Re: People are still trying to get away with disagreeing with the semantics of the x86 language

On 2024-07-01 12:44:57 +0000, olcott said:

On 7/1/2024 1:05 AM, Mikko wrote:

On 2024-06-30 17:18:09 +0000, olcott said:

 Richard just said that he affirms that when DDD correctly
simulated by HHH calls HHH(DDD) that this call returns even
though the semantics of the x86 language disagrees.
 On 6/30/2024 7:34 AM, Richard Damon wrote:
 > It is still true that the xemantics of the x86
 > language define the behavior of a set of bytes,
 > as the behavior when you ACTUALLY RUN THEM,
 > and nothing else.
 _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]
 Richard thinks that he can get away with disagreeing with this
verified fact:
 The call from DDD to HHH(DDD) when N steps of DDD are correctly
emulated by any pure function x86 emulator HHH cannot possibly
return.

 It is your HHH so you should know whether it returns. Others may
have wrong impression about it if they have trusted your lies.

 I have never lied about this.

At least you have claimed more than proven.

_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]
 DDD is correctly emulated by HHH which calls an
emulated HHH(DDD) to repeat the process until aborted.

The correctness remain unproven.

Once aborted the DDD emulated by HHH immediately stops.

No, it does not. DDD stops when it executes the ret instruction at 2183
or the function at 15d2 calls exit.

At no point in this emulation does the call from DDD
correctly emulated by HHH to HHH(DDD) ever return.

Everyone who knows x86 machine or assembly language can see that
DDD terminates normally if and only if HHH terminates normally.
--
Mikko