Re: DDD correctly emulated by H0

On 6/25/2024 4:22 AM, joes wrote:

Am Sat, 22 Jun 2024 13:47:24 -0500 schrieb olcott:

On 6/22/2024 1:39 PM, Fred. Zwarts wrote:

Op 21.jun.2024 om 15:21 schreef olcott:

 

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.

Yes. Which is wrong, because H0 should terminate.
 

typedef void (*ptr)();
int H0(ptr P);
void DDD()
{
   H0(DDD);
}
int main()
{
   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.
Until you acknowledge this is true, this is the
only thing that I am willing to talk to you about.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer