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Cleaned up OPFLAG handling

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Jeff 2016-09-29 14:02:37 -07:00 • committed by Jeff Parsons
commit 98d3700fa4
8 changed files with 381 additions and 361 deletions
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12
modules/pdp11/lib/cpuops.js
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@ -1527,7 +1527,7 @@ PDP11.opSPL = function(opCode)
this.assert(opCode & 0x08);
if (!(this.regPSW & PDP11.PSW.CMODE)) {
this.regPSW = (this.regPSW & ~(PDP11.PSW.UNUSED | PDP11.PSW.PRI)) | ((opCode & 0x7) << PDP11.PSW.SHIFT.PRI);
this.priorityReview = 1;
this.opFlags |= PDP11.OPFLAG.INTQ_SPL;
}
this.nStepCycles -= 1;
};
@ -1618,13 +1618,21 @@ PDP11.opWAIT = function(opCode)
* However, the PCjs approach requires the CPU to continue running. One simple solution to this dilemma:
*
* 1) opWAIT() sets a new opFlags bit (OPFLAG.WAIT)
* 2) When stepCPU() sees OPFLAG.WAIT, it checks for interrupts; if none, it rewinds the PC back to the WAIT
* 2) Rewind PC back to WAIT
* 3) Whenever stepCPU() detects OPFLAG.WAIT, call checkInterruptQueue()
* 4) If checkInterruptQueue() detects an interrupt, advance PC past WAIT before dispatching it
*
* Technically, the PC is already exactly where it's supposed to be, so why are we wasting time with steps
* 2 and 4? It's largely for the Debugger's sake, so that as long as execution is "blocked" by a WAIT, that's
* what you'll see in the Debugger. I could make those steps conditioned on the presence of the Debugger,
* but I feel it's better to keep all code paths the same.
*
* NOTE: It's almost always a bad idea to add more checks to the inner stepCPU() loop, because every additional
* check can have a measurable (negative) impact on performance. Which is why it's important to use opFlags bits
* whenever possible, since we can test for multiple (up to 32) exceptional conditions with a single check.
*/
this.opFlags |= PDP11.OPFLAG.WAIT;
this.advancePC(-2);
this.nStepCycles -= 1;
};