Re: DDD correctly emulated by H0

On 6/22/2024 2:22 PM, Richard Damon wrote:

On 6/22/24 2:53 PM, olcott wrote:

On 6/22/2024 1:50 PM, Fred. Zwarts wrote:

Op 22.jun.2024 om 15:11 schreef olcott:

>
It is a verified fact that the behavior that finite string DDD presents
to HH0 is that when DDD correctly simulated by HH0 calls HH0(DDD) that
this call DOES NOT RETURN.
>
It is a verified fact that the behavior that finite string DDD presents
to HH1 is that when DDD correctly simulated by HH0 calls HH1(DDD) that
this call DOES RETURN.
>
I don't get why people here insist on lying about verified facts.
>

>
We know that 'verified fact' for you means 'my wish'.

>
Ignoramus?
>
When we stipulate that the only measure of a correct emulation is the semantics of the x86 programming language then we see that when DDD is correctly emulated by H0 that its call to H0(DDD) cannot possibly return.
>
_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]
>
When we define H1 as identical to H0 except that DDD does not call H1 then we see that when DDD is correctly emulated by H1 that its call to H0(DDD) does return. This is the same behavior as the directly executed DDD().
>

>
By a strict interpreation of your measure, this input has UNDEFIINED BEHAVIOR, so it is improper to ask about it.
>

 That is a stupid thing to say. The behavior of THE INPUT
is specified by the semantics of the x86 programming language.