Re: Any honest person that knows the x86 language can see... predict correctly

On 2024-08-01 04:03:36 +0000, wij said:

int main()
{
   HHH(DDD);
}

That is not very useful. Running this program may demostrate that
HHH answers but does not reveal the answer. The program should be:
void output(char *); // in Olcott's library
int main () {
  output("Asking HHH whether DDD will halt.");
  int ans = HHH(DDD);
  if (ans) {
    output("HHH predicts that DDD will halt.");
  } else {
    output("HHH predicts that DDD will not halt.");
  }
  output("Running DDD");
  DDD();
  output("DDD halted");
  return ans;
}
--
Mikko