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Op 09.apr.2025 om 19:29 schreef olcott:Simply over your level of technical competence.>If HHH would correctly simulate DD (and the functions called by DD) then the simulated HHH would return to DD and DD would halt.
On 4/8/2025 10:31 AM, Fred. Zwarts wrote:Op 08.apr.2025 om 17:13 schreef olcott:>On 4/8/2025 2:45 AM, Fred. Zwarts wrote:Everyone with a little bit of C knowledge understands that if HHH returns with a value 0, then DDD halts.Op 08.apr.2025 om 06:33 schreef olcott:>>>
typedef void (*ptr)();
int HHH(ptr P);
>
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
>
int main()
{
HHH(DD);
}
>
*Simulating termination analyzer Principle*
It is always correct for any simulating termination
analyzer to stop simulating and reject any input that
would otherwise prevent its own termination.
>
In this case there is nothing to prevent, because the finite string specifies a program that halts.
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
>
This stuff is simply over-your-head.
HHH(DD) meets the above: *Simulating termination analyzer Principle*
Anyone with sufficient competence with the C programming language
will understand this.
>
DDD CORRECTLY SIMULATED BY HHH
NOT ANY OTHER DAMN DDD IN THE UNIVERSE NITWIT.
>
But HHH failed to complete the simulation of the halting program,HHH is only required to report on the behavior of its
because the programmer was dreaming of an infinite recursion.I really think that you may simply be a troll playing head games.
If I didn't have to tell you this hundreds of times and you didn't
persist in the straw-man deception I would not have called you a nitwit.
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