Use backtick code blocks for syntax highlighting
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James Gregory
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10-04.md
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10-04.md
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@ -12,93 +12,95 @@ pages: 200-203
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**LISTING 10.5 L10-5.ASM**
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; Finds and returns the greatest common divisor of two integers.
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; Uses Euclid's algorithm: divides the larger integer by the
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; smaller; if the remainder is 0, the smaller integer is the GCD,
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; otherwise the smaller integer becomes the larger integer, the
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; remainder becomes the smaller integer, and the process is
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; repeated. Avoids code recursion.
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;
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;
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;
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; C near-callable as:
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; unsigned int gcd(unsigned int int1, unsigned int int2);
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```nasm
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; Finds and returns the greatest common divisor of two integers.
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; Uses Euclid's algorithm: divides the larger integer by the
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; smaller; if the remainder is 0, the smaller integer is the GCD,
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; otherwise the smaller integer becomes the larger integer, the
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; remainder becomes the smaller integer, and the process is
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; repeated. Avoids code recursion.
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;
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;
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;
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; C near-callable as:
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; unsigned int gcd(unsigned int int1, unsigned int int2);
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; Parameter structure:
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parms struc
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dw ? ;pushed BP
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dw ? ;pushed return address
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int1 dw ? ;integers for which to find
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int2 dw ? ; the GCD
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parms ends
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; Parameter structure:
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parms struc
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dw ? ;pushed BP
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dw ? ;pushed return address
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int1 dw ? ;integers for which to find
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int2 dw ? ; the GCD
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parms ends
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.model small
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.code
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public _gcd
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align 2
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_gcd proc near
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push bp ;preserve caller's stack frame
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mov bp,sp ;set up our stack frame
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push si ;preserve caller's register variables
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push di
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.model small
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.code
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public _gcd
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align 2
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_gcd proc near
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push bp ;preserve caller's stack frame
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mov bp,sp ;set up our stack frame
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push si ;preserve caller's register variables
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push di
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;Swap if necessary to make sure that int1 >= int2
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mov ax,int1[bp]
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mov bx,int2[bp]
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cmp ax,bx ;is int1 >= int2?
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jnb IntsSet ;yes, so we're all set
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xchg ax,bx ;no, so swap int1 and int2
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IntsSet:
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;Swap if necessary to make sure that int1 >= int2
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mov ax,int1[bp]
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mov bx,int2[bp]
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cmp ax,bx ;is int1 >= int2?
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jnb IntsSet ;yes, so we're all set
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xchg ax,bx ;no, so swap int1 and int2
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IntsSet:
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; Now loop, dividing int1 by int2 and checking the remainder, until
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; the remainder is 0. At each step, if the remainder isn't 0, assign
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; int2 to int1, and the remainder to int2, then repeat.
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GCDLoop:
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;if the remainder of int1 divided by
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; int2 is 0, then int2 is the gcd
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sub dx,dx ;prepare int1 in DX:AX for division
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div bx ;int1/int2; remainder is in DX
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and dx,dx ;is the remainder zero?
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jz Done ;yes, so int2 (BX) is the gcd
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;no, so move int2 to int1 and the
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; remainder to int2, and repeat the
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; process
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mov ax,bx ;int1 = int2;
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mov bx,dx ;int2 = remainder from DIV
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; Now loop, dividing int1 by int2 and checking the remainder, until
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; the remainder is 0. At each step, if the remainder isn't 0, assign
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; int2 to int1, and the remainder to int2, then repeat.
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GCDLoop:
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;if the remainder of int1 divided by
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; int2 is 0, then int2 is the gcd
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sub dx,dx ;prepare int1 in DX:AX for division
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div bx ;int1/int2; remainder is in DX
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and dx,dx ;is the remainder zero?
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jz Done ;yes, so int2 (BX) is the gcd
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;no, so move int2 to int1 and the
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; remainder to int2, and repeat the
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; process
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mov ax,bx ;int1 = int2;
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mov bx,dx ;int2 = remainder from DIV
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;—start of loop unrolling; the above is repeated three times—
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sub dx,dx ;prepare int1 in DX:AX for division
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div bx ;int1/int2; remainder is in DX
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and dx,dx ;is the remainder zero?
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jz Done ;yes, so int2 (BX) is the gcd
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mov ax,bx ;int1 = int2;
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mov bx,dx ;int2 = remainder from DIV
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;—
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sub dx,dx ;prepare int1 in DX:AX for division
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div bx ;int1/int2; remainder is in DX
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and dx,dx ;is the remainder zero?
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jz Done ;yes, so int2 (BX) is the gcd
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mov ax,bx ;int1 = int2;
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mov bx,dx ;int2 = remainder from DIV
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;—
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sub dx,dx ;prepare int1 in DX:AX for division
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div bx ;int1/int2; remainder is in DX
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and dx,dx ;is the remainder zero?
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jz Done ;yes, so int2 (BX) is the gcd
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mov ax,bx ;int1 = int2;
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mov bx,dx ;int2 = remainder from DIV
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;—end of loop unrolling—
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jmp GCDLoop
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;—start of loop unrolling; the above is repeated three times—
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sub dx,dx ;prepare int1 in DX:AX for division
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div bx ;int1/int2; remainder is in DX
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and dx,dx ;is the remainder zero?
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jz Done ;yes, so int2 (BX) is the gcd
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mov ax,bx ;int1 = int2;
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mov bx,dx ;int2 = remainder from DIV
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;—
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sub dx,dx ;prepare int1 in DX:AX for division
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div bx ;int1/int2; remainder is in DX
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and dx,dx ;is the remainder zero?
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jz Done ;yes, so int2 (BX) is the gcd
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mov ax,bx ;int1 = int2;
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mov bx,dx ;int2 = remainder from DIV
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;—
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sub dx,dx ;prepare int1 in DX:AX for division
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div bx ;int1/int2; remainder is in DX
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and dx,dx ;is the remainder zero?
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jz Done ;yes, so int2 (BX) is the gcd
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mov ax,bx ;int1 = int2;
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mov bx,dx ;int2 = remainder from DIV
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;—end of loop unrolling—
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jmp GCDLoop
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align2
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Done:
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mov ax,bx ;return the GCD
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pop di ;restore caller's register variables
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pop si
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pop bp ;restore caller's stack frame
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ret
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_gcd endp
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end
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align2
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Done:
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mov ax,bx ;return the GCD
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pop di ;restore caller's register variables
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pop si
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pop bp ;restore caller's stack frame
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ret
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_gcd endp
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end
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```
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Assembly language optimization is pattern matching on a local scale.
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Frankly, it's also the sort of boring, brute-force work that people are
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