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On 4/9/2025 1:58 PM, Fred. Zwarts wrote:That would make HHH to never halt. But HHH aborts and halts. By this abort it is no longer an emulation of DD by HHH that obeys the semantics of the C or the x86utm programming languages respectfully.Op 09.apr.2025 om 19:29 schreef olcott:No way, José:>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.
>
By "correctly simulate" I mean a simulation or an emulation
of DD by HHH that obeys the semantics of the C or the x86utm
programming languages respectfully. This expressly includes
that HHH simulates/emulates itself simulating/emulating DD.
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