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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Anyone that claims this is not telling the truth
Date: Mon, 19 Aug 2024 10:26:14 +0300
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On 2024-08-18 12:48:32 +0000, olcott said:

> On 8/18/2024 3:58 AM, Mikko wrote:
>> On 2024-08-17 15:09:01 +0000, olcott said:
>> 
>>> On 8/17/2024 10:06 AM, Richard Damon wrote:
>>>> On 8/17/24 10:58 AM, olcott wrote:
>>>>> On 8/17/2024 9:10 AM, Richard Damon wrote:
>>>>>> On 8/17/24 8:29 AM, olcott wrote:
>>>>>>> void DDD()
>>>>>>> {
>>>>>>>    HHH(DDD);
>>>>>>> }
>>>>>>> 
>>>>>>> _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]
>>>>>>> 
>>>>>>> *It is a basic fact that DDD emulated by HHH according to*
>>>>>>> *the semantics of the x86 language cannot possibly stop*
>>>>>>> *running unless aborted* (out of memory error excluded)
>>>>>>> 
>>>>>> 
>>>>>> No, anyone saying that the above is something that CAN be correctly 
>>>>>> emulated by the semantics of the x86 language is just a LIAR.
>>>>>> 
>>>>> 
>>>>> You are inserting a word that I did not say.
>>>>> 
>>>> 
>>>> To say that DDD is emulated by HHH means that it must be possible to 
>>>> validly do that act.
>>>> 
>>> 
>>> You are not going to get very far with any claim that
>>> emulating a sequence of x86 machine-code bytes is impossible.
>> 
>> It is impossible to emulate a call if the memory content at the
>> called address is not kown (or even whether there is any memory
>> at that address). The byte sequence above contains a call to
>> an address that is not a part of the shown sequence.
>> 
> 
> x86utm takes the compiled Halt7.obj file of this c program
> https://github.com/plolcott/x86utm/blob/master/Halt7.c
> Thus making all of the code of HHH directly available to
> DDD and itself. HHH emulates itself emulating DDD.

It is not an emulation of DDD if the execution differs from a
real execution of DDD.

-- 
Mikko