9355 035024 404 00 0 00 035133 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9356 035025 444 00 0 00 035133 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9357 035026 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9358 035027 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9359 STOP^ + 9360 035030 254 04 0 00 035031 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9361 035031 324 00 0 00 035032 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9362 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9363 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9364 + 9365 ;********** + 9366 + 9367 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9368 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9369 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9370 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9371 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9372 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9373 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9374 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9375 013042 SN=SN+1 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-15 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0195 + + 9376 100000 000000 ZZ=ZZ+ZZ + 9377 IFE ZZ, + 9378 035032 200 00 0 00 035134 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9379 035033 404 00 0 00 035134 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9380 035034 444 00 0 00 035134 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9381 035035 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9382 035036 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9383 STOP^ + 9384 035037 254 04 0 00 035040 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9385 035040 324 00 0 00 035041 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9386 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9387 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9388 + 9389 ;********** + 9390 + 9391 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9392 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9393 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9394 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9395 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9396 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9397 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9398 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9399 013043 SN=SN+1 + 9400 200000 000000 ZZ=ZZ+ZZ + 9401 IFE ZZ, + 9402 035041 200 00 0 00 035135 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9403 035042 404 00 0 00 035135 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9404 035043 444 00 0 00 035135 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9405 035044 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9406 035045 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 + 9407 STOP^ + 9408 035046 254 04 0 00 035047 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9409 035047 324 00 0 00 035050 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9410 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9411 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9412 + 9413 ;********** + 9414 + 9415 ;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. + 9416 ;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN + 9417 ;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). + 9418 ;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD + 9419 ;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, + 9420 ;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. + 9421 ;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO + 9422 ;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. + 9423 013044 SN=SN+1 + 9424 400000 000000 ZZ=ZZ+ZZ + 9425 IFE ZZ, + 9426 035050 200 00 0 00 035065 MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + 9427 035051 404 00 0 00 035065 AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + 9428 035052 444 00 0 00 035065 EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + 9429 035053 444 00 0 00 035062 EQV [0] ;RESULTS IN C(AC)=0 + 9430 035054 332 00 0 00 000000 SKIPE ;PASS TEST IF C(AC)=0 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 31-16 +DAKAAM MAC 19-JAN-77 13:42 TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) SEQ 0196 + + 9431 STOP^ + 9432 035055 254 04 0 00 035056 HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + 9433 035056 324 00 0 00 035057 JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) TO + 9434 ;JUMPA X (X IS THE ADDRESS OF THE FIRST INSTRUCTION + 9435 ;IN THE SUBTEST) TO LOOP ON ERROR^ + 9436 + 9437 ;********** + 9438 + 9439 035057 254 00 0 00 030057 ENDXX: JRST BEGEND ;LOOP PROGRAM +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 1 +STOR KLM 18-JAN-77 11:42 *STOR* RESERVED STORAGE, JAN 18,1977 SEQ 0197 + + 9440 SUBTTL *STOR* RESERVED STORAGE, JAN 18,1977 + 9441 + 9442 ;PROGRAM LITERALS + 9443 + 9444 XLIST + 9445 IFNDEF $LPAPER, + 9446 035060 LIT + 9447 035060 000001 000001 + 9448 035061 254 00 0 00 030741 + 9449 035062 000000 000000 + 9450 035063 777777 777777 + 9451 035064 377777 777777 + 9452 035065 400000 000000 + 9453 035066 777000 000000 + 9454 035067 000377 000000 + 9455 035070 000000 177000 + 9456 035071 000000 000777 + 9457 035072 000000 600000 + 9458 035073 000400 000000 + 9459 035074 000000 000001 + 9460 035075 000000 000002 + 9461 035076 000000 000004 + 9462 035077 000000 000010 + 9463 035100 000000 000020 + 9464 035101 000000 000040 + 9465 035102 000000 000100 + 9466 035103 000000 000200 + 9467 035104 000000 000400 + 9468 035105 000000 001000 + 9469 035106 000000 002000 + 9470 035107 000000 004000 + 9471 035110 000000 010000 + 9472 035111 000000 020000 + 9473 035112 000000 040000 + 9474 035113 000000 100000 + 9475 035114 000000 200000 + 9476 035115 000000 400000 + 9477 035116 000001 000000 + 9478 035117 000002 000000 + 9479 035120 000004 000000 + 9480 035121 000010 000000 + 9481 035122 000020 000000 + 9482 035123 000040 000000 + 9483 035124 000100 000000 + 9484 035125 000200 000000 + 9485 035126 001000 000000 + 9486 035127 002000 000000 + 9487 035130 004000 000000 + 9488 035131 010000 000000 + 9489 035132 020000 000000 + 9490 035133 040000 000000 + 9491 035134 100000 000000 + 9492 035135 200000 000000 + 9493 035136 000000 001234 + 9494 035137 000000 777777 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 1-1 +STOR KLM 18-JAN-77 11:42 *STOR* RESERVED STORAGE, JAN 18,1977 SEQ 0198 + + 9495 035140 777777 000000 + 9496 035141 707070 707070 + 9497 035142 777777 777776 + 9498 035143 777777 777775 + 9499 035144 777777 777773 + 9500 035145 777777 777767 + 9501 035146 777777 777757 + 9502 035147 777777 777737 + 9503 035150 777777 777677 + 9504 035151 777777 777577 + 9505 035152 777777 777377 + 9506 035153 777777 776777 + 9507 035154 777777 775777 + 9508 035155 777777 773777 + 9509 035156 777777 767777 + 9510 035157 777777 757777 + 9511 035160 777777 737777 + 9512 035161 777777 677777 + 9513 035162 777777 577777 + 9514 035163 777777 377777 + 9515 035164 777776 777777 + 9516 035165 777775 777777 + 9517 035166 777773 777777 + 9518 035167 777767 777777 + 9519 035170 777757 777777 + 9520 035171 777737 777777 + 9521 035172 777677 777777 + 9522 035173 777577 777777 + 9523 035174 777377 777777 + 9524 035175 776777 777777 + 9525 035176 775777 777777 + 9526 035177 773777 777777 + 9527 035200 767777 777777 + 9528 035201 757777 777777 + 9529 035202 737777 777777 + 9530 035203 677777 777777 + 9531 035204 577777 777777 + 9532 LIST + 9533 035205 000000 000000 ENDSLD: 0 + 9534 + 9535 IFDEF DEBUG,< + 9536 PATCH: BLOCK DEBUG ;PATCHING AREA + 9537 > + 9538 + 9539 ;PROGRAM VARIABLES + 9540 035206 VAR + 9541 + 9542 IFDEF PGMEND,< + 9543 035206 000000 000000 END: 0 + 9544 030000 END BEGIN > + +NO ERRORS DETECTED + +PROGRAM BREAK IS 000000 +ABSLUTE BREAK IS 035207 +DAKAA PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) 0,2 MACRO %52(537) 13:54 19-JAN-77 PAGE 1-2 +STOR KLM 18-JAN-77 11:42 *STOR* RESERVED STORAGE, JAN 18,1977 SEQ 0199 + +CPU TIME USED 00:40.863 + +11K CORE USED + +A00 54 55 56 57 58 59 60 61 62 63 64 889# +A100 944# SEQ 0200 +A1000 1049# +A10000 7679# +A10100 8097# +A10200 8520# +A10300 8540# +A1100 1064# +A1200 1077# +A12500 890# +A12600 934# +A12700 926# +A1300 1091# +A13000 8554# +A13100 2021# +A1400 1105# +A1500 1119# +A15000 905# +A15100 916# +A1600 1133# +A1700 1147# +A200 957# +A2000 1161# +A2100 1175# +A2200 1188# +A2300 1202# +A2400 1216# +A2500 1230# +A2600 1243# +A2700 1257# +A300 971# +A3000 1271# +A3100 1283# +A3200 1988# +A3300 2006# +A3400 2034# +A3500 2059# +A3600 2077# +A3700 2095# +A400 987# +A4000 2146# +A4100 2162# +A4200 2307# +A4300 2325# +A4400 2343# +A4500 2110# +A4600 2125# +A4700 2362# +A500 1002# +A5000 2378# +A5100 2394# +A5200 2409# +A5300 2426# +A5400 2446# +A5500 2459# SEQ 0201 +A5600 2819# +A5700 3183# +A600 1021# +A6000 3199# +A6100 3219# +A6200 3237# +A6300 3258# +A6400 3271# +A6500 4340# +A6600 5120# +A6700 5138# +A700 1036# +A7000 5158# +A7100 5175# +A7200 5191# +A7300 5205# +A7400 5945# +A7500 6870# +A7600 7648# +A7700 7665# +AABCLR 353# +AADRBK 353# +AAPRC1 353# +ABORT 85# +AC 2160# 2193 2193# 2194 2194# 2196 2197 2198 2222 2222# 2223 2225 2226 2227 + 2251 2251# 2252 2254 2255 2256 2280 2280# 2281 2283 2284 2285 +AC0 680# +AFOCLR 353# +AFODIS 353# +AFOENB 353# +AFOV 353# +ALTMGO 439# +AMPCLR 353# +AMPV 353# +ANXCLR 353# +ANXM 353# +AOVCLR 353# +AOVDIS 353# +AOVENB 353# +AOVFLO 353# +APAREN 353# +APARER 353# +APDCLR 353# +APDLOV 353# +APR 777 778 779 +APWRFL 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000000431100 +ET0 0 030660 0 0 000000431100 0 000000431100 +ST1A 0 030660 0 0 000000431100 0 000000431100 +ST9 0 030660 0 0 000000431100 0 000000431100 + diff --git a/apps/pdp10/tapes/diags/klad/dakaa/README.md b/apps/pdp10/tapes/diags/klad/dakaa/README.md new file mode 100644 index 000000000..f25681992 --- /dev/null +++ b/apps/pdp10/tapes/diags/klad/dakaa/README.md @@ -0,0 +1,1291 @@ +--- +layout: page +title: PDP-10 KA10 Basic Instruction Diagnostic #1 +permalink: /apps/pdp10/tapes/diags/klad/dakaa/ +machines: + - id: testka10 + type: pdp10 + config: /devices/pdp10/machine/ka10/test/debugger/machine.xml + debugger: true +--- + +PDP-10 KA10 Basic Instruction Diagnostic #1 +------------------------------------------- + +The *PDP-10 KA10 Basic Instruction Diagnostic #1* (MAINDEC-10-DAKAA-B-D) test code has been extracted from +[DAKAA.MAC](DAKAA.MAC) [[original file](http://pdp-10.trailing-edge.com/klad_sources/01/klad.sources/dakaam.mac.html)] +for use with the [PDP-10 Test Machine with Debugger](/devices/pdp10/machine/ka10/test/debugger/) below. + +Resources for this test include: + +- [DAKAA.TXT](#dakaatxt) +- [DAKAA.MAC](#dakaamac) +- [DAKAA.SEQ](#dakaaseq) + +{% include machine.html id="testka10" %} + +The Debugger's assemble ("a") command can be used to test the new built-in +[MACRO-10 mini-assembler](/modules/pdp10/lib/macro10.js), which supports a subset +of the [MACRO-10](http://archive.pcjs.org/pubs/dec/pdp10/tops10/02_1973AsmRef_macro.pdf) assembly language. +This command: + + a 30724 /apps/pdp10/tapes/diags/klad/dakaa/TEST.MAC + +will automatically read the [TEST.MAC](TEST.MAC) source file (a slightly modified copy of [DAKAA.MAC](DAKAA.MAC)), assemble it, +and then load the binary output at the specified address. + +--- + +DAKAA.TXT +--------- + +``` +MAINDEC-10-DAKAA + + + + + + + + IDENTIFICATION + -------------- + + PRODUCT CODE: MAINDEC-10-DAKAA-B-D + + PRODUCT NAME: DECSYSTEM10 PDP-10 KA10 BASIC + INSTRUCTION DIAGNOSTIC (1) + + FUNCTION: BASIC INSTRUCTIONS + + VERSION: 0.2 + + DATE RELEASED: JANUARY 1977 + + MAINTAINED BY: DIAGNOSTIC ENGINEERING GROUP + + AUTHOR: JOHN R. KIRCHOFF + +COPYRIGHT(C) 1976,1977 +DIGITAL EQUIPMENT CORPORATION +MARLBORO, MASS. 01752 + +THIS SOFTWARE IS FURNISHED UNDER A LICENSE FOR USE ONLY +ON A SINGLE COMPUTER SYSTEM AND MAY BE COPIED ONLY WITH +THE INCLUSION OF THE ABOVE COPYRIGHT NOTICE. THIS SOFTWARE, +OR ANY OTHER COPIES THEREOF, MAY NOT BE PROVIDED OR OTHERWISE +MADE AVAILABLE TO ANY OTHER PERSON EXECPT FOR USE ON SUCH SYSTEM +AND TO ONE WHO AGREES TO THESE LICENSE TERMS. TITLE TO AND +OWNERSHIP OF THE SOFTWARE SHALL AT ALL TIMES REMAIN IN DEC. + +THE INFORMATION IN THIS DOCUMENT IS SUBJECT TO CHANGE WITHOUT +NOTICE AND SHOULD NOT BE CONSTRUED AS A COMMITMENT BY DIGITAL +EQUIPMENT CORPORATION. + +DEC ASSUMES NO RESPONSIBILITY FOR THE USE OR RELIABILITY OF ITS +SOFTWARE ON EQUIPMENT WHICH IS NOT SUPPLIED BY DEC. + MAINDEC-10-DAKAA.TXT + PAGE 2 + + + TABLE OF CONTENTS + ----------------- + +1.0 ABSTRACT + +2.0 REQUIREMENTS + +2.1 EQUIPMENT + +2.2 STORAGE + +2.3 PRELIMINARY PROGRAMS + +3.0 PROGRAM PROCEDURES + +3.1 LOADING PROCEDURE + +3.2 STARTING PROCEDURE + +3.3 OPERATING PROCEDURE + +4.0 ERRORS + +5.0 ITERATION COUNTER + +6.0 CYCLE TIME + +7.0 OPERATIONAL VARIATIONS + +8.0 MISCELLANEOUS + +9.0 LISTING + MAINDEC-10-DAKAA.TXT + PAGE 3 + + +1.0 ABSTRACT + + THIS PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC IS THE FIRST IN A + SERIES OF PDP-10 KA10 PROCESSOR DIAGNOSTICS. IT IS THE MOST + BASIC OF THE PROCESSOR DIAGNOSTICS. THIS DIAGNOSTIC + ASSUMES THE HALT INSTRUCTION AND THE COMPUTER CONSOLE + TO BE OPERATIVE. IT MAKES NO FURTHER ASSUMPTIONS EXCEPT + THAT IT IS LOADED INTO MEMORY CORRECTLY. + + THE DIAGNOSTIC TESTS SOME OF THE MOVE AND SKIP INSTRUCTIONS. + IT ALSO TESTS THE ADDERS AND OTHER LOGIC WITH AND, XOR, EQV, + AND ALL BOOLE INSTRUCTIONS. + +2.0 REQUIREMENTS + +2.1 EQUIPMENT + + A PDP-10 KA10 WITH A MINIMUM OF 32K OF MEMORY + + PAPER TAPE READER + DECTAPE (OPTIONAL) + CONSOLE TELETYPE + +2.2 STORAGE + + THE PROGRAM RUNS WITHIN 32K OF MEMORY. + +2.3 PRELIMINARY PROGRAMS + + CONSOLE FUNCTIONS WORKING PROPERLY + PAPER TAPE OR DECTAPE READ-IN WORKING PROPERLY + MAINDEC-10-DAKAA.TXT + PAGE 4 + + +3.0 PROGRAM PROCEDURES + +3.1 LOADING PROCEDURE + + PAPER TAPE - HARDWARE READ-IN (READER DEVICE CODE 104) + DECTAPE - LOAD WITH DIAMON (DECTAPE DEVICE CODE 320) + +3.2 STARTING PROCEDURE + + STAND-ALONE STARTING ADDRESS IS 30000. + + IF THE DIAGNOSTIC FAILS TO START CORRECTLY TRY STARTING AT THE + FIRST TEST INSTEAD OF AT THE BEGINNING OF THE CONTROL SEQUENCE. + (SEE LISTING). + +3.3 OPERATING PROCEDURE + + ONCE STARTED THE PROGRAM WILL CYCLE CONTINUALLY UNTIL STOPPED + OR AN ERROR OCCURS. + +4.0 ERRORS + + ERRORS ARE IN THE FORM OF HALT INSTRUCTIONS. THE LISTING + SHOULD BE CONSULTED TO DETERMINE THE CAUSE OF THE ERROR. A + NO OPERATION (JUMP) INSTRUCTION FOLLOWS EACH HALT. THIS + MAY BE USEFUL IN CONSTRUCTING A SCOPE LOOP TO CYCLE ON THE + FAILING INSTRUCTION. + +5.0 ITERATION COUNTER + + THE ITERATION COUNT OF THE PROGRAM IS DISPLAYED IN THE MEM- + ORY INDICATORS (MI). THIS COUNT IS A DECREMENTING COUNT AND + INITIALLY STARTS AT -1 IN STAND-ALONE OPERATION. + +6.0 CYCLE TIME + + THE CYCLE TIME OF THE PROGRAM IS IN THE MILLISECOND RANGE AND + IS THEREFORE SUITABLE FOR TAKING MARGINS, VIBRATION TESTS, ETC. + + MAINDEC-10-DAKAA.TXT + PAGE 5 + + +7.0 OPERATIONAL VARIATIONS + + A. DIAGNOSTIC MONITOR + + THE PROGRAM IS USABLE WITH THE DIAGNOSTIC MONITOR TO PRO- + VIDE RELIABILITY TESTS, ACCEPTANCE TESTS, AND/OR TO PRO- + VIDE A QUICK METHOD OF ISOLATION OF A FAULT TO A PARTICULAR + AREA OF THE PROCESSOR. CERTAIN PROCEDURES ARE USED WHEN + THE PROGRAM IS USED IN THIS MANNER. THEY ARE: + + 1. THE DIAGNOSTIC MONITOR TRANSFERS CONTROL TO THE PRO- + GRAM AND STARTS IT AT LOCATION 30002. + + 2. MONCTL - LOCATION 30043 IS USED AS THE DIAGNOSTIC MON- + ITOR CONTROL FLAG WORD. + + B. USER MODE + + THE PROGRAM WILL OPERATE IN USER MODE AND AS SUCH PROVIDES + ASSURANCE THAT THE PROCESSOR IS PERFORMING ALL FUNCTIONS + CORRECTLY. USER MODE STARTING ADDRESS IS 30000. + + C. SYSTEM EXERCISER + + STARTING ADDRESS IS 30003. NO DATA SWITCHES ARE USED BY + THIS PROGRAM. + +8.0 MISCELLANEOUS + + NONE + +9.0 LISTING + +``` + +DAKAA.MAC +--------- + +``` +SUBTTL DIAGNOSTIC SECTION + +START: SETZM USER# ;CLEAR USER CONTROL WORD + JSP 0,.+1 ;GET FLAGS + TLNE USERF ;IN USER MODE? + SETOM USER ;YES, SET USER CONTROL WORD + SKIPN MONFLG ;SPECIAL USER MODE? + SETZM USER ;YES, CLEAR USER CONTROL WORD + SKIPN USER + JRST STARTA + SKIPL MONCTL + TTCALL 3,PGMNAM ;MENTION OUR NAME + JRST STARTA + +PGMNAM: ASCIZ/ +PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) [DAKAA] +/ + +;BASIC INSTRUCTION TEST (3) +;THE TEST IS DESIGNED FOR INITIAL DEBUGGING OF +;PROCESSOR HARDWARE AND TO DETECT (SOLID) FAILURES +;IN THE FIELD. + + +STARTA: JRST .+1 +SUBTTL TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS + +;NOTE: AN "*" IN THE COMMENT FIELD OF AN INSTRUCTION INDICATES THAT IT IS THE TESTED +;INSTRUCTION. +;********** + +;THIS TEST VERIFIES THAT SKIPA ALWAYS SKIPS THE NEXT INSTRUCTION + +A00=. +A12500: SKIPA ;*SKIPA SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + HALT ;IF PROGRAM HALTS, SKIPA DID NOT SKIP + +;IF PROGRAM HANGS UP ON SKIPA INSTRUCTION, CHECK AB PC EN [ABC], AB PC F/F [ABC], +;AB PC B [ABC OR AB], AB PC (FETCH) EN [ABC], IR SKIPS [IR3], IR SKIPX [IR1], +;IR 3XX [IR1], IR BITS 00 - 08 [IR1], F CYC ACT EN C [FI], FT6 F/F [F2], +;PC CLOCK EN [PCC], ET2 J F/F [E], PC CLK (ET2) EN [PCC], ADZ COND P [ADZ] AND +;ADZ COND R [ADZ], PC CLOCK A [PCC], PC CLOCK B [PCC] + +;IF PROGRAM HALTED ON HALT INSTRUCTION, CHECK ST1 COND [ST2] + +;********** + +;THIS TEST VERIFIES THAT JUMP NEVER JUMPS + +A15000: JUMP .+1 ;*JUMP SHOULD NEVER JUMP + SKIPA ;SKIP HALT INSTRUCTION IF JUMP IS SUCCESSFUL + HALT ;HALT IF JUMP FAILS + +;IF PROGRAM HANGS UP ON JUMP .+1 INSTRUCTION, THE JUMP INSTRUCTION +;MAY HAVE FAILED. CHECK ST1 COND [ST2], ST INST FET ST1 EN [ST2], ET2 D F/F [E], +;IR JUMPS [IR1], IR JUMPX [IR1] + +;********** +;THIS TEST VERIFIES THAT JUMP NEVER JUMPS + +A15100: JUMP .+2 ;*JUMP SHOULD NEVER JUMP + SKIPA ;SKIP HALT INSTRUCTION IF JUMP IS SUCCESSFUL + HALT ;HALT IF JUMP FAILS + +;AB PC EN [ABC] + +;********** + +;THIS TEST VERIFIES THAT JUMPA .+1 ALWAYS JUMPS TO THE NEXT INSTRUCTION + +A12700: JUMPA .+1 ;*JUMPA .+1 SHOULD NEVER JUMP + SKIPA ;SKIP HALT INSTRUCTION IF JUMPA IS SUCCESSFUL + HALT ;HALT IF JUMPA .+1 FAILED + +;********** + +;THIS TEST VERIFIES THAT JUMPA JUMPS TO THE LOCATION SPECIFIED BY E WHEN E=.+2 + +A12600: JUMPA .+2 ;*JUMPA .+2 SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + HALT ;PROGRAM HALTS HERE IF JUMPA .+2 FAILS + +;ST1 COND [ST2] + +;********** +;THIS TEST VERIFIES THAT SKIPA IS DATA INDEPENDENT. HENCE, IT ALWAYS SKIPS THE NEXT +;INSTRUCTION. +;THE MOVE [0] INSTRUCTION IS USED TO TEST THE DATA INDEPENDENCE OF SKIPA. + +A100: MOVE [0] ;PRESET LOCATION 0 TO ALL ZEROS + SKIPA ;*SKIPA SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST INSURES THAT SKIPA IS DATA INDEPENDENT. HENCE, IT ALWAYS SKIPS THE NEXT +;INSTRUCTION. +;THE MOVE [-1] INSTRUCTION IS USED TO TEST THE DATA INDEPENDENCE OF SKIPA. + +A200: MOVE [-1] ;PRESET LOCATION 0 TO ALL ONES + SKIPA ;*SKIPA SKOULD ALWAYS SKIP THE NEXT INSTRUCTION + STOP + +;COND P [ADZ] + +;********** +;THIS TEST VERIFIES THAT SKIPGE SKIPS THE NEXT INSTRUCTION WHEN C(E) +;IS ALL ZEROS. THE MOVE INSTRUCTION IS USED TO LOAD E WITH ALL ZEROS. + +A300: MOVE [0] ;PRESET E TO ALL ZEROS + SKIPGE ;*SKIPGE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;COND R [ADZ], AD00 S-A-1, AD AR + EN [ADAP], AD LT AR + EN [ADAP], +;AD RT AR + EN [ADAP], AD AR + EN C [ADAP], F CYC ACT EN A [F1] + +;********** + +;THIS TEST VERIFIES THAT SKIP IS DATA INDEPENDENT. HENCE, SKIP NEVER SKIPS. IT ALWAYS +;CONTINUES ON TO THE NEXT INSTRUCTION. + +A400: MOVE [-1] ;PRESET E TO ALL ONES + SKIP ;*SKIP SHOULD NEVER SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;PC CHANGE [PCC], PC CHANGE A [PCC], SEE MEMORY CONTROL DWGS [MC1-4] + +;********** +;THIS TEST VERIFIES THAT SKIPGE WILL NOT SKIP THE NEXT INSTRUCTION WHEN C(E) IS +;NEGATIVE. THE MOVE INSTRUCTION LOADS E WITH -1, A NEGATIVE NUMBER. + +A500: MOVE [-1] ;PRESET E WITH ALL ONES, A NEGATIVE NUMBER + SKIPGE ;*SKIPGE SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;IF MOVE INSTRUCTION FAILED, CHECK MOVX-: FCE [F1], -IR XXXI [IR1], -IR XXXM [IR1], +;IR HWTFWT [IR3], IR FWT [IR2], IR2XX [IR1] + +;IF SKIPGE FAILED, CHECK SKIPX: FCE [F1] + +;********** + +;THIS TEST VERIFIES THAT SKIPN WILL NOT SKIP THE NEXT INSTRUCTION WHEN +;C(E) IS ZERO. THE MOVE INSTRUCTION LOADS E WITH ALL ZEROS. + +A600: MOVE [0] ;PRESET E WITH ALL ZEROS + SKIPN ;*SKIPN SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT SHOULD SKIP +;THE NEXT INSTRUCTION WHEN C(E) IS NON-ZERO. THE MOVE INSTRUCTION LOADS +;E WITH -1, A NON-ZERO NUMBER. + +A700: MOVE [-1] ;PRESET E WITH ALL ONES, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPE IS DATA DEPENDENT. HENCE, IT SHOULD NOT +;SKIP THE NEXT INSTRUCTION WHEN C(E) IS NON-ZERO. THE MOVE INSTRUCTION LOADS E +;WITH -1, A NON-ZERO NUMBER. + +A1000: MOVE [-1] ;PRESET E WITH ALL ONES, A NON-ZERO NUMBER + SKIPE ;*SKIPE SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPE IS DATA DEPENDENT. HENCE, IT SHOULD SKIP +;THE NEXT INSTRUCTION WHEN C(E) IS ZERO. THE MOVE INSTRUCTION LOADS ALL +;ZEROS INTO E. + +A1100: MOVE [0] ;PRESET E TO ALL ZEROS + SKIPE ;*SKIPE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIP IS DATA INDEPENDENT. HENCE, IT NEVER SKIPS. IT ALWAYS +;CONTINUES ON TO THE NEXT INSTRUCTION. + +A1200: MOVE [0] ;PRESET E TO ALL ZEROS + SKIP ;*SKIP SHOULD NEVER SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NOT NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD +;E WITH ALL ZEROS, A NON-NEGATIVE NUMBER. + +A1300: MOVE [0] ;PRESET E T0 ZERO, A NON-NEGATIVE NUMBER + SKIPL ;*SKIPL SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD E WITH -1, +;A NEGATIVE NUMBER. + +A1400: MOVE [-1] ;PRESET E TO -1, A NEGATIVE NUMBER + SKIPL ;*SKIPL SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPG IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS GREATER THAN ZERO. THE MOVE INSTRUCTION IS USED TO LOAD +;E WITH A NON-ZERO POSITIVE NUMBER. + +A1500: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A NON-ZERO POSITIVE NUMBER + SKIPG ;*SKIPG SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPG IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NOT GREATER THAN ZERO. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH -1, A NUMBER NOT GREATER THAN ZERO. + +A1600: MOVE [-1] ;PRESET E TO -1, A NEGATIVE NUMBER + SKIPG ;*SKIPG SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NON-NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD E WITH +;377777777777, A NON-NEGATIVE NUMBER. + +A1700: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A NON-NEGATIVE NUMBER + SKIPL ;*SKIPL SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS ZERO. THE MOVE INSTRUCTION LOADS E WITH ALL ZEROS. + +A2000: MOVE [0] ;PRESET E TO ALL ZEROS + SKIPLE ;*SKIPLE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NEGATIVE NUMBER. THE MOVE INSTRUCTION IS USED TO LOAD +;E WITH 400000000000, A NEGATIVE NUMBER. + +A2100: MOVE [XWD 400000,0] ;PRESET E TO 400000000000, A NEGATIVE NUMBER. + SKIPLE ;*SKIPLE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A POSITIVE NUMBER. THE MOVE INSTRUCTION LOADS +;E WITH 377777777777, A POSITIVE NUMBER. + +A2200: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A POSITIVE NUMBER + SKIPLE ;*SKIPLE SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********* +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION ;WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD 777000000000, A NON-ZERO NUMBER INTO E. + +A2300: MOVE [XWD 777000,0] ;PRESET E WITH 777000000000, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD A NON-ZERO NUMBER, 377000000 INTO E. + +A2400: MOVE [XWD 377,0] ;PRESET E TO 377000000, A NON ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH A NON-ZERO NUMBER, 177000. + +A2500: MOVE [177000] ;PRESET E WITH 177000, A NON ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD 777, A NON-ZERO NUMBER INTO E. + +A2600: MOVE [777] ;PRESET E WITH 777, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH 600000, A NON-ZERO NUMBER. + +A2700: MOVE [600000] ;PRESET E WITK 600000, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH 400000000, A NON-ZERO NUMBER. + +A3000: MOVE [XWD 400,000] ;PRESET E WITH 400000000, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +SN=3100 + ZZ=0 + +A3100: REPEAT ^D36, +<;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST +;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE +;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO +;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). +;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +> +SUBTTL TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS + +;THIS TEST VERIFIES THAT MOVEI LOADS E INTO THE AC. SKIPG IS USED TO CHECK THAT +;THE AC WAS INDEED MODIFIED BY MOVEI. + +A3200: MOVEI [-1] ;*MOVEI SHOULD LOAD A POSITIVE NUMBER INTO AC0 + SKIPG ;SKIP HALT INSTRUCTION IF MOVEI LOADED AC CORRECTLY + STOP + +;IF PROGRAM HANGS UP ON MOVEI INSTRUCTION, CHECK AD FM +EN [ADFP], AD FM + EN A [ADFP], +;AD FM + F/F [ADFP], F CYC ACT EN C [F1] + +;********** + +;THIS TEST VERIFIES THAT SKIPL IS SENSITIVE TO THE DATA IN E AND NOT SENSITIVE TO +;THE DATA IN THE AC. HENCE, SKIPL WILL SKIP THE NEXT INSTRUCTION ONLY IF +;C(E) IS LESS THAN ZERO. E IS SET TO -1, A NEGATIVE NUMBER; AND +;THE AC IS SET TO 0 IN THIS TEST. THEREFORE SKIPL SHOULD SKIP. + +A3300: MOVE [0] ;SET THE AC TO ALL ZEROS + SKIPL [-1] ;*SKIPL SHOULD SKIP BECAUSE C(E) IS NEGATIVE + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPL IS SENSITIVE TO THE SIGN BIT IN E. SKIPL SHOULD +;SKIP THE NEXT INSTRUCTION WHEN AND ONLY WHEN THE SIGN BIT IS SET. IN THIS +;TEST, C(AC) IS PRESET TO ALL ZEROS, THEN SKIPL IS EXECUTED WITH C(E)=400000,,0. +;A FAILURE UCCURS IF SKIPL DOES NOT SKIP. + +A13100: MOVE [0] ;SET THE AC TO ALL ZEROS + SKIPL [400000,,0] ;*SKIPL SHOULD SKIP BECAUSE C(E) IS NEGATIVE + STOP + +;********** +;THIS TEST VERIFIES THAT THE CAM INSTRUCTION DOES NOT MODIFY C(AC). THE AC IS +;INITIALLY LOADED WITH ALL ZEROS; THEN CAM IS EXECUTED WITH C(E) NON-ZERO. +;THE AC IS THEN CHECKED FOR ALL ZEROS. THIS TEST FAILS IF THE AC WAS MODIFIED BY CAM. + +A3400: MOVE [0] ;PRESET THE AC TO ALL ZEROS + CAM [1234] ;*CAM SHOULD NOT MODIFY THE AC + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION SERVES AS A NO-OP. + ;IT IS A PROGRAM FILLER IN CASE CAM SKIPS.) + SKIPE ;SKIP HALT INSTRUCTION IF AC WAS NOT MODIFIED BY CAM + STOP + +;IF TEST HANGS ON CAM INSTRUCTION, AB PC EN [ABC], AB PC (FETCH) EN [ABC], +;IR CAXX [IR3] + +;IF TEST LOOPS ON CAM INSTRUCTION, MC INSTR FETCH EN [MC1], MC INST FET ET2 EN [MC1], +;ET2 J F/F [E], IR CAXX [IR3] + +;IF TEST HALTS AT HALT INSTRUCTION, SAC INH [ST2], IR CAXX [IR3], IR CAMX [IR1] + +;********** + +;THIS TEST VERIFIES THAT THE CAI INSTRUCTION DOES NOT MODIFY C(AC). THE AC IS +;INITIALLY LOADED WITH ALL ZEROS; THEN CAI IS EXECUTED WITH E NON-ZERO. THE AC IS +;THEN CHECKED FOR ALL ZEROS. THIS TEST FAILS IF THE AC WAS MODIFIED BY CAM. + +A3500: MOVE [0] ;PRESET THE AC TO ALL ZEROS + CAI [1234] ;*CAI SHOULD NOT MODIFY THE AC + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A PROGRAM FILLER + ;IN CASE CAI MODIFIES THE AC. IT SERVES AS A NO-OP) + SKIPE ;SKIP HALT INSTRUCTION IF CAI DID NOT MODIFY THE AC + STOP + +;********** +;THIS TEST VERIFIES THAT CAI DOES NOT MODIFY C(AC0) WHEN AN AC OTHER THAN AC0 IS +;SPECIFIED BY THE CAI INSTRUCTION. +;THIS IS THE FIRST TEST THAT USES AN AC OTHER THAN AC0. +;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, THE CAI INSTRUCTION IS EXECUTED WITH +;A NON-ZERO E. AC 0 IS THEN TESTED F0R ALL ZEROS. + +A3600: MOVE [0] ;LOAD AC0 WITH ALL ZEROS + CAI 17,[1234] ;*CAI SHOULD NOT MODIFY AC0 + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + ;CAI SKIPS THE NEXT INSTRUCTION. IT ACTS AS A NO-OP) + SKIPE ;SKIP HALT INSTRUCTION IF C(AC0) WAS NOT MODIFIED + STOP + +;********** + +;THIS TEST VERIFIES THAT CAM DOES NOT MODIFY C(AC0) WHEN AN AC OTHER THAN AC0 IS +;SPECIFIED BY THE CAM INSTRUCTION. +;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, THE CAM INSTRUCTION IS EXECUTED WITH +;A NON-ZERO E. AC 0 IS THEN TESTED F0R ALL ZEROS. + +A3700: MOVE [0] ;PRESET C(AC0) TO ALL ZEROS + CAM 17,[1234] ;*CAM SHOULD NOT MODIFY AC0 + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + ;CAM SKIPS THE NEXT INSTRUCTION. IT SERVES AS A NO-OP) + SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS NOT MODIFIED BY CAM + STOP + +;********** +;THIS TEST VERIFIES THAT MOVE WILL PLACE C(E) INTO THE AC. AC, E AND C(E) ARE NON-ZERO +;IN THIS TEST. THIS TEST PASSES IF FINAL C(AC) OF THE MOVE INSTRUCTION IS NON-ZERO. + +A4500: MOVE 1,[0] ;LOAD THE AC WITH ALL ZEROS + MOVE 1,[-1] ;*MOVE SHOULD PLACE ALL ONES IN THE AC + SKIPN 1 ;SKIP HALT INSTRUCTION IF MOVE MODIFIED THE AC + STOP + +;********** + +;THIS TEST VERIFIES THAT THE MOVE INSTRUCTION WILL PLACE C(E) INTO THE AC. +;AC, E ARE NON-ZERO AND C(E)=0. +;THIS TEST FAILS IF THE FINAL C(AC) IS NOT EQUAL TO ZERO, THE ORIGINAL C(E) + +A4600: MOVE 1,[0] ;*PLACE ALL ZEROS INTO THE AC + SKIPE 1 ;SKIP HALT INSTRUCTION IF MOVE PLACED ALL ZEROS INTO AC + STOP + +;IF JUMP .+1 OF PREVIOUS TEST FAILED, CHECK JUMPS: AD FM + EN [ADFP], IR JUMPS [IR1], +;IR JUMPX [IR1], AR AD EN [ARMA], AR AD EN A [ARMA], AR LT AD EN A [ARMA], +;AR LT ADD EN B [ARMA], AR RT AD EN A [ARMA], +;AR RT AD EN B [ARMA], AR A CLK INPUT [ARMA], +;ET2 C F/F [E], AR AD ET2 EN [ARMA], AR AD BR (ET2) B [ARMA] + +;********** +;THIS TEST VERIFIES THAT SOS DOES NOT MODIFY C(AC) WHEN THE AC IS AC0. +;FIRST, AC0 IS LOADED WITH ALL ZEROS AND E IS LOADED WITH A POSITIVE NUMBER; +;THEN, SOS IS EXECUTED. SOS SHOULD SUBTRACT ONE FROM C((E) +;AND EXECUTE THE NEXT INSTRUCTION. IT SHOULD NOT MODIFY AC0. +;THE TEST PASSES IF AC0 WAS NOT MODIFIED. + +A4000: MOVE [0] ;PRESET AC0 TO ALL ZEROS + MOVE 7,[1234] ;PRESET E WITH A POSITIVE NUMBER + SOS 7 ;*SOS SHOULD NOT MODIFY C(AC0) + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + ;SOS SKIPS THE NEXT INSTRUCTION. IT SERVES AS A NO-OP) + SKIPE ;SKIP THE HALT INSTRUCTION IF C(AC0) WAS NOT MODIFIED + STOP + +;********** +SN=4100 + AC=0 + +A4100: REPEAT 4, +<;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL PLACE THE CONTENTS OF E INTO THE AC +;WHEN THE AC IS NOT AC0. THIS TEST IS PERFORMED FOR AC1, AC2, AC4 AND AC10. +;FIRST, THE AC IS LOADED WITH ALL ZEROS; THEN, THE SKIP INSTRUCTION IS EXECUTED +;WITH AC1, AC2, AC4 OR AC10. C(E) IS EQUAL TO -1. THE SKIP INSTRUCTION SHOULD CAUSE +;THE AC TO CONTAIN -1. THE AC IS THEN CHECKED FOR +;NON-ZERO CONTENTS. THE TEST FAILS IF C(AC)=0. +;THIS TEST ALSO VERIFIES THAT 'SKIP AC,[-1]' DOES NOT CLOBBER AC0. + +SN=SN+1 + AC=AC+AC + IFE AC, + MOVE 0,[0] ;PRESET AC0 TO ALL ZEROS + MOVE AC,[0] ;PRESET THE AC TO ALL ZEROS + SKIP AC,[-1] ;*SKIP SHOULD PLACE -1 INTO THE AC + SKIPN AC ;SKIP THE HALT INSTRUCTION IF THE AC IS NON-ZERO + STOP + SKIPE 0 ;VERIFY THAT AC0 DID NOT GET CLOBBERED BY SKIP + STOP + +;********** +> +SUBTTL TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS + +;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL NOT MODIFY THE CONTENTS OF AC0. +;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, +;THE SKIP INSTRUCTION IS EXECUTED WITH C(E)=-1 AND AC0. AC0 IS THEN TESTED FOR ALL +;ZEROS. THE TEST FAILS IF AC0 WAS MODIFIED BY THE SKIP INSTRUCTION. + +A4200: MOVE [0] ;PRESET AC0 TO ALL ZEROS + SKIP [-1] ;*SKIP SHOULD NOT MODIFY AC0 + SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS NOT MODIFIED BY SKIP + STOP + +;AC=0: SAC INH, IR BIT 9, 10, 11 OR 12 S-A-1 + +;********** + +;THIS TEST VERIFIES THAT MOVSS DOES NOT MODIFY C(AC0) WHEN AC0 IS SPECIFIED AS THE AC. +;FIRST, E IS LOADED WITH ALL ONES AND THE AC (AC0) +;IS LOADED WITH ALL ZEROS; THEN, MOVSS IS EXECUTED. MOVSS SHOULD NOT MODIFY AC0. +;AC0 IS THEN CHECKED TO INSURE THAT IT WAS NOT MODIFIED BY MOVSS. + +A4300: MOVE 7,[-1] ;LOAD E OF THE MOVSS INSTRUCTION WITH ALL ONES + MOVE [0] ;PRESET AC0 TO ALL ZEROS + MOVSS 7 ;*MOVSS SHOULD NOT MODIFY AC0 + SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS INTACT + STOP + +;MOVX & AC=0: SAC INH [ST2], IR XXXS [IR1] + +;********** +;THIS TEST VERIFIES THAT HRRS DOES NOT MODIFY C(AC0) WHEN AC0 IS THE SPECIFIED AC. +;FIRST, E IS LOADED WITH ALL ONES AND AC0 IS LOADED WITH ALL ZEROS; +;THEN, HRRS IS EXECUTED WITH AC=AC0. AC0 IS THEN CHECKED FOR ALL ZEROS. +;IF AC0 WAS MODIFIED BY THE HRRS INSTRUCTION, THE TEST FAIL. + +A4400: MOVE 7,[-1] ;PRESET E OF THE HRRS INSTRUCTION TO ALL ONES + MOVE [0] ;PRESET AC0 TO ALL ZEROS + HRRS 7 ;*HRRS SHOULD NOT MODIFY THE AC WHEN THE AC IS AC0 + SKIPE ;SKIP THE HALT INSTRUCTION IF AC0 WAS INTACT + STOP + +;(HXXXS&AC=0): SAC INH [ST2], IR HWTFWT [IR3], IR XXXS [IR1], IR HWT [IR1], +;IR BITS S-A-1 + +;********** +;THIS TEST VERIFIES THAT THE JUMP INSTRUCTION DOES NOT MODIFY C(AC). +;THE AC IS FIRST LOADED WITH ALL ZEROS; +;THEN JUMP .+1 IS EXECUTED. THE AC IS THEN CHECKED TO INSURE THAT IT WAS NOT MODIFIED +;BY JUMP. THE TEST FAILS IF THE AC WAS MODIFIED + +A4700: MOVE [0] ; PRESET THE AC TO ALL ZEROS + JUMP .+1 ;*JUMP SHOULD NOT MODIFY THE AC + SKIPE ;SKIP HALT INSTRUCTION IF AC WAS INTACT + STOP + +;********** + +;THIS TEST VERIFIES THAT MOVS WILL NOT PICK UP ANY ONES WHEN SWAPPING A WORD +;OF ALL ZEROS. FIRST, E IS LOADED WITH A WORD OF ALL ZEROS; +;THEN, MOVS IS EXECUTED. THE AC IS THEN CHECKED TO INSURE THAT THE RESULT DOES NOT +;CONTAIN ANY ONES. + +A5000: MOVE [0] ;PRESET E WITH ALL ZEROS + MOVS ;*MOVS SHOULD RESULT WITH AN AC 0F ALL ZEROS + SKIPE ;SKIP HALT INSTRUCTION IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT A MOVS EXECUTED ON AN E CONTAINING A NON-ZERO NUMBER +;RESULTS IN A NON-ZERO NUMBER IN THE AC. FIRST A NON-ZERO WORD +;IS LOADED INTO E; THEN, MOVS IS EXECUTED. THE AC IS CHECKED FOR A NON-ZERO RESULT. +;IF C(AC)=0, THE TEST FAILS. + +A5100: MOVE [XWD 0,-1] ;LOAD E WITH A NON-ZERO VALUE + MOVS ;*MOVS SHOULD RESULT IN C(AC) NON-ZERO + SKIPN ;SKIP HALT INSTRUCTION IF C(AC) OF MOVS IS NON-ZERO + STOP + +;********** +;THIS TEST VERIFIES THAT A MOVS EXECUTED ON AN E CONTAINING A NON-ZERO NUMBER +;RESULTS IN A NON-ZERO NUMBER IN THE AC. FIRST A NON-ZERO WORD +;IS LOADED INTO E; THEN, MOVS IS EXECUTED. THE AC IS CHECKED FOR A NON-ZERO +;RESULT. IF C(AC)=0, THE TEST FAILS. + +A5200: MOVE [XWD -1,0] ;LOAD E OF MOVS WITH A NON-ZERO NUMBER + MOVS ;*MOVS SHOULD RESULT WITH C(AC) NON-ZERO + SKIPN ;SKIP HALT INSTRUCTION IF C(AC) OF MOVS IS NON-ZERO + STOP + +;********** + +;THIS TEST VERIFIES THAT THE MOVS INSTRUCTION SWAPS BITS 0 AND 188 OF E +;AND STORES THEM IN THE AC UNLESS AR BIT 0 IS S-A-0. FIRST, A WORD +;WITH BIT 0 SET TO A ONE AND BIT 18 CLEARED IS LOADED INTO E. THEN, MOVS +;IS EXECUTED. THE AC IS THEN CHECKED FOR A 0 IN BIT 0, WHICH IS THE +;CORRECT RESULT. + +A5300: MOVE [XWD -1,0] ;LOAD E OF MOVS WITH 777777000000 + MOVS ;*MOVS SHOULD RESULT WITH AC BIT 0 EQUAL TO 0 + SKIPG ;SKIP HALT INSTRUCTION IF BIT 0 OF AC IS 0 + STOP + +;MOVSX: AR SWAP [ARMB], AR LT AR RT A [ARMB], AR LT AR RT B [ARMB], +;AR RT AR LT A [ARMB], AR RT AR LT B [ARMB], ET2 A F/F [E], IR MOVSX [IR1], +;IR MOV(E,S)X [IR1] + +;********** + +;THIS TEST VERIFIES THAT THE MOVS INSTRUCTION SWAPS BITS 0 AND 18 OF E AND +;STORES THEM IN THE AC UNLESS AR BIT 0 IS S-A-1. FIRST, A WORD WITH BIT 0 +;SET TO A ONE AND BIT 18 CLEARED IS LOADED INTO E. THEN, MOVS IS EXECUTED. +;THE AC IS CHECKED FOR A ZERO IN BIT ZERO, WHICH IS THE CORRECT RESULT. + +A5400: MOVE [XWD 0,-1] ;LOAD E OF MOVS WITH 777777 + MOVS ;*MOVS SHOULD RESULT WITH BIT 0 OF THE AC SET + SKIPL ;PASS IF AC BIT 0 IS SET + STOP + +;********** +SN=5500 + ZZ=0 + +A5500: REPEAT ^D18, +<;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, +;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. +;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES +;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT +;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [0] ;CLEAR AC + MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + STOP + +;********** +> +SN=5600 + +A5600: REPEAT ^D18, +<;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL +;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS +;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS +;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT +;HALF OF THE AC IS TESTED WITH A 1. + +SN=SN+1 + ZZ=ZZ+ZZ + MOVE [0] ;CLEAR AC + MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + STOP + +;********** +> +SUBTTL TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) + +;THIS TEST VERIFIES THAT SETM LOADS AC FROM E. THE AC IS FIRST SET +;TO ZERO; THEN, SETM IS EXECUTED WITH ALL ONES IN E. BIT 0 OF THE AC +;IS THEN CHECKED FOR A 1. + +A5700: MOVE [0] ;PRELOAD AC WITH ALL ZEROS + SETM [-1] ;*SETM SHOULD LOAD AC WITH ALL ONES + SKIPL ;PASS TEST IF AC BIT 0 IS A ONE + STOP +;FCE [F1], IR BOOLE [IR1], IRXX(0,5,12,17) [IR2] + +;********** + +;THIS TEST VERIFIES THAT SETM LOADS AC FROM E. THE AC IS FIRST SET TO +;ALL ONES; THEN, SETM IS EXECUTED WITH ALL ZEROS IN E. THE AC IS CHECKED +;FOR ALL ZEROS + +A6000: MOVE [-1] ;PRESET AC TO ALL ONES + SETM [0] ;*SETM SHOULD LOAD AC WITH ALL ONES + SKIPE ;PASSES TEST IF C(AC) = 0 + STOP + +;AD AR + EN [ADAP], AD AR + EN C[ADAP], +;F CYC ACT EN A[F1], IR BOOLE AD A +F[IR2], +;IR BOOLE [IR1], IRXX(1-3,6,11,15,16) [IR2] + +;********** + +;THIS TEST VERIFIES THAT SETZ LOADS TH AC WITH ALL ZEROS. FIRST, A WORD +;OF ALL ZEROS IS LOADED INTO THE AC VIA A MOVE INSTC. THEN, SETZ IS +;EXECUTED. THE AC IS CHECKED FOR ALL ZEROS. +;IF ANY BITS IN THE AC ARE SET, THE TEST FAILS. + +A6100: MOVE [0] ;RESET AC TO ALL ZEROS + SETZ ;*SETZ SHOULD CLEAR THE AC + SKIPE ;PASS TEST IF C(AC) = 0 + STOP + +;AD MASK GEN EN [ADC1], AD MASK GEN F/F'S [ADC1] +;AD MASK EN B; F CYC ACT EN B[F1]; IRXX00 [IR1]; +;IR BOOLE [IR1] + +;********** +;THIS TEST VERIFIES THAT SETZ LOADS THE AC WITH ALL ZEROS. FIRST, +;THE AC IS SET TO ALL ONES; THEN, SETZ IS EXECUTED. THE AC IS THEN +;CHECKED FOR ALL ZEROS. IF ANY BITS IN THE AC ARE SET, THE TEST FAILS. + +A6200: MOVE [-1] ;PRESET AC TO ALL ONES + SETZ . ;*SETZ SHOULD CLEAR THE AC + SKIPE ;PASS TEST IF C(AC) = 0 + STOP + +;AR AD EN [ARMA], AR AD ET2 EN [ARMA], IR BOOLE [IR] + +;********** + +;THIS TEST VERIFIES THAT THE AND INSTRUCTION OPERATING ON AN AC +;CONTAINING ALL ZEROS AND AN E CONTAINING ALL ZEROS RESULTS IN AN AC +;OF ALL ZEROS. THE AC AND E ARE THE SAME IN THIS TEST. +;AC, E ARE LOADED WITH ZEROS; THEN, AND IS EXECUTED. THE AC IS THEN +;CHECKED FOR ALL ZEROS. IF ANY BITS IN THE AC ARE SET BY THE AND INSTRUCTION, +;THE TEST FAILS. + + +A6300: MOVE [0] ;PRESET AC, E TO ALL ZEROS + AND ;*AND SHOULD RESULT IN C(AC) = 0 + SKIPE ;PASS TEST IF C(AC) = 0 + STOP + +;********** +SN=6400 + ZZ=0 + +A6400: REPEAT ^D36, +<;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION +;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH +;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X +;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. +;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, +;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS +;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + SETZ ;PRESET E TO ALL ZEROS + MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + ;WHERE X VARIES FROM 0 THRU 35 + AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + SKIPE 1 ;FAIL IF C(AC) NON-ZERO + STOP + SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + STOP +;********** +> +SN=6500 + ZZ=0 + +A6500: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF +;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE +;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS +;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED +;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + ;POSITION WITH ITSELF SHOULD RESULT WITH + ;C(AC) NONZERO + SKIPN ;PASS TEST IF C(AC) NONZERO + STOP + +;********** +> +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE +;C(AC)=0 AND C(E)=0 AND E OUTSIDE OF THE AC RANGE IS ALL ZEROS IN THE AC. +;FIRST, THE AC IS CLEARED; THEN, THE C(AC) IS ANDED WITH C(E) WHEN +;E GREATER THAN 20 AND C(E)=0. THE RESULT SHOULD BE C(AC)=0. +;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A6600: MOVE [0] ;PRESET AC TO ALL ZEROS + AND [0] ;*AND OF TWO WORDS OF ALL ZEROS + ;SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(AC)=0 +;AND C(E) IS ALL ONES IS AN AC CONTAINING ALL ZEROS. FIRST, THE AC IS +;CLEARED; THEN THE C(AC) IS ANDED WITH AN E CONTAINING ALL ONES, WHERE +;E IS BEYOND THE AC RANGE. THE RESULT SHOULD BE C(AC)=0. +;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A6700: MOVE [0] ;PRESET AC TO ALL ZEROS + AND [-1] ;*AND OF TWO WORDS WHERE C(AC)=0 SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;AD FM+EN [ADFP],AD FM+EN A [ADFP], +;F CYC ACT EN C [F1],IR BOOLE AD FM+F [IR2], +;IR BOOLE [IR1],IRXX(1,4,5,11,13,16) [IR2] + +;********** +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(E)=0 +;AND C(AC) IS ALL ONES IS AN AC CONTAINING ALL ZEROS. FIRST, THE +;AC IS SET TO ALL ONES; THEN, THE AC IS ANDED WITH AN E CONTAINING ALL +;ZEROS, WHERE E IS BEYOND THE AC RANGE. THE RESULT SHOULD BE C(AC)=0. +;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A7000: MOVE [-1] ;PRESET AC TO ALL ONES + AND [0] ;*AND OF TWO WORDS WHERE C(E)=0 SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(E) IS +;ALL ONES AND C(AC) IS ALL ONES IS AN AC CONTAINING ALL ONES. FIRST, THE +;AC IS SET TO ALL ONES; THEN, C(AC) IS ANDED WITH AN E CONTAINING ALL +;ONES. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +;IF C(AC) IS ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A7100: MOVE [-1] ;PRELOAD AC WITH ALL ONES + AND [-1] ;*AND OF TWO WORDS OF ALL ONES SHOULD + ;RESULT IN C(AC)=ALL ONES + SKIPN ;PASS TEST IF C(AC) IS NON ZERO + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHERE CONTENTS IS NON-ZERO. +;FIRST, THE AC IS CLEARED, THEN, XOR IS EXECUTED WHERE C(E) IS NON-ZERO. +;THE AC IS THEN CHECKED FOR NON-ZERO CONTENTS. +A7200: MOVE [0] ;CLEAR AC + XOR [707070707070] ;*XOR WITH C(AC)=0 AND C(E)=7070707070 SHOULD + ;RESTULT IN C(AC)=707070707070 + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +SN=7300 + ZZ=0 + +A7300: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS +;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED +;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. +;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [0] ;CLEAR AC + XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + ;RESULT IN C(AC) NONZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +> +SN=7400 + ZZ=0 + +A7400: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE +;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS +;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE +;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E +;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + ;RESULT IN C(AC) NON-ZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], +;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], +;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], +;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + +;********** +> +SN=7500 + ZZ=0 + +A7500: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION +;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, +;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT +;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME +;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST +;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY +;INDICATE AN ILLEGAL CARRY GENERATION. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +> +;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION WHERE +;C(AC)=C(E)=0 IS AN AC CONTAINING ALL ZEROS; FIRST THE AC IS CLEARED; THEN +;XOR IS EXECUTED WITH C(E)=0. THE RESULT IN THE AC IS CHECKED FOR ALL ZEROS + +A7600: MOVE [0] ;INITIALIZE AC TO ALL ZEROS + XOR [0] ;*XOR WITH C(AC)=C(E)=0 SHOULD RESULT INC(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE +;C(AC)=0 AND C(E) ALL ONES IS AN AC CONTAINING ALL ZEROS SINCE +;THE EQUIVALENCE FUNCTION SHOULD SET ONLY LIKE BITS. FIRST, THE AC +;SHOULD BE CLEARED; THEN, EQV IS EXECUTED WITH E CONTAINING ALL ONES. +;THE AC IS THEN CHECKED FOR ALL ZEROS + +A7700: MOVE [0] ;CLEAR THE AC + EQV [-1] ;*EQV WITH C(AC)=0 AND C(E) ALL ONES SHOULD + ;RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +SN=10000 + ZZ=0 + +A10000: REPEAT ^D18, +<;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN +;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE +;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT +;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS +;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE +;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO +SN=SN+1 + ZZ=ZZ+ZZ+1 + IFE , + MOVE [0] ;PRESET AC TO ALL ZEROS + EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + ;RESULT IN C(AC) NON-ZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +> +SN=10100 + ZZ=0 + +A10100: REPEAT ^D18, +<;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN +;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE +;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT +;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS +;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE +;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO +SN=SN+1 + ZZ=ZZ+ZZ+1 + IFE , + MOVE [0] ;PRESET AC TO ALL ZEROS + EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + ;RESULT IN C(AC) NON-ZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +> +;THIS TEST VERIFIES THAT THE RESULT OF AN AC CONTAINING ALL ZEROS +;EQUIVALENCED WITH AN E CONTAINING ALL ZEROS IS AN AC CONTAINING ALL +;ONES; AND FUTHERMORE, IF THIS AC CONTAINING ALL ONES IS EQUIVALANCED +;WITH AN E CONTAINING ALL ZEROS, THE FINAL RESULT IN THE AC +;SHOULD BE ALL ZEROS. FIRST, THE AC IS INITIALIZED TO ALL ZEROS, THEN, +;THIS AC IS EQUIVALENCED TWICE WITH AN E CONTAINING ALL ZEROS. THE AC IS +;THEN CHECKED FOR ALL ZEROS + +A10200: MOVE [0] ;CLEAR THE AC + EQV [0] ;*EQV SHOULD RESULT IN C(AC)=ALL ONES + EQV [0] ;*EQV SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF AN AC CONTAINING ALL ONES +;EQUIVALENCED WITH AN E CONTAINING ALL ONES IS AN AC CONTAINING ALL ONES; +;AND FUTHERMORE, IF THIS AC CONTAINING ALL ONES ISEQUIVALENCED WITH AN E +;CONTAINING ALL ZEROS, THE FINAL RESULT IN THE AC SHOULD BE ALL ZEROS. +;FIRST, THE AC IS INITIALIZED TO ALL ONES; THEN, THIS AC IS +;EQUIVALENCED WITH AN E OF ALL ONES. NEXT, THIS SAME AC IS EQUIVALENCED +;WITH AN E OF ALL ZEROS. THE AC IS THEN CHECKED FOR ALL ZEROS. + +A10300: MOVE [-1] ;SET C(AC) TO ALL ONES + EQV [-1] ;*EQV SHOULD RESULT IN C(AC) OF ALL ONES + EQV [0] ;*EQV SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +SN=13000 + ZZ=0 + +A13000: REPEAT ^D36, +<;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. +;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN +;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). +;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD +;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, +;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. +;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO +;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + EQV [0] ;RESULTS IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +> +ENDXX: JRST BEGEND ;LOOP PROGRAM + +``` + +DAKAA.SEQ +--------- + +To view the contents of **DAKAA.SEQ** (a combination of [DAKAA.TXT](#dakaatxt), a MACRO-10 assembly listing, and run data), +click [here](DAKAA.SEQ.txt). diff --git a/apps/pdp10/tapes/diags/klad/dakaa/TEST.MAC b/apps/pdp10/tapes/diags/klad/dakaa/TEST.MAC new file mode 100644 index 000000000..818244639 --- /dev/null +++ b/apps/pdp10/tapes/diags/klad/dakaa/TEST.MAC @@ -0,0 +1,1054 @@ +;MACROS + +; STOP - USED FOR SCOPE LOOP, IF INSTRUCTION FAILS, CHANGE (JUMPA .+1) +; TO A (JUMPA X) TO CYCLE ON FAILING INSTRUCTION + +DEFINE STOP (A)< + HALT .+1 ;TEST FAILED IF PROGRAM HALTS HERE + JUMPA .+1 ;IF TEST FAILS, CHANGE THIS INSTRUCTION (JUMPA .+1) + ;TO JUMPA X(X IS THE ADDRESS OF THE FIRST + ;INSTRUCTION IN THE SUBTEST) TO LOOP ON ERROR + ;AND CHANGE HALT INSTRUCTION TO JUMPA .+1> + +SUBTTL DIAGNOSTIC SECTION + +START: +; SETZM USER# ;CLEAR USER CONTROL WORD +; JSP 0,.+1 ;GET FLAGS +; TLNE USERF ;IN USER MODE? +; SETOM USER ;YES, SET USER CONTROL WORD +; SKIPN MONFLG ;SPECIAL USER MODE? +; SETZM USER ;YES, CLEAR USER CONTROL WORD +; SKIPN USER +; JRST STARTA +; SKIPL MONCTL +; TTCALL 3,PGMNAM ;MENTION OUR NAME + JRST STARTA + +PGMNAM: ASCIZ/ +PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC (1) [DAKAA] +/ + +;BASIC INSTRUCTION TEST (3) +;THE TEST IS DESIGNED FOR INITIAL DEBUGGING OF +;PROCESSOR HARDWARE AND TO DETECT (SOLID) FAILURES +;IN THE FIELD. + + +STARTA: JRST .+1 +SUBTTL TEST OF JUMP, JUMPA AND SKIPX INSTRUCTIONS + +;NOTE: AN "*" IN THE COMMENT FIELD OF AN INSTRUCTION INDICATES THAT IT IS THE TESTED +;INSTRUCTION. +;********** + +;THIS TEST VERIFIES THAT SKIPA ALWAYS SKIPS THE NEXT INSTRUCTION + +A00=. +A12500: SKIPA ;*SKIPA SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + HALT ;IF PROGRAM HALTS, SKIPA DID NOT SKIP + +;IF PROGRAM HANGS UP ON SKIPA INSTRUCTION, CHECK AB PC EN [ABC], AB PC F/F [ABC], +;AB PC B [ABC OR AB], AB PC (FETCH) EN [ABC], IR SKIPS [IR3], IR SKIPX [IR1], +;IR 3XX [IR1], IR BITS 00 - 08 [IR1], F CYC ACT EN C [FI], FT6 F/F [F2], +;PC CLOCK EN [PCC], ET2 J F/F [E], PC CLK (ET2) EN [PCC], ADZ COND P [ADZ] AND +;ADZ COND R [ADZ], PC CLOCK A [PCC], PC CLOCK B [PCC] + +;IF PROGRAM HALTED ON HALT INSTRUCTION, CHECK ST1 COND [ST2] + +;********** + +;THIS TEST VERIFIES THAT JUMP NEVER JUMPS + +A15000: JUMP .+1 ;*JUMP SHOULD NEVER JUMP + SKIPA ;SKIP HALT INSTRUCTION IF JUMP IS SUCCESSFUL + HALT ;HALT IF JUMP FAILS + +;IF PROGRAM HANGS UP ON JUMP .+1 INSTRUCTION, THE JUMP INSTRUCTION +;MAY HAVE FAILED. CHECK ST1 COND [ST2], ST INST FET ST1 EN [ST2], ET2 D F/F [E], +;IR JUMPS [IR1], IR JUMPX [IR1] + +;********** +;THIS TEST VERIFIES THAT JUMP NEVER JUMPS + +A15100: JUMP .+2 ;*JUMP SHOULD NEVER JUMP + SKIPA ;SKIP HALT INSTRUCTION IF JUMP IS SUCCESSFUL + HALT ;HALT IF JUMP FAILS + +;AB PC EN [ABC] + +;********** + +;THIS TEST VERIFIES THAT JUMPA .+1 ALWAYS JUMPS TO THE NEXT INSTRUCTION + +A12700: JUMPA .+1 ;*JUMPA .+1 SHOULD NEVER JUMP + SKIPA ;SKIP HALT INSTRUCTION IF JUMPA IS SUCCESSFUL + HALT ;HALT IF JUMPA .+1 FAILED + +;********** + +;THIS TEST VERIFIES THAT JUMPA JUMPS TO THE LOCATION SPECIFIED BY E WHEN E=.+2 + +A12600: JUMPA .+2 ;*JUMPA .+2 SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + HALT ;PROGRAM HALTS HERE IF JUMPA .+2 FAILS + +;ST1 COND [ST2] + +;********** +;THIS TEST VERIFIES THAT SKIPA IS DATA INDEPENDENT. HENCE, IT ALWAYS SKIPS THE NEXT +;INSTRUCTION. +;THE MOVE [0] INSTRUCTION IS USED TO TEST THE DATA INDEPENDENCE OF SKIPA. + +A100: MOVE [0] ;PRESET LOCATION 0 TO ALL ZEROS + SKIPA ;*SKIPA SHOULD ALWAYS SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST INSURES THAT SKIPA IS DATA INDEPENDENT. HENCE, IT ALWAYS SKIPS THE NEXT +;INSTRUCTION. +;THE MOVE [-1] INSTRUCTION IS USED TO TEST THE DATA INDEPENDENCE OF SKIPA. + +A200: MOVE [-1] ;PRESET LOCATION 0 TO ALL ONES + SKIPA ;*SKIPA SKOULD ALWAYS SKIP THE NEXT INSTRUCTION + STOP + +;COND P [ADZ] + +;********** +;THIS TEST VERIFIES THAT SKIPGE SKIPS THE NEXT INSTRUCTION WHEN C(E) +;IS ALL ZEROS. THE MOVE INSTRUCTION IS USED TO LOAD E WITH ALL ZEROS. + +A300: MOVE [0] ;PRESET E TO ALL ZEROS + SKIPGE ;*SKIPGE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;COND R [ADZ], AD00 S-A-1, AD AR + EN [ADAP], AD LT AR + EN [ADAP], +;AD RT AR + EN [ADAP], AD AR + EN C [ADAP], F CYC ACT EN A [F1] + +;********** + +;THIS TEST VERIFIES THAT SKIP IS DATA INDEPENDENT. HENCE, SKIP NEVER SKIPS. IT ALWAYS +;CONTINUES ON TO THE NEXT INSTRUCTION. + +A400: MOVE [-1] ;PRESET E TO ALL ONES + SKIP ;*SKIP SHOULD NEVER SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;PC CHANGE [PCC], PC CHANGE A [PCC], SEE MEMORY CONTROL DWGS [MC1-4] + +;********** +;THIS TEST VERIFIES THAT SKIPGE WILL NOT SKIP THE NEXT INSTRUCTION WHEN C(E) IS +;NEGATIVE. THE MOVE INSTRUCTION LOADS E WITH -1, A NEGATIVE NUMBER. + +A500: MOVE [-1] ;PRESET E WITH ALL ONES, A NEGATIVE NUMBER + SKIPGE ;*SKIPGE SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;IF MOVE INSTRUCTION FAILED, CHECK MOVX-: FCE [F1], -IR XXXI [IR1], -IR XXXM [IR1], +;IR HWTFWT [IR3], IR FWT [IR2], IR2XX [IR1] + +;IF SKIPGE FAILED, CHECK SKIPX: FCE [F1] + +;********** + +;THIS TEST VERIFIES THAT SKIPN WILL NOT SKIP THE NEXT INSTRUCTION WHEN +;C(E) IS ZERO. THE MOVE INSTRUCTION LOADS E WITH ALL ZEROS. + +A600: MOVE [0] ;PRESET E WITH ALL ZEROS + SKIPN ;*SKIPN SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT SHOULD SKIP +;THE NEXT INSTRUCTION WHEN C(E) IS NON-ZERO. THE MOVE INSTRUCTION LOADS +;E WITH -1, A NON-ZERO NUMBER. + +A700: MOVE [-1] ;PRESET E WITH ALL ONES, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPE IS DATA DEPENDENT. HENCE, IT SHOULD NOT +;SKIP THE NEXT INSTRUCTION WHEN C(E) IS NON-ZERO. THE MOVE INSTRUCTION LOADS E +;WITH -1, A NON-ZERO NUMBER. + +A1000: MOVE [-1] ;PRESET E WITH ALL ONES, A NON-ZERO NUMBER + SKIPE ;*SKIPE SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPE IS DATA DEPENDENT. HENCE, IT SHOULD SKIP +;THE NEXT INSTRUCTION WHEN C(E) IS ZERO. THE MOVE INSTRUCTION LOADS ALL +;ZEROS INTO E. + +A1100: MOVE [0] ;PRESET E TO ALL ZEROS + SKIPE ;*SKIPE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIP IS DATA INDEPENDENT. HENCE, IT NEVER SKIPS. IT ALWAYS +;CONTINUES ON TO THE NEXT INSTRUCTION. + +A1200: MOVE [0] ;PRESET E TO ALL ZEROS + SKIP ;*SKIP SHOULD NEVER SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NOT NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD +;E WITH ALL ZEROS, A NON-NEGATIVE NUMBER. + +A1300: MOVE [0] ;PRESET E T0 ZERO, A NON-NEGATIVE NUMBER + SKIPL ;*SKIPL SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD E WITH -1, +;A NEGATIVE NUMBER. + +A1400: MOVE [-1] ;PRESET E TO -1, A NEGATIVE NUMBER + SKIPL ;*SKIPL SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPG IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS GREATER THAN ZERO. THE MOVE INSTRUCTION IS USED TO LOAD +;E WITH A NON-ZERO POSITIVE NUMBER. + +A1500: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A NON-ZERO POSITIVE NUMBER + SKIPG ;*SKIPG SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPG IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NOT GREATER THAN ZERO. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH -1, A NUMBER NOT GREATER THAN ZERO. + +A1600: MOVE [-1] ;PRESET E TO -1, A NEGATIVE NUMBER + SKIPG ;*SKIPG SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPL IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS NON-NEGATIVE. THE MOVE INSTRUCTION IS USED TO LOAD E WITH +;377777777777, A NON-NEGATIVE NUMBER. + +A1700: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A NON-NEGATIVE NUMBER + SKIPL ;*SKIPL SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN C(E) IS ZERO. THE MOVE INSTRUCTION LOADS E WITH ALL ZEROS. + +A2000: MOVE [0] ;PRESET E TO ALL ZEROS + SKIPLE ;*SKIPLE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NEGATIVE NUMBER. THE MOVE INSTRUCTION IS USED TO LOAD +;E WITH 400000000000, A NEGATIVE NUMBER. + +A2100: MOVE [XWD 400000,0] ;PRESET E TO 400000000000, A NEGATIVE NUMBER. + SKIPLE ;*SKIPLE SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPLE IS DATA DEPENDENT. HENCE, IT WILL NOT SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A POSITIVE NUMBER. THE MOVE INSTRUCTION LOADS +;E WITH 377777777777, A POSITIVE NUMBER. + +A2200: MOVE [XWD 377777,-1] ;PRESET E TO 377777777777, A POSITIVE NUMBER + SKIPLE ;*SKIPLE SHOULD NOT SKIP THE NEXT INSTRUCTION + SKIPA ;SKIP HALT INSTRUCTION IF TEST PASSES + STOP + +;********* +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION ;WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD 777000000000, A NON-ZERO NUMBER INTO E. + +A2300: MOVE [XWD 777000,0] ;PRESET E WITH 777000000000, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD A NON-ZERO NUMBER, 377000000 INTO E. + +A2400: MOVE [XWD 377,0] ;PRESET E TO 377000000, A NON ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH A NON-ZERO NUMBER, 177000. + +A2500: MOVE [177000] ;PRESET E WITH 177000, A NON ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD 777, A NON-ZERO NUMBER INTO E. + +A2600: MOVE [777] ;PRESET E WITH 777, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH 600000, A NON-ZERO NUMBER. + +A2700: MOVE [600000] ;PRESET E WITK 600000, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPN IS DATA DEPENDENT. HENCE, IT WILL SKIP THE NEXT +;INSTRUCTION WHEN E CONTAINS A NON-ZERO NUMBER. THE MOVE INSTRUCTION IS USED TO +;LOAD E WITH 400000000, A NON-ZERO NUMBER. + +A3000: MOVE [XWD 400,000] ;PRESET E WITH 400000000, A NON-ZERO NUMBER + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +SN=3100 + ZZ=0 + +A3100: REPEAT ^D36, +<;THIS TEST VERIFIES THAT SKIPN IS SENSITIVE TO A ONE IN EVERY BIT POSITION. THIS TEST +;IS REPEATED 36 TIMES. EACH TIME A WORD OF ALL ZEROS EXCEPT FOR A ONE IN ONE +;BIT POSITION IS MOVED INTO E. THEN SKIPN IS EXECUTED. THIS TEST ALSO +;CHECKS ADDER GATING. IF THIS TEST FAILS, CHECK C(E). +;IF C(E)=0, MOVE FAILED; OTHERWISE SKIPN FAILED. + +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;MOVE THE CURRENT VALUE OF ZZ INTO E. ZZ IS NON-ZERO + SKIPN ;*SKIPN SHOULD SKIP THE NEXT INSTRUCTION + STOP + +;********** +> +SUBTTL TEST OF MOVE, SKIP AND COMPARE INSTRUCTIONS + +;THIS TEST VERIFIES THAT MOVEI LOADS E INTO THE AC. SKIPG IS USED TO CHECK THAT +;THE AC WAS INDEED MODIFIED BY MOVEI. + +A3200: MOVEI [-1] ;*MOVEI SHOULD LOAD A POSITIVE NUMBER INTO AC0 + SKIPG ;SKIP HALT INSTRUCTION IF MOVEI LOADED AC CORRECTLY + STOP + +;IF PROGRAM HANGS UP ON MOVEI INSTRUCTION, CHECK AD FM +EN [ADFP], AD FM + EN A [ADFP], +;AD FM + F/F [ADFP], F CYC ACT EN C [F1] + +;********** + +;THIS TEST VERIFIES THAT SKIPL IS SENSITIVE TO THE DATA IN E AND NOT SENSITIVE TO +;THE DATA IN THE AC. HENCE, SKIPL WILL SKIP THE NEXT INSTRUCTION ONLY IF +;C(E) IS LESS THAN ZERO. E IS SET TO -1, A NEGATIVE NUMBER; AND +;THE AC IS SET TO 0 IN THIS TEST. THEREFORE SKIPL SHOULD SKIP. + +A3300: MOVE [0] ;SET THE AC TO ALL ZEROS + SKIPL [-1] ;*SKIPL SHOULD SKIP BECAUSE C(E) IS NEGATIVE + STOP + +;********** + +;THIS TEST VERIFIES THAT SKIPL IS SENSITIVE TO THE SIGN BIT IN E. SKIPL SHOULD +;SKIP THE NEXT INSTRUCTION WHEN AND ONLY WHEN THE SIGN BIT IS SET. IN THIS +;TEST, C(AC) IS PRESET TO ALL ZEROS, THEN SKIPL IS EXECUTED WITH C(E)=400000,,0. +;A FAILURE UCCURS IF SKIPL DOES NOT SKIP. + +A13100: MOVE [0] ;SET THE AC TO ALL ZEROS + SKIPL [400000,,0] ;*SKIPL SHOULD SKIP BECAUSE C(E) IS NEGATIVE + STOP + +;********** +;THIS TEST VERIFIES THAT THE CAM INSTRUCTION DOES NOT MODIFY C(AC). THE AC IS +;INITIALLY LOADED WITH ALL ZEROS; THEN CAM IS EXECUTED WITH C(E) NON-ZERO. +;THE AC IS THEN CHECKED FOR ALL ZEROS. THIS TEST FAILS IF THE AC WAS MODIFIED BY CAM. + +A3400: MOVE [0] ;PRESET THE AC TO ALL ZEROS + CAM [1234] ;*CAM SHOULD NOT MODIFY THE AC + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION SERVES AS A NO-OP. + ;IT IS A PROGRAM FILLER IN CASE CAM SKIPS.) + SKIPE ;SKIP HALT INSTRUCTION IF AC WAS NOT MODIFIED BY CAM + STOP + +;IF TEST HANGS ON CAM INSTRUCTION, AB PC EN [ABC], AB PC (FETCH) EN [ABC], +;IR CAXX [IR3] + +;IF TEST LOOPS ON CAM INSTRUCTION, MC INSTR FETCH EN [MC1], MC INST FET ET2 EN [MC1], +;ET2 J F/F [E], IR CAXX [IR3] + +;IF TEST HALTS AT HALT INSTRUCTION, SAC INH [ST2], IR CAXX [IR3], IR CAMX [IR1] + +;********** + +;THIS TEST VERIFIES THAT THE CAI INSTRUCTION DOES NOT MODIFY C(AC). THE AC IS +;INITIALLY LOADED WITH ALL ZEROS; THEN CAI IS EXECUTED WITH E NON-ZERO. THE AC IS +;THEN CHECKED FOR ALL ZEROS. THIS TEST FAILS IF THE AC WAS MODIFIED BY CAM. + +A3500: MOVE [0] ;PRESET THE AC TO ALL ZEROS + CAI [1234] ;*CAI SHOULD NOT MODIFY THE AC + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A PROGRAM FILLER + ;IN CASE CAI MODIFIES THE AC. IT SERVES AS A NO-OP) + SKIPE ;SKIP HALT INSTRUCTION IF CAI DID NOT MODIFY THE AC + STOP + +;********** +;THIS TEST VERIFIES THAT CAI DOES NOT MODIFY C(AC0) WHEN AN AC OTHER THAN AC0 IS +;SPECIFIED BY THE CAI INSTRUCTION. +;THIS IS THE FIRST TEST THAT USES AN AC OTHER THAN AC0. +;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, THE CAI INSTRUCTION IS EXECUTED WITH +;A NON-ZERO E. AC 0 IS THEN TESTED F0R ALL ZEROS. + +A3600: MOVE [0] ;LOAD AC0 WITH ALL ZEROS + CAI 17,[1234] ;*CAI SHOULD NOT MODIFY AC0 + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + ;CAI SKIPS THE NEXT INSTRUCTION. IT ACTS AS A NO-OP) + SKIPE ;SKIP HALT INSTRUCTION IF C(AC0) WAS NOT MODIFIED + STOP + +;********** + +;THIS TEST VERIFIES THAT CAM DOES NOT MODIFY C(AC0) WHEN AN AC OTHER THAN AC0 IS +;SPECIFIED BY THE CAM INSTRUCTION. +;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, THE CAM INSTRUCTION IS EXECUTED WITH +;A NON-ZERO E. AC 0 IS THEN TESTED F0R ALL ZEROS. + +A3700: MOVE [0] ;PRESET C(AC0) TO ALL ZEROS + CAM 17,[1234] ;*CAM SHOULD NOT MODIFY AC0 + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + ;CAM SKIPS THE NEXT INSTRUCTION. IT SERVES AS A NO-OP) + SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS NOT MODIFIED BY CAM + STOP + +;********** +;THIS TEST VERIFIES THAT MOVE WILL PLACE C(E) INTO THE AC. AC, E AND C(E) ARE NON-ZERO +;IN THIS TEST. THIS TEST PASSES IF FINAL C(AC) OF THE MOVE INSTRUCTION IS NON-ZERO. + +A4500: MOVE 1,[0] ;LOAD THE AC WITH ALL ZEROS + MOVE 1,[-1] ;*MOVE SHOULD PLACE ALL ONES IN THE AC + SKIPN 1 ;SKIP HALT INSTRUCTION IF MOVE MODIFIED THE AC + STOP + +;********** + +;THIS TEST VERIFIES THAT THE MOVE INSTRUCTION WILL PLACE C(E) INTO THE AC. +;AC, E ARE NON-ZERO AND C(E)=0. +;THIS TEST FAILS IF THE FINAL C(AC) IS NOT EQUAL TO ZERO, THE ORIGINAL C(E) + +A4600: MOVE 1,[0] ;*PLACE ALL ZEROS INTO THE AC + SKIPE 1 ;SKIP HALT INSTRUCTION IF MOVE PLACED ALL ZEROS INTO AC + STOP + +;IF JUMP .+1 OF PREVIOUS TEST FAILED, CHECK JUMPS: AD FM + EN [ADFP], IR JUMPS [IR1], +;IR JUMPX [IR1], AR AD EN [ARMA], AR AD EN A [ARMA], AR LT AD EN A [ARMA], +;AR LT ADD EN B [ARMA], AR RT AD EN A [ARMA], +;AR RT AD EN B [ARMA], AR A CLK INPUT [ARMA], +;ET2 C F/F [E], AR AD ET2 EN [ARMA], AR AD BR (ET2) B [ARMA] + +;********** +;THIS TEST VERIFIES THAT SOS DOES NOT MODIFY C(AC) WHEN THE AC IS AC0. +;FIRST, AC0 IS LOADED WITH ALL ZEROS AND E IS LOADED WITH A POSITIVE NUMBER; +;THEN, SOS IS EXECUTED. SOS SHOULD SUBTRACT ONE FROM C((E) +;AND EXECUTE THE NEXT INSTRUCTION. IT SHOULD NOT MODIFY AC0. +;THE TEST PASSES IF AC0 WAS NOT MODIFIED. + +A4000: MOVE [0] ;PRESET AC0 TO ALL ZEROS + MOVE 7,[1234] ;PRESET E WITH A POSITIVE NUMBER + SOS 7 ;*SOS SHOULD NOT MODIFY C(AC0) + MOVE ;MOVE C(0) INTO 0 (THIS INSTRUCTION IS A FILLER IN CASE + ;SOS SKIPS THE NEXT INSTRUCTION. IT SERVES AS A NO-OP) + SKIPE ;SKIP THE HALT INSTRUCTION IF C(AC0) WAS NOT MODIFIED + STOP + +;********** +SN=4100 + AC=0 + +A4100: REPEAT 4, +<;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL PLACE THE CONTENTS OF E INTO THE AC +;WHEN THE AC IS NOT AC0. THIS TEST IS PERFORMED FOR AC1, AC2, AC4 AND AC10. +;FIRST, THE AC IS LOADED WITH ALL ZEROS; THEN, THE SKIP INSTRUCTION IS EXECUTED +;WITH AC1, AC2, AC4 OR AC10. C(E) IS EQUAL TO -1. THE SKIP INSTRUCTION SHOULD CAUSE +;THE AC TO CONTAIN -1. THE AC IS THEN CHECKED FOR +;NON-ZERO CONTENTS. THE TEST FAILS IF C(AC)=0. +;THIS TEST ALSO VERIFIES THAT 'SKIP AC,[-1]' DOES NOT CLOBBER AC0. + +SN=SN+1 + AC=AC+AC + IFE AC, + MOVE 0,[0] ;PRESET AC0 TO ALL ZEROS + MOVE AC,[0] ;PRESET THE AC TO ALL ZEROS + SKIP AC,[-1] ;*SKIP SHOULD PLACE -1 INTO THE AC + SKIPN AC ;SKIP THE HALT INSTRUCTION IF THE AC IS NON-ZERO + STOP + SKIPE 0 ;VERIFY THAT AC0 DID NOT GET CLOBBERED BY SKIP + STOP + +;********** +> +SUBTTL TEST OF SKIP, FULL WORD TRANSFER AND HALF WORD TRANSFER INSTRUCTIONS + +;THIS TEST VERIFIES THAT THE SKIP INSTRUCTION WILL NOT MODIFY THE CONTENTS OF AC0. +;FIRST, AC0 IS LOADED WITH ALL ZEROS; THEN, +;THE SKIP INSTRUCTION IS EXECUTED WITH C(E)=-1 AND AC0. AC0 IS THEN TESTED FOR ALL +;ZEROS. THE TEST FAILS IF AC0 WAS MODIFIED BY THE SKIP INSTRUCTION. + +A4200: MOVE [0] ;PRESET AC0 TO ALL ZEROS + SKIP [-1] ;*SKIP SHOULD NOT MODIFY AC0 + SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS NOT MODIFIED BY SKIP + STOP + +;AC=0: SAC INH, IR BIT 9, 10, 11 OR 12 S-A-1 + +;********** + +;THIS TEST VERIFIES THAT MOVSS DOES NOT MODIFY C(AC0) WHEN AC0 IS SPECIFIED AS THE AC. +;FIRST, E IS LOADED WITH ALL ONES AND THE AC (AC0) +;IS LOADED WITH ALL ZEROS; THEN, MOVSS IS EXECUTED. MOVSS SHOULD NOT MODIFY AC0. +;AC0 IS THEN CHECKED TO INSURE THAT IT WAS NOT MODIFIED BY MOVSS. + +A4300: MOVE 7,[-1] ;LOAD E OF THE MOVSS INSTRUCTION WITH ALL ONES + MOVE [0] ;PRESET AC0 TO ALL ZEROS + MOVSS 7 ;*MOVSS SHOULD NOT MODIFY AC0 + SKIPE ;SKIP HALT INSTRUCTION IF AC0 WAS INTACT + STOP + +;MOVX & AC=0: SAC INH [ST2], IR XXXS [IR1] + +;********** +;THIS TEST VERIFIES THAT HRRS DOES NOT MODIFY C(AC0) WHEN AC0 IS THE SPECIFIED AC. +;FIRST, E IS LOADED WITH ALL ONES AND AC0 IS LOADED WITH ALL ZEROS; +;THEN, HRRS IS EXECUTED WITH AC=AC0. AC0 IS THEN CHECKED FOR ALL ZEROS. +;IF AC0 WAS MODIFIED BY THE HRRS INSTRUCTION, THE TEST FAIL. + +A4400: MOVE 7,[-1] ;PRESET E OF THE HRRS INSTRUCTION TO ALL ONES + MOVE [0] ;PRESET AC0 TO ALL ZEROS + HRRS 7 ;*HRRS SHOULD NOT MODIFY THE AC WHEN THE AC IS AC0 + SKIPE ;SKIP THE HALT INSTRUCTION IF AC0 WAS INTACT + STOP + +;(HXXXS&AC=0): SAC INH [ST2], IR HWTFWT [IR3], IR XXXS [IR1], IR HWT [IR1], +;IR BITS S-A-1 + +;********** +;THIS TEST VERIFIES THAT THE JUMP INSTRUCTION DOES NOT MODIFY C(AC). +;THE AC IS FIRST LOADED WITH ALL ZEROS; +;THEN JUMP .+1 IS EXECUTED. THE AC IS THEN CHECKED TO INSURE THAT IT WAS NOT MODIFIED +;BY JUMP. THE TEST FAILS IF THE AC WAS MODIFIED + +A4700: MOVE [0] ; PRESET THE AC TO ALL ZEROS + JUMP .+1 ;*JUMP SHOULD NOT MODIFY THE AC + SKIPE ;SKIP HALT INSTRUCTION IF AC WAS INTACT + STOP + +;********** + +;THIS TEST VERIFIES THAT MOVS WILL NOT PICK UP ANY ONES WHEN SWAPPING A WORD +;OF ALL ZEROS. FIRST, E IS LOADED WITH A WORD OF ALL ZEROS; +;THEN, MOVS IS EXECUTED. THE AC IS THEN CHECKED TO INSURE THAT THE RESULT DOES NOT +;CONTAIN ANY ONES. + +A5000: MOVE [0] ;PRESET E WITH ALL ZEROS + MOVS ;*MOVS SHOULD RESULT WITH AN AC 0F ALL ZEROS + SKIPE ;SKIP HALT INSTRUCTION IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT A MOVS EXECUTED ON AN E CONTAINING A NON-ZERO NUMBER +;RESULTS IN A NON-ZERO NUMBER IN THE AC. FIRST A NON-ZERO WORD +;IS LOADED INTO E; THEN, MOVS IS EXECUTED. THE AC IS CHECKED FOR A NON-ZERO RESULT. +;IF C(AC)=0, THE TEST FAILS. + +A5100: MOVE [XWD 0,-1] ;LOAD E WITH A NON-ZERO VALUE + MOVS ;*MOVS SHOULD RESULT IN C(AC) NON-ZERO + SKIPN ;SKIP HALT INSTRUCTION IF C(AC) OF MOVS IS NON-ZERO + STOP + +;********** +;THIS TEST VERIFIES THAT A MOVS EXECUTED ON AN E CONTAINING A NON-ZERO NUMBER +;RESULTS IN A NON-ZERO NUMBER IN THE AC. FIRST A NON-ZERO WORD +;IS LOADED INTO E; THEN, MOVS IS EXECUTED. THE AC IS CHECKED FOR A NON-ZERO +;RESULT. IF C(AC)=0, THE TEST FAILS. + +A5200: MOVE [XWD -1,0] ;LOAD E OF MOVS WITH A NON-ZERO NUMBER + MOVS ;*MOVS SHOULD RESULT WITH C(AC) NON-ZERO + SKIPN ;SKIP HALT INSTRUCTION IF C(AC) OF MOVS IS NON-ZERO + STOP + +;********** + +;THIS TEST VERIFIES THAT THE MOVS INSTRUCTION SWAPS BITS 0 AND 188 OF E +;AND STORES THEM IN THE AC UNLESS AR BIT 0 IS S-A-0. FIRST, A WORD +;WITH BIT 0 SET TO A ONE AND BIT 18 CLEARED IS LOADED INTO E. THEN, MOVS +;IS EXECUTED. THE AC IS THEN CHECKED FOR A 0 IN BIT 0, WHICH IS THE +;CORRECT RESULT. + +A5300: MOVE [XWD -1,0] ;LOAD E OF MOVS WITH 777777000000 + MOVS ;*MOVS SHOULD RESULT WITH AC BIT 0 EQUAL TO 0 + SKIPG ;SKIP HALT INSTRUCTION IF BIT 0 OF AC IS 0 + STOP + +;MOVSX: AR SWAP [ARMB], AR LT AR RT A [ARMB], AR LT AR RT B [ARMB], +;AR RT AR LT A [ARMB], AR RT AR LT B [ARMB], ET2 A F/F [E], IR MOVSX [IR1], +;IR MOV(E,S)X [IR1] + +;********** + +;THIS TEST VERIFIES THAT THE MOVS INSTRUCTION SWAPS BITS 0 AND 18 OF E AND +;STORES THEM IN THE AC UNLESS AR BIT 0 IS S-A-1. FIRST, A WORD WITH BIT 0 +;SET TO A ONE AND BIT 18 CLEARED IS LOADED INTO E. THEN, MOVS IS EXECUTED. +;THE AC IS CHECKED FOR A ZERO IN BIT ZERO, WHICH IS THE CORRECT RESULT. + +A5400: MOVE [XWD 0,-1] ;LOAD E OF MOVS WITH 777777 + MOVS ;*MOVS SHOULD RESULT WITH BIT 0 OF THE AC SET + SKIPL ;PASS IF AC BIT 0 IS SET + STOP + +;********** +SN=5500 + ZZ=0 + +A5500: REPEAT ^D18, +<;THIS TEST VERIFIES THAT AR LT INPUT IS FUNCTIONING. A WORD OF ALL ZER0S, +;EXCEPT FOR ONE ONE IN THE RIGHT HALF IS SWAPPED. +;THE AC IS CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES +;IF C(AC) IS NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT +;OF THE LEFT HALF OF THE AC IS TESTED WITH A 1. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [0] ;CLEAR AC + MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + STOP + +;********** +> +SN=5600 + +A5600: REPEAT ^D18, +<;THIS TEST VERIFIES THAT AR RT INPUT IS FUNCTIONING. A WORD OF ALL +;ZEROS, EXCEPT FOR ONE ONE IN THE LEFT HALF IS SWAPPED. THE AC IS +;CHECKED FOR ITS CONTENTS NON-ZERO. THE TEST PASSES IF C(AC) IS +;NON-ZERO. THIS TEST IS REPEATED 18 TIMES SO THAT EACH BIT OF THE RIGHT +;HALF OF THE AC IS TESTED WITH A 1. + +SN=SN+1 + ZZ=ZZ+ZZ + MOVE [0] ;CLEAR AC + MOVS [ZZ] ;*MOVS SHOULD RESULT WITH A NON-ZERO WORD IN THE AC + SKIPN ;SKIP HALT INSTRUCTION IF SET BIT IS NOT LOST IN SWAP. + STOP + +;********** +> +SUBTTL TEST OF BOOLEAN INSTRUCTIONS (SETM, SETZ, AND, XOR, EQV) + +;THIS TEST VERIFIES THAT SETM LOADS AC FROM E. THE AC IS FIRST SET +;TO ZERO; THEN, SETM IS EXECUTED WITH ALL ONES IN E. BIT 0 OF THE AC +;IS THEN CHECKED FOR A 1. + +A5700: MOVE [0] ;PRELOAD AC WITH ALL ZEROS + SETM [-1] ;*SETM SHOULD LOAD AC WITH ALL ONES + SKIPL ;PASS TEST IF AC BIT 0 IS A ONE + STOP +;FCE [F1], IR BOOLE [IR1], IRXX(0,5,12,17) [IR2] + +;********** + +;THIS TEST VERIFIES THAT SETM LOADS AC FROM E. THE AC IS FIRST SET TO +;ALL ONES; THEN, SETM IS EXECUTED WITH ALL ZEROS IN E. THE AC IS CHECKED +;FOR ALL ZEROS + +A6000: MOVE [-1] ;PRESET AC TO ALL ONES + SETM [0] ;*SETM SHOULD LOAD AC WITH ALL ONES + SKIPE ;PASSES TEST IF C(AC) = 0 + STOP + +;AD AR + EN [ADAP], AD AR + EN C[ADAP], +;F CYC ACT EN A[F1], IR BOOLE AD A +F[IR2], +;IR BOOLE [IR1], IRXX(1-3,6,11,15,16) [IR2] + +;********** + +;THIS TEST VERIFIES THAT SETZ LOADS TH AC WITH ALL ZEROS. FIRST, A WORD +;OF ALL ZEROS IS LOADED INTO THE AC VIA A MOVE INSTC. THEN, SETZ IS +;EXECUTED. THE AC IS CHECKED FOR ALL ZEROS. +;IF ANY BITS IN THE AC ARE SET, THE TEST FAILS. + +A6100: MOVE [0] ;RESET AC TO ALL ZEROS + SETZ ;*SETZ SHOULD CLEAR THE AC + SKIPE ;PASS TEST IF C(AC) = 0 + STOP + +;AD MASK GEN EN [ADC1], AD MASK GEN F/F'S [ADC1] +;AD MASK EN B; F CYC ACT EN B[F1]; IRXX00 [IR1]; +;IR BOOLE [IR1] + +;********** +;THIS TEST VERIFIES THAT SETZ LOADS THE AC WITH ALL ZEROS. FIRST, +;THE AC IS SET TO ALL ONES; THEN, SETZ IS EXECUTED. THE AC IS THEN +;CHECKED FOR ALL ZEROS. IF ANY BITS IN THE AC ARE SET, THE TEST FAILS. + +A6200: MOVE [-1] ;PRESET AC TO ALL ONES + SETZ . ;*SETZ SHOULD CLEAR THE AC + SKIPE ;PASS TEST IF C(AC) = 0 + STOP + +;AR AD EN [ARMA], AR AD ET2 EN [ARMA], IR BOOLE [IR] + +;********** + +;THIS TEST VERIFIES THAT THE AND INSTRUCTION OPERATING ON AN AC +;CONTAINING ALL ZEROS AND AN E CONTAINING ALL ZEROS RESULTS IN AN AC +;OF ALL ZEROS. THE AC AND E ARE THE SAME IN THIS TEST. +;AC, E ARE LOADED WITH ZEROS; THEN, AND IS EXECUTED. THE AC IS THEN +;CHECKED FOR ALL ZEROS. IF ANY BITS IN THE AC ARE SET BY THE AND INSTRUCTION, +;THE TEST FAILS. + + +A6300: MOVE [0] ;PRESET AC, E TO ALL ZEROS + AND ;*AND SHOULD RESULT IN C(AC) = 0 + SKIPE ;PASS TEST IF C(AC) = 0 + STOP + +;********** +SN=6400 + ZZ=0 + +A6400: REPEAT ^D36, +<;THIS TEST VERIFIES THAT IF ONE OF THE OPERANDS OF THE AND INSTRUCTION +;IS ZERO, THE RESULT WILL BE ZERO. THIS TEST IS REPEATED 36 TIMES WITH +;ALL ZEROS IN THE AC EXCEPT FOR ONE ONE IN BIT X +;WHERE X TAKES ON INTEGER VALUES 0 THRU 35. E CONTAINS ALL ZEROS. +;THE RESULT SHOULD BE ALL ZEROS IN THE AC. IF C(AC) IS NON-ZERO, +;'AND' FAILED. C(0) IS ALSO CHECKED FOR ALL ZEROS. IF C(0) IS +;NON-ZERO, AC0 WAS CLOBBERED BY 'AND'. + +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + SETZ ;PRESET E TO ALL ZEROS + MOVE 1,[ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN BIT X + ;WHERE X VARIES FROM 0 THRU 35 + AND 1,0 ;*AND SHOULD RESULT IN C(AC) = 0 + SKIPE 1 ;FAIL IF C(AC) NON-ZERO + STOP + SKIPE 0 ;FAIL IF C(0) IS NON-ZERO + STOP +;********** +> +SN=6500 + ZZ=0 + +A6500: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF A WORD ANDED WITH ITSELF +;SHOULD BE THAT SAME WORD. A WORD CONTAINING ALL ZEROS EXCEPT FOR A ONE +;IN ONLY ONE BIT IS ANDED WITH ITSELF. THE AC IS THEN CHECKED FOR ITS +;CONTENTS NONZERO. IF C(AC)=0, THE 'AND' INSTRUCTION FAILED. THIS TEST IS REPEATED +;36 TIMES TO CHECK EACH BIT INDIVIDUALLY. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;LOAD AC WITH ALL ZEROS EXCEPT FOR A ONE IN ONE BIT + AND ;*AND OF WORD CONTAINING A ONE IN ONE BIT + ;POSITION WITH ITSELF SHOULD RESULT WITH + ;C(AC) NONZERO + SKIPN ;PASS TEST IF C(AC) NONZERO + STOP + +;********** +> +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE +;C(AC)=0 AND C(E)=0 AND E OUTSIDE OF THE AC RANGE IS ALL ZEROS IN THE AC. +;FIRST, THE AC IS CLEARED; THEN, THE C(AC) IS ANDED WITH C(E) WHEN +;E GREATER THAN 20 AND C(E)=0. THE RESULT SHOULD BE C(AC)=0. +;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A6600: MOVE [0] ;PRESET AC TO ALL ZEROS + AND [0] ;*AND OF TWO WORDS OF ALL ZEROS + ;SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(AC)=0 +;AND C(E) IS ALL ONES IS AN AC CONTAINING ALL ZEROS. FIRST, THE AC IS +;CLEARED; THEN THE C(AC) IS ANDED WITH AN E CONTAINING ALL ONES, WHERE +;E IS BEYOND THE AC RANGE. THE RESULT SHOULD BE C(AC)=0. +;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A6700: MOVE [0] ;PRESET AC TO ALL ZEROS + AND [-1] ;*AND OF TWO WORDS WHERE C(AC)=0 SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;AD FM+EN [ADFP],AD FM+EN A [ADFP], +;F CYC ACT EN C [F1],IR BOOLE AD FM+F [IR2], +;IR BOOLE [IR1],IRXX(1,4,5,11,13,16) [IR2] + +;********** +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(E)=0 +;AND C(AC) IS ALL ONES IS AN AC CONTAINING ALL ZEROS. FIRST, THE +;AC IS SET TO ALL ONES; THEN, THE AC IS ANDED WITH AN E CONTAINING ALL +;ZEROS, WHERE E IS BEYOND THE AC RANGE. THE RESULT SHOULD BE C(AC)=0. +;IF C(AC) IS NON-ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A7000: MOVE [-1] ;PRESET AC TO ALL ONES + AND [0] ;*AND OF TWO WORDS WHERE C(E)=0 SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF THE AND INSTRUCTION WHERE C(E) IS +;ALL ONES AND C(AC) IS ALL ONES IS AN AC CONTAINING ALL ONES. FIRST, THE +;AC IS SET TO ALL ONES; THEN, C(AC) IS ANDED WITH AN E CONTAINING ALL +;ONES. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +;IF C(AC) IS ZERO, 'AND' FAILED WITH THE MEMORY OPERAND. + +A7100: MOVE [-1] ;PRELOAD AC WITH ALL ONES + AND [-1] ;*AND OF TWO WORDS OF ALL ONES SHOULD + ;RESULT IN C(AC)=ALL ONES + SKIPN ;PASS TEST IF C(AC) IS NON ZERO + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHERE CONTENTS IS NON-ZERO. +;FIRST, THE AC IS CLEARED, THEN, XOR IS EXECUTED WHERE C(E) IS NON-ZERO. +;THE AC IS THEN CHECKED FOR NON-ZERO CONTENTS. +A7200: MOVE [0] ;CLEAR AC + XOR [707070707070] ;*XOR WITH C(AC)=0 AND C(E)=7070707070 SHOULD + ;RESTULT IN C(AC)=707070707070 + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +SN=7300 + ZZ=0 + +A7300: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS +;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS CLEARED AND XOR IS EXECUTED +;WITH ALL BITS OF E CLEAR EXCEPT FOR ONE FLOATING BIT WHICH IS SET TO A ONE. +;THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [0] ;CLEAR AC + XOR [ZZ] ;*XOR WITH C(AC)=0 AND ONE BIT OF C(E) SET SHOULD + ;RESULT IN C(AC) NONZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +> +SN=7400 + ZZ=0 + +A7400: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF THE XOR INSTRUCTION WHERE +;C(E)=0 AND C(AC) IS NON-ZERO IS AN AC WHOSE CONTENTS IS NON-ZERO. THIS +;TEST IS REPEATED 36 TIMES. EACH TIME THE AC IS INITIALIZED SO THAT ONE +;AND ONLY ONE FLOATING BIT IS SET TO ONE AND XOR IS EXECUTED WITH E +;CONTAINING ALL ZEROS. THE AC IS THEN CHECKED FOR A NON-ZERO RESULT. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;PRELOAD AC WITH ONE FLOATING BIT SET + XOR [0] ;*XOR WITH C(E)=0 AND ONE BIT OF C(AC) SET SHOULD + ;RESULT IN C(AC) NON-ZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;AD FM- EN [ADFM], AD FM- F/F'S[ADFM],F CYC ACT EN D[R1], +;IR BOOLE AD FM- F [IR2],IR BOOLE[IR1],IR XX (2,6,7,10,12,15)[IR2], +;AD EQV[ADCR],AD EQV F/F[ADC2],FCYC ACT EN A[F1], +;IR BOOLE (6,11)[IR2],IR BOOLE[IR1],IRXX(6,11)[IR2] + +;********** +> +SN=7500 + ZZ=0 + +A7500: REPEAT ^D36, +<;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION +;WHERE C(AC) = C(E) IS AN AC CONTAINING ALL ZEROS. FIRST, +;THE AC IS INITIALIZED WITH A WORD CONTAINING ALL ZEROS EXCEPT +;FOR ONE FLOATING ONE; THEN, XOR IS EXECUTED WITH E CONTAINING THE SAME +;NUMBER. THE RESULT SHOULD BE AN AC CONTAINING ALL ZEROS. THIS TEST +;IS REPEATED 36 TIMES. IF ANY BIT IN THE RESULT IS NON-ZERO, IT MAY +;INDICATE AN ILLEGAL CARRY GENERATION. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;INITIALIZE AC WITH ONE FLOATING BIT SET + XOR [ZZ] ;*XOR WITH C(AC)=C(E) SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +> +;THIS TEST VERIFIES THAT THE RESULT OF AN XOR INSTRUCTION WHERE +;C(AC)=C(E)=0 IS AN AC CONTAINING ALL ZEROS; FIRST THE AC IS CLEARED; THEN +;XOR IS EXECUTED WITH C(E)=0. THE RESULT IN THE AC IS CHECKED FOR ALL ZEROS + +A7600: MOVE [0] ;INITIALIZE AC TO ALL ZEROS + XOR [0] ;*XOR WITH C(AC)=C(E)=0 SHOULD RESULT INC(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE +;C(AC)=0 AND C(E) ALL ONES IS AN AC CONTAINING ALL ZEROS SINCE +;THE EQUIVALENCE FUNCTION SHOULD SET ONLY LIKE BITS. FIRST, THE AC +;SHOULD BE CLEARED; THEN, EQV IS EXECUTED WITH E CONTAINING ALL ONES. +;THE AC IS THEN CHECKED FOR ALL ZEROS + +A7700: MOVE [0] ;CLEAR THE AC + EQV [-1] ;*EQV WITH C(AC)=0 AND C(E) ALL ONES SHOULD + ;RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +SN=10000 + ZZ=0 + +A10000: REPEAT ^D18, +<;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN +;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE +;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT +;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS +;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE +;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO +SN=SN+1 + ZZ=ZZ+ZZ+1 + IFE , + MOVE [0] ;PRESET AC TO ALL ZEROS + EQV [XWD -1,ZZ] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + ;RESULT IN C(AC) NON-ZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +> +SN=10100 + ZZ=0 + +A10100: REPEAT ^D18, +<;THIS TEST VERIFIES THAT THE RESULT OF AN EQV INSTRUCTION WHERE +;C(AC)=0 AND C(E) IS ALL ONES EXCEPT FOR A ZERO IN ONLY ONE BIT IS AN +;AC WHOSE CONTENTS IS NON-ZERO (A ONE RESULTS IN THE BIT WHERE +;E CONTAINS A ZERO). THIS TEST IS REPEATED 36 TIMES SO THAT EACH BIT +;OF THE RESULT IS A ONE ONCE AND ONLY ONCE. FIRST, THE AC IS +;CLEARED; THEN,EQV IS EXECUTED WITH ONE BIT OF E EQUAL TO ZERO. THE +;RESULT IN THE AC IS THEN CHECKED FOR ITS CONTENTS NON-ZERO +SN=SN+1 + ZZ=ZZ+ZZ+1 + IFE , + MOVE [0] ;PRESET AC TO ALL ZEROS + EQV [XWD ZZ,-1] ;*EQV WITH C(AC)=0 AND ONE BIT OF C(E)=0 SHOULD + ;RESULT IN C(AC) NON-ZERO + SKIPN ;PASS TEST IF C(AC) NON-ZERO + STOP + +;********** +> +;THIS TEST VERIFIES THAT THE RESULT OF AN AC CONTAINING ALL ZEROS +;EQUIVALENCED WITH AN E CONTAINING ALL ZEROS IS AN AC CONTAINING ALL +;ONES; AND FUTHERMORE, IF THIS AC CONTAINING ALL ONES IS EQUIVALANCED +;WITH AN E CONTAINING ALL ZEROS, THE FINAL RESULT IN THE AC +;SHOULD BE ALL ZEROS. FIRST, THE AC IS INITIALIZED TO ALL ZEROS, THEN, +;THIS AC IS EQUIVALENCED TWICE WITH AN E CONTAINING ALL ZEROS. THE AC IS +;THEN CHECKED FOR ALL ZEROS + +A10200: MOVE [0] ;CLEAR THE AC + EQV [0] ;*EQV SHOULD RESULT IN C(AC)=ALL ONES + EQV [0] ;*EQV SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** + +;THIS TEST VERIFIES THAT THE RESULT OF AN AC CONTAINING ALL ONES +;EQUIVALENCED WITH AN E CONTAINING ALL ONES IS AN AC CONTAINING ALL ONES; +;AND FUTHERMORE, IF THIS AC CONTAINING ALL ONES ISEQUIVALENCED WITH AN E +;CONTAINING ALL ZEROS, THE FINAL RESULT IN THE AC SHOULD BE ALL ZEROS. +;FIRST, THE AC IS INITIALIZED TO ALL ONES; THEN, THIS AC IS +;EQUIVALENCED WITH AN E OF ALL ONES. NEXT, THIS SAME AC IS EQUIVALENCED +;WITH AN E OF ALL ZEROS. THE AC IS THEN CHECKED FOR ALL ZEROS. + +A10300: MOVE [-1] ;SET C(AC) TO ALL ONES + EQV [-1] ;*EQV SHOULD RESULT IN C(AC) OF ALL ONES + EQV [0] ;*EQV SHOULD RESULT IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +SN=13000 + ZZ=0 + +A13000: REPEAT ^D36, +<;THIS TEST CHECKS THE 'AND' INSTRUCTION AND THE ADDER ON A ONE-BIT-AT-A-TIME BASIS. +;THE AC IS LOADED WITH ALL ZEROS EXCEPT FOR ONE BIT EQUAL TO ONE; THE AC IS THEN +;ANDED WITH C(E) WHERE C(E)=C(AC). THE NET RESULT SHOULD BE NO CHANGE TO C(AC). +;NEXT, THE AC IS EQUIVALENCED WITH C(E) WHERE C(E)=C(AC). THIS RESULT SHOULD +;BE ALL ONES IN THE AC. THE AC IS THEN EQUIVALENCED WITH A WORD OF ALL ZEROS, +;WHICH SHOULD RESULT IN C(AC)=0. THE AC IS CHECKED FOR ALL ZEROS. +;IF C(AC) IS NON-ZERO, THE TEST FAILS. THIS TEST IS REPEATED 36 TIMES TO +;RIPPLE A ONE THROUGH EVERY BIT OF THE AC. +SN=SN+1 + ZZ=ZZ+ZZ + IFE ZZ, + MOVE [ZZ] ;SETUP AC WITH A ONE IN ONLY ONE BIT + AND [ZZ] ;*AND FLOATING ONE WITH ITSELF SHOULD NOT CHANGE C(AC) + EQV [ZZ] ;RESULTS IN C(AC)=ALL ONES + EQV [0] ;RESULTS IN C(AC)=0 + SKIPE ;PASS TEST IF C(AC)=0 + STOP + +;********** +> +ENDXX: +; JRST BEGEND ;LOOP PROGRAM diff --git a/apps/pdp10/tapes/diags/klad/dakad/README.md b/apps/pdp10/tapes/diags/klad/dakad/README.md index c46022aab..12dcd434a 100644 --- a/apps/pdp10/tapes/diags/klad/dakad/README.md +++ b/apps/pdp10/tapes/diags/klad/dakad/README.md @@ -1,6 +1,6 @@ --- layout: page -title: PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC +title: PDP-10 KA10 Basic Instruction Diagnostic #4 permalink: /apps/pdp10/tapes/diags/klad/dakad/ machines: - id: testka10 @@ -9,12 +9,12 @@ machines: debugger: true --- -PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC ----------------------------------------- +PDP-10 KA10 Basic Instruction Diagnostic #4 +------------------------------------------- -The *PDP-10 KA10 BASIC INSTRUCTION DIAGNOSTIC* (MAINDEC-10-DAKAD-B-D) test code has been extracted from [DAKAD.MAC](DAKAD.MAC) -[[original file](http://pdp-10.trailing-edge.com/klad_sources/01/klad.sources/dakadm.mac.html)] for use with the -[PDP-10 Test Machine with Debugger](/devices/pdp10/machine/ka10/test/debugger/) below. +The *PDP-10 KA10 Basic Instruction Diagnostic #4* (MAINDEC-10-DAKAD-B-D) test code has been extracted from +[DAKAD.MAC](DAKAD.MAC) [[original file](http://pdp-10.trailing-edge.com/klad_sources/01/klad.sources/dakadm.mac.html)] +for use with the [PDP-10 Test Machine with Debugger](/devices/pdp10/machine/ka10/test/debugger/) below. Resources for this test include: diff --git a/devices/pdp10/machine/ka10/test/debugger/machine.xml b/devices/pdp10/machine/ka10/test/debugger/machine.xml index 7ef8645a0..e2048bc3b 100644 --- a/devices/pdp10/machine/ka10/test/debugger/machine.xml +++ b/devices/pdp10/machine/ka10/test/debugger/machine.xml @@ -8,5 +8,5 @@ - +