Re: Anyone that disagrees with this is not telling the truth V4

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Sujet : Re: Anyone that disagrees with this is not telling the truth V4
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 19. Aug 2024, 13:08:51
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v9vckj$2rjt1$4@dont-email.me>
References : 1 2 3 4
User-Agent : Mozilla Thunderbird
On 8/19/2024 2:30 AM, Mikko wrote:
On 2024-08-18 12:25:05 +0000, olcott said:
 
*Everything that is not expressly stated below is*
*specified as unspecified*
void DDD()
{
   HHH(DDD);
   return;
}
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
*It is a basic fact that DDD emulated by HHH according to*
*the semantics of the x86 language cannot possibly stop*
*running unless aborted* (out of memory error excluded)
X = DDD emulated by HHH∞ according to the semantics of the x86 language
Y = HHH∞ never aborts its emulation of DDD
Z = DDD never stops running
My claim boils down to this: (X ∧ Y) ↔ Z
void EEE()
{
   HERE: goto HERE;
}
HHHn predicts the behavior of DDD the same
way that HHHn predicts the behavior of EEE.

>
That HHH <is> and x86 emulator <is> sufficient to
determine exactly what the behavior of DDD emulated by HHH
according to the semantics of the x86 language would be.
 The last "would be" means that the clause is conterfactual.
But why would anybody care about the conterfactual behaviour?
 
It is not counter-factual.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

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