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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:>void 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.
>
Nope, it is just the correct PARTIAL emulation of the first N instructions of DDD, and not of all of DDD,
That is what I said dufuss.
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.
Show the exact machine code trace of how DDD emulated>Remember how English works:>>A correct simulation of N instructions of DDD by HHH is>
sufficient to correctly predict the behavior of an unlimited
simulation.
Nope, if a HHH returns to its caller,
*Try to show exactly how DDD emulated by HHH returns to its caller*
(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".
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