Re: Every sufficiently competent C programmer knows --- Very Stupid Mistake or Liars

Op 14.mrt.2025 om 04:53 schreef olcott:

On 3/13/2025 10:03 PM, Richard Damon wrote:

On 3/13/25 10:07 PM, olcott wrote:

On 3/13/2025 6:09 PM, Richard Damon wrote:

On 3/13/25 9:41 AM, olcott wrote:

On 3/13/2025 6:18 AM, Dan Cross wrote:

In article <vqud4e$36e14$3@dont-email.me>,
Fred. Zwarts <F.Zwarts@HetNet.nl> wrote:

Op 12.mrt.2025 om 16:31 schreef olcott:

[snip]
When N steps of DDD are correctly emulated by every element
of the set of C functions named HHH that do x86 emulation and
>
N is each element of the set of natural numbers
>
then no DDD of the set of HHH/DDD pairs ever reaches its
"return" instruction and terminates normally.

>
In other words no HHH of the set of HHH/DDD pairs ever succeeds to
complete the simulation of a halting program. Failure to reach the end
of a halting program is not a great success. If all HHH in this set
fail, it would be better to change your mind and start working on
something more useful.

>
He seems to think that he's written a program that detects that
his thing hasn't 'reached its "return" instruction and
terminate[d] normally', given some number of steps, where that
number is ... the cardinality of the natural numbers.
>
I wonder if he knows that the set of natural numbers is
infintite, though I suspect he'd say something like, "but it's
countable!"  To which I'd surmise that he has no idea what that
means.
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
Everyone here knows that when N steps of DDD are correctly
simulated by HHH that DDD never reaches its own "return"
instruction and terminates normally thus never halts.
*AND THEY LIE ABOUT IT BY ENDLESSLY CHANGING THE SUBJECT*

>
>
No, the PARTIAL EMULATION done by HHH can't reach that point,

>
But a complete emulation can?
>

>
Yes, but an HHH that gives an answer doesn't do one, due to the pathological design of the template used to build DD to the HHH it calls (which is the only HHH that can exist, or you have violated the basic rules of programing and logic).
>
We have two basic cases,
>
1) if HHH does the partial emulation you describe, then the complete emulation of DD will see that DD call HHH, and it will emulate its input for a while, then abort and theu return 0 to DD which will then halt.
>

 int main()
{
   HHH(DDD); // No DDD can possibly ever return.
}
 

        int main() {
          return HHH(main);
        }
HHH returns reporting that it will not halt. HHH produces false negatives.