Liste des Groupes | Revenir à c theory |
On 7/20/24 4:02 PM, olcott wrote:Yet again trying to get away with saying that when every elementOn 7/20/2024 2:36 PM, Fred. Zwarts wrote:And the problem you ignore is that each HHH is given the DDD that calls itself, and not some other HHH, and thus you can't look at the otherOp 20.jul.2024 om 21:09 schreef olcott:>On 7/20/2024 2:00 PM, Fred. Zwarts wrote:>Op 20.jul.2024 om 17:28 schreef olcott:>void DDD()>
{
HHH(DDD);
}
>
int main()
{
DDD();
}
>
(a) Termination Analyzers / Partial Halt Deciders must halt
this is a design requirement.
>
(b) Every simulating termination analyzer HHH either
aborts the simulation of its input or not.
>
(c) Within the hypothetical case where HHH does not abort
the simulation of its input {HHH, emulated DDD and executed DDD}
never stop running.
>
This violates the design requirement of (a) therefore HHH must
abort the simulation of its input.
And when it aborts, the simulation is incorrect. When HHH aborts and halts, it is not needed to abort its simulation, because it will halt of its own.
So you are trying to get away with saying that no HHH
ever needs to abort the simulation of its input and HHH
will stop running?
>
No, you try to get away with saying that a HHH that is coded to abort and halt, will never stop running, only because you are dreaming of *another* HHH that does not abort.
>
*You know that I didn't say anything like that*
>
Unless I refer to the infinite set of every possible
HHH my reviewers try to get away with saying that I am
referring to the wrong HHH.
>
void DDD()
{
HHH(DDD);
return;
}
>
DDD correctly simulated by pure function HHH cannot
possibly reach its own return instruction.
>
>
Les messages affichés proviennent d'usenet.