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On 6/7/2025 10:32 AM, olcott wrote:*That is not an instruction of DDD*The execution trace of HHH1(DDD) shows the divergenceFalse.
of DDD emulated by HHH from DDD emulated by HHH1.
>
int main()
{
HHH1(DDD);
}
>
Shows that DDD emulated by HHH and DDD emulated by
HHH1 diverges as soon as HHH begins emulating itself
emulating DDD.
>
*From the execution trace of HHH1(DDD) shown below*
DDD emulated by HHH1 DDD emulated by HHH
[00002183] push ebp [00002183] push ebp
[00002184] mov ebp,esp [00002184] mov ebp,esp
[00002186] push 00002183 ; DDD [00002186] push 00002183 ; DDD
[0000218b] call 000015c3 ; HHH [0000218b] call 000015c3 ; HHH
*HHH1 emulates DDD once then HHH emulates DDD once, these match*
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The next instruction of DDD that HHH emulates is at
the machine address of 00002183.
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The next instruction of DDD that HHH1 emulates is at
the machine address of 00002190.
The next instruction of DDD that both HHH and HHH1 emulates is at the machine address of 000015c3,
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