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

On 7/1/2024 6:08 AM, Richard Damon wrote:

On 6/30/24 11:25 PM, olcott wrote:

On 6/30/2024 9:16 PM, Richard Damon wrote:

On 6/30/24 9:38 PM, olcott wrote:

On 6/30/2024 8:24 PM, Richard Damon wrote:

On 6/30/24 9:03 PM, olcott wrote: >> On 6/30/2024 7:44 PM, Richard Damon wrote:

>
I had to dumb this down because even the smartest
people here were overwhelmed:
>
The call from DDD to HHH(DDD) when N steps of DDD are
correctly emulated by any pure function x86 emulator
HHH at machine address 0000217a cannot possibly return.

>
But that is NOT the "behavior of the input", and CAN NOT BE SO DEFINED.
>

>
I don't understand why you so stupidly lie about this.
>
_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.
>

>
And, since the HHH that DDD calls will abort is emulation, it WILL return to DDD and it will return also.
>

>
How can stopping the emulation the first four
instructions of DDD possibly do anything besides
cause them to stop?
>

>
The emulation stops, and the emulating behavor of HHH stops, but not the behavior of the input.
>

While your hand is still in the cash register stealing the
money you say there is no hand and there is no money.