Sujet : Re: People are still trying to get away with disagreeing with the semantics of the x86 language
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 03. Jul 2024, 18:03:29
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v63soh$28dpi$4@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12
User-Agent : Mozilla Thunderbird
On 7/3/2024 10:59 AM, Fred. Zwarts wrote:
Op 03.jul.2024 om 15:21 schreef olcott:
On 7/3/2024 3:26 AM, Fred. Zwarts wrote:
Op 02.jul.2024 om 21:48 schreef olcott:
On 7/2/2024 2:22 PM, Fred. Zwarts wrote:
Op 02.jul.2024 om 20:43 schreef olcott:
On 7/2/2024 1:59 AM, Mikko wrote:
On 2024-07-01 12:44:57 +0000, olcott said:
>
On 7/1/2024 1:05 AM, Mikko wrote:
On 2024-06-30 17:18:09 +0000, olcott said:
>
Richard just said that he affirms that when DDD correctly
simulated by HHH calls HHH(DDD) that this call returns even
though the semantics of the x86 language disagrees.
>
On 6/30/2024 7:34 AM, Richard Damon wrote:
> It is still true that the xemantics of the x86
> language define the behavior of a set of bytes,
> as the behavior when you ACTUALLY RUN THEM,
> and nothing else.
>
_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]
>
Richard thinks that he can get away with disagreeing with this
verified fact:
>
The call from DDD to HHH(DDD) when N steps of DDD are correctly
emulated by any pure function x86 emulator HHH cannot possibly
return.
>
It is your HHH so you should know whether it returns. Others may
have wrong impression about it if they have trusted your lies.
>
I have never lied about this.
>
At least you have claimed more than proven.
>
_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 the process until aborted.
>
The correctness remain unproven.
>
>
IT IS PROVEN BY THE SEMANTICS OF THE X86 LANGUAGE
THAT YOU REMAIN WILLFULLY IGNORANT OF SEMANTICS OF
THE X86 LANGUAGE DOES NOT MEAN IT HAS NOT BEEN PROVEN.
>
>
Please, point to the paragraph in the specification of the X86 language that says that a two cycle recursion should be aborted after one cycle.
Claiming that the abort is related to the x86 language is apparently wilfully incorrect.
>
>
>
I am not going to show you the trace of the Peano axioms
that prove the 2 + 3 = 5, if you disagree you are a liar
or an ignoramus.
>
>
This change of subject does not hide that your claim that the x86 language proves your claim is incorrect as a verified fact.
>
>
So you opted for liar then:
>
_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 correctly emulated by HHH calls an emulated HHH(DDD)
that emulates DDD that calls an emulated HHH(DDD)
in a cycle that cannot end unless aborted.
>
HHH aborts after two cycles.
That is not what I said. I will make in more clear.
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.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer