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

On 6/30/24 10:27 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 stop?
 

The emulation stops, and the emulating behavor of HHH stops, but not the behavior of the input.
You don't seem to understand that fact, perhaps because you confuse Truth with Knowledge.
Emulation doesn't create or define the behavior of the input, that existed at the moment the program prepresented by the input was created.
Emulation REVEALS that behvior to the emulator. The only behavior it creates is that of the emulator, not the emulated.
You seem to not understand that property of programs, perhaps because you don't understand determinism, which requires a good understanding of what Truth is.