Re: Anyone that claims this is not telling the truth

On 8/17/24 1:24 PM, olcott wrote:

On 8/17/2024 11:16 AM, Richard Damon wrote:

On 8/17/24 11:35 AM, olcott wrote:

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]
>
*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.
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And thus, ALL of memory is the "input" and thus any change in it renders that answer possibly different.
>

 There are no other words that can be added to my
words that change the immutable fact of my words.
 

Which are just meaningless garbage and a LIE.
The DDD given can not be "emulated" by HHH, as it isn't complete, and to look at other non-inputs is a violation of the meaning of the terms. That emulation MUST stop at the call HHH, as no HHH has been provided as part of the input to HHH.
If you want to include the code of HHH as the input, then you must say so, and that means that every different HHH is seeig a DIFFERENT DDD, and any HHH that doesn't run until it hits an end hasn't done a correct x86 emulation.
THAT IS THE SEMANTICS OF THE x86 LANGUAGE.
Note, the fact that EVERY such HHH *WILL* hit the out of memory case if it is actually correctly emulating the input that includes itself, means that you have a null set of cases to look at, and thus a meaningless proposition.
You ar just proving your stupidity.