Re: DDD correctly emulated by H0 --- Why Lie?

On 6/24/24 7:38 PM, olcott wrote:

On 6/24/2024 6:17 PM, Richard Damon wrote:

On 6/24/24 9:39 AM, olcott wrote:

On 6/24/2024 6:02 AM, Richard Damon wrote:

On 6/23/24 11:26 PM, olcott wrote:

>
You tried to get away with the lie that P correctly simulated
by H had to have the same behavior as the directly executed
P(P) for three years.

>
Different statement. The instruction seen in the "correct" emulation of P by H will see EXACTLY the same instructions as seen in the direct execution of P, up to the point that H stops simulating.
>

>
*That is not true*
I just ran it to make sure that my current code does this.

>
So, what instruction CORRECTLY emulated differed in behavior to the directly executed instruction.
>

 It is much clearer to compare apples to apples thus
see that DDD correctly simulated by H1 has different
behavior than DDD correctly simulated by H0.

Dodge and Weave, Dodge and Weave.
The emulation of the two are IDENTICAL to the point that H0 aborts its emulation.
Of course an aborted simulation won't make it to the end.

 void DDD()
{
   H0(DDD);
}
 _DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
 The call from DDD to H0(DDD) when DDD is correctly emulated
by H0 cannot possibly return.