Sujet : Re: People are still trying to get away with disagreeing with the semantics of the x86 language
De : F.Zwarts (at) *nospam* HetNet.nl (Fred. Zwarts)
Groupes : comp.theoryDate : 03. Jul 2024, 20:25:18
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v6452f$29pag$2@dont-email.me>
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Op 03.jul.2024 om 19:58 schreef olcott:
On 7/3/2024 12:51 PM, Fred. Zwarts wrote:
Op 03.jul.2024 om 18:03 schreef olcott:
>
_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]
DDD is correctly emulated by HHH which calls an emulated HHH(DDD)
to repeat this process an endless number of times until aborted
or out-of-memory error.
Anyone knowing the x86 language knows that a program cannot be programmed to do two different things
It cannot do both run out of memory *and* abort.
DDD correctly emulated by any element of the infinite
set of every pure function HHH cannot possibly reach
its own ret instruction and halt.
Exactly! Well done! This proves that HHH cannot possibly correctly simulate itself. If it aborts, it does so one cycle too soon.
That HHH aborts its
emulation at some point or never aborts its emulation
cannot possibly change this.
Indeed, an abort does not make the simulation correct and an infinite simulation is not correct either.
So make up your mind. What does it do?
And what does "endless number of times until aborted" mean? Does it abort after an infinite number of steps?
Make clear what you mean. After how many cycles is the simulation aborted?