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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: People are still trying to get away with disagreeing with the
 semantics of the x86 language
Date: Thu, 4 Jul 2024 11:03:09 -0500
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On 7/4/2024 10:06 AM, joes wrote:
> Am Thu, 04 Jul 2024 08:41:22 -0500 schrieb olcott:
>> On 7/4/2024 8:26 AM, joes wrote:
>>> Am Thu, 04 Jul 2024 07:46:15 -0500 schrieb olcott:
>>>> On 7/4/2024 5:15 AM, joes wrote:
>>>>> Am Wed, 03 Jul 2024 09:45:57 -0500 schrieb olcott:
>>>>>> On 7/3/2024 9:39 AM, joes wrote:
>>>>>>> Am Wed, 03 Jul 2024 08:21:40 -0500 schrieb 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.
>>>>>>> Which semantics?
>>>>> I repeat.
>>> What x86 semantics say that HHH can’t return?
> Hello?
> 
> 
> 
> 
> 
>>>>>>>> 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.
>>>>>>> But HHH aborts, so the cycle does end.
>>>>>> As long as it is impossible for DDD correctly emulated by HHH to
>>>>>> reach its own ret instruction then DDD never halts even when its
>>>>>> stops running because its emulation was aborted.
>>>>> HHH halts by definition. Why can’t DDD?
>>>> By definition DDD calls its simulator.
>>> Yes, and nothing else. So when HHH returns, so does DDD.
>> *Machine address 00002174 of DDD is never reached*
> Why not? Clearly HHH halts. Does it not return or what?
> 

Why does 2 + 3 = 5 and and not a bucket of rusted bolts?

The semantics of the x86 language proves that DDD
correctly emulated by HHH cannot possibly reach its
own machine address 00002183.

*Failing to comprehend this is less than no rebuttal at all*

_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]



-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer