Re: Anyone that claims this is not telling the truth

On 8/17/24 4:10 PM, olcott wrote:

On 8/17/2024 2:27 PM, Richard Damon wrote:
 > On 8/17/24 3:00 PM, olcott wrote:>> On 8/17/2024 1:50 PM, Richard Damon wrote:
 >>>>>> And thus ALL of memory is part of the input,

>
Any additional details have no effect what-so-ever on my claim.
>

>
Suure it does.
>
Since your argument tries to say that since DDD is the same to all of them, so its the behavior.
>
You are just admitting to being a LIAR.
>

 *Calling me a liar admits that insults is all that you have*
*If I made a mistake then show that*

I did.

 FOR THREE YEARS YOU ALWAYS CHEAT
BY CHANGING MY WORDS AND REBUTTING THESE CHANGED WORDS
 *Everything that is not expressly stated below is*
*specified as unspecified*

Ok.

 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, DDD can NOT be emulated accoreding to the semantics of the x86 langauge, because the contents of the location 000015d2 is not provided to be emulated, and will need to be emulated after emulating the call instruction.
Remember, you said that anything not expressly stated was unspecified, and thus we can't let HHH use anything not specified as the input.
Sorry, you are just proving your ignorance.