Re: DDD correctly emulated by H0 --- Why Lie? -- Repeat until Closure

On 6/24/2024 8:13 PM, Richard Damon wrote:

On 6/24/24 9:05 PM, olcott wrote:

>
D correctly simulated by H does not have the same behavior
as the directly executed D(D) because
>
the call from D to H(D,D) cannot possibly return when D
is correctly simulated by H.

 Which just means that the D correctly simulated by H trace stops in the middle of the simulation of H,

Why risk your salvation on head games?
Anyone that totally understands the following code knows
that you are lying about that.
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.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer