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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: DDD correctly emulated by H0
Date: Sat, 22 Jun 2024 16:10:02 -0400
Organization: i2pn2 (i2pn.org)
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On 6/22/24 3:45 PM, olcott wrote:
> On 6/22/2024 2:22 PM, Richard Damon wrote:
>> On 6/22/24 2:53 PM, olcott wrote:
>>> On 6/22/2024 1:50 PM, Fred. Zwarts wrote:
>>>> Op 22.jun.2024 om 15:11 schreef olcott:
>>>>>
>>>>> It is a verified fact that the behavior that finite string DDD 
>>>>> presents
>>>>> to HH0 is that when DDD correctly simulated by HH0 calls HH0(DDD) that
>>>>> this call DOES NOT RETURN.
>>>>>
>>>>> It is a verified fact that the behavior that finite string DDD 
>>>>> presents
>>>>> to HH1 is that when DDD correctly simulated by HH0 calls HH1(DDD) that
>>>>> this call DOES RETURN.
>>>>>
>>>>> I don't get why people here insist on lying about verified facts.
>>>>>
>>>>
>>>> We know that 'verified fact' for you means 'my wish'.
>>>
>>> Ignoramus?
>>>
>>> When we stipulate that the only measure of a correct emulation is the 
>>> semantics of the x86 programming language then we see that when DDD 
>>> is correctly emulated by H0 that its call to H0(DDD) cannot possibly 
>>> return.
>>>
>>> _DDD()
>>> [00002172] 55               push ebp      ; housekeeping
>>> [00002173] 8bec             mov ebp,esp   ; housekeeping
>>> [00002175] 6872210000       push 00002172 ; push DDD
>>> [0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
>>> [0000217f] 83c404           add esp,+04
>>> [00002182] 5d               pop ebp
>>> [00002183] c3               ret
>>> Size in bytes:(0018) [00002183]
>>>
>>> When we define H1 as identical to H0 except that DDD does not call H1 
>>> then we see that when DDD is correctly emulated by H1 that its call 
>>> to H0(DDD) does return. This is the same behavior as the directly 
>>> executed DDD().
>>>
>>
>> By a strict interpreation of your measure, this input has UNDEFIINED 
>> BEHAVIOR, so it is improper to ask about it.
>>
> 
> That is a stupid thing to say. The behavior of THE INPUT
> is specified by the semantics of the x86 programming language.
> 

Right, so what does a call to a location of memory that doesn't exist do?