Re: Every sufficiently competent C programmer knows

On 3/11/2025 5:36 AM, Mikko wrote:

On 2025-03-11 08:55:22 +0000, Fred. Zwarts said:
 

Op 11.mrt.2025 om 00:41 schreef olcott:

typedef void (*ptr)();
int HHH(ptr P);
>
void Infinite_Loop()
{
  HERE: goto HERE;
  return;
}
>
void Infinite_Recursion()
{
  Infinite_Recursion();
  return;
}
>
void DDD()
{
  HHH(DDD);
  return;
}
>
int DD()
{
  int Halt_Status = HHH(DD);
  if (Halt_Status)
    HERE: goto HERE;
  return Halt_Status;
}
>
That when HHH correctly emulates N steps of the
above functions that none of these functions can
possibly reach their own "return" instruction
and terminate normally.

>

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

>
All competent C programmers see that HHH correctly reports that it cannot possibly reach the 'return' instruction.

 How may competent C programmers did you ask?
 

Two C programmers with masters degrees in computer
science both agreed that DDD correctly emulated by HHH
cannot possibly reach its own "return" instruction and
terminate normally.

The information given is clearly insufficient to determine whether HHH
reports at all or what it reports. That HHH returns an int is given but
not how that int (or anything else) relates to "HHH cannot possibly reach
the return instruction".
 

HHH simply sees that same non-terminating pattern that
every competent C programmer sees.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer