Re: Every sufficiently competent C programmer knows --- Correct Emulation Specifies Behavior

Am Sun, 16 Mar 2025 18:48:11 -0500 schrieb olcott:

On 3/16/2025 3:21 PM, joes wrote:

Am Sun, 16 Mar 2025 14:36:11 -0500 schrieb olcott:

*Every sufficiently competent C programmer knows*
>
typedef void (*ptr)();
int HHH(ptr P);
>
void DDD()
{
    HHH(DDD);
    return;
}
int DD()
{
    int Halt_Status = HHH(DD); if (Halt_Status)
      HERE: goto HERE;
    return Halt_Status;
}
>
When HHH correctly emulates N steps of the above functions none of
them can possibly reach their own "return" instruction and terminate
normally.

 

The fun thing is that DDD does terminate, but HHH can't simulate it.
 

The semantic behavioral property of THE INPUT TO HHH is non halting as
proven by DDD [...]

No, DDD halts.

Disagreeing with the semantics of the x86 language is the same as
disagreeing with the semantics of arithmetic. The semantics of the x86
language specifies what a correct emulation is.

No, the direct execution does.

Disagreeing that a correct emulation of the input specifies the behavior
of this input disavows the whole notion of a UTM.

If only the simulation agreed with the direct execution.

Since HHH does see that same pattern that competent C programmers see
it correctly aborts its emulation and rejects these inputs as non
terminating.

It could just reject them as terminating.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.