Re: Anyone that claims this is not telling the truth

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