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On 7/4/2024 8:56 PM, Richard Damon wrote:So, you don't understand what the semantics of the x86 language actually is?On 7/4/24 9:35 PM, olcott wrote:OK liar I give up.>>
_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]
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*In other words you are denying the verified fact*
That when DDD emulated by HHH according to the
semantics of the x86 language calls HHH(DDD) that
this call cannot possibly return.
No, if HHH(DDD) returns, then by the semantics of the x86 language, and the fact that DDD calls the exact same code sequence as the call from main calls, the call to HHH will return, just after HHH stops its emulation.
>
And by the x86 language, the "behavior" doesn't stop just because the HHH stopped emulating the bytes, because the x86 langugage the byte specify doesn't know that will happen.
>>>
*By denying this verified fact you are affirming*
That when DDD emulated by HHH according to the
semantics of the x86 language calls HHH(DDD) that
*THIS CALL CAN RETURN*
Yes, just not in the emulation that HHH does.
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