Re: The execution trace of HHH1(DDD) shows the divergence

Op 07.jun.2025 om 16:32 schreef olcott:

The execution trace of HHH1(DDD) shows the divergence
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*
 The next instruction of DDD that HHH emulates is at
the machine address of 00002183.
 The next instruction of DDD that HHH1 emulates is at
the machine address of 00002190.
 00002183 != 00002190
 

Then both are wrong. The next instruction should be at 000015c3, the address specified in the call.
You skipped instructions. You do not show which instruction causes that HHH arrives at 00002183 and HHH1 at 00002190.