Re: Anyone that claims this is not telling the truth

On 2024-08-18 13:00:19 +0000, olcott said:
 

On 8/18/2024 4:02 AM, Mikko wrote:

On 2024-08-17 15:35:33 +0000, olcott said:
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On 8/17/2024 10:30 AM, Richard Damon wrote:

On 8/17/24 11:09 AM, olcott wrote:

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]
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*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)
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No, anyone saying that the above is something that CAN be correctly emulated by the semantics of the x86 language is just a LIAR.
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You are inserting a word that I did not say.
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To say that DDD is emulated by HHH means that it must be possible to validly do that act.
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You are not going to get very far with any claim that
emulating a sequence of x86 machine-code bytes is impossible.
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How do you emulate dthe CALL HHH instruction without the code that follows?
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Who is the silly one now?
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No it has moved up to a ridiculous and utterly
baseless false assumption that is directly contradicted
by the verified fact that x86utm takes Halt7.obj as
its input data, thus having all of the machine code
of HHH directly available to DDD.

>
Now you switched the topic. Earlier you were not talking about x86utm
but emulation, both specifically of DDD by HHH and generally.
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It is stupid to assume that HHH and DDD are not in the
same shared memory space.

 It is stupid to assume that HHH and DDD can be in only one memory space.