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On 8/13/24 11:48 PM, olcott wrote:The trace is regular enough that we could define a formal language forOn 8/13/2024 10:21 PM, Richard Damon wrote:On 8/13/24 10:38 PM, olcott wrote:A complete emulation of one instruction isOn 8/13/2024 9:29 PM, Richard Damon wrote:Nope. You didn't. I added clairifying words, pointing out why you claim is incorrect.On 8/13/24 8:52 PM, olcott wrote:That is what I said dufuss.void DDD()Nope, it is just the correct PARTIAL emulation of the first N instructions of DDD, and not of all of DDD,
{
HHH(DDD);
return;
}
_DDD()
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
A simulation of N instructions of DDD by HHH according to
the semantics of the x86 language is necessarily correct.
For an emulation to be "correct" it must be complete, as partial emulations are only partially correct, so without the partial modifier, they are not correct.
a complete emulation of one instructionNo. The trace is to long, and since you HHH doesn't meet your requirements (since it isn't a pure function) you can't give me a compldte input to trace.Show the exact machine code trace of how DDD emulatedRemember how English works:*Try to show exactly how DDD emulated by HHH returns to its caller*A correct simulation of N instructions of DDD by HHH isNope, if a HHH returns to its caller,
sufficient to correctly predict the behavior of an unlimited
simulation.
(the first one doesn't even have a caller)
Use the above machine language instructions to show this.
When you ask "How DDD emulated by HHH returns to its callers".
by HHH (according to the semantics of the x86 language)
reaches its own machine address 00002183
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