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

On 7/1/2024 11:01 AM, joes wrote:

Am Mon, 01 Jul 2024 07:44:57 -0500 schrieb olcott:

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.

The x86 semantics say that an aborted simulation returns.
 

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.

You are inconsistent about whether HHH returns or not.
 

_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.

DDD is emulated by HHH which calls an emulated HHH(DDD) to
repeat the process until aborted. Once aborted the DDD emulated by HHH
immediately stops.

DDD running by itself does not stop. HHH stops simulating it.

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

Where does the outer call to HHH get stuck after aborting?
 

HHH(DDD) returns 0 to main();
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer