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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: People are still trying to get away with disagreeing with the
 semantics of the x86 language
Date: Sun, 30 Jun 2024 22:25:25 -0500
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On 6/30/2024 9:16 PM, Richard Damon wrote:
> On 6/30/24 9:38 PM, olcott wrote:
>> On 6/30/2024 8:24 PM, Richard Damon wrote:
>>> On 6/30/24 9:03 PM, olcott wrote: >> On 6/30/2024 7:44 PM, Richard 
>>> Damon wrote:
>>>>
>>>> I had to dumb this down because even the smartest
>>>> people here were overwhelmed:
>>>>
>>>> The call from DDD to HHH(DDD) when N steps of DDD are
>>>> correctly emulated by any pure function x86 emulator
>>>> HHH at machine address 0000217a cannot possibly return.
>>>
>>> But that is NOT the "behavior of the input", and CAN NOT BE SO DEFINED.
>>>
>>
>> I don't understand why you so stupidly lie about this.
>>
>> _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.
>>
> 
> And, since the HHH that DDD calls will abort is emulation, it WILL 
> return to DDD and it will return also.
> 

How can stopping the emulation the first four
instructions of DDD possibly do anything besides
cause them to stop?

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