Re: Every sufficiently competent C programmer knows

On 3/10/25 10:27 PM, olcott wrote:

On 3/10/2025 9:21 PM, Richard Damon wrote:

On 3/10/25 7:41 PM, olcott wrote:

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.
>

>
Problem: DD Isn't a program, and if you try to compile it, you will get an undiefined symbol HHH.
>

 HHH need not be a program for this correct thought experiment.
The only detail required to know about HHH is that it correctly
emulates N steps of DD.
 

Sure it does, otherwise it can not be a decider.
Where do you get that it doesn't need to be a program? Your FRAUD again?
And DD needs to be a program, but it isn't, especially if the it calls HHH isn't one.
So, you logic seems to be that you can prove something about programs when you are studing something that isn't a program at all.
Thats like saying you can determine a taxominy of Cats, but studying Flowers.
You are just showing how stupid and ignorant you are.
Your "logic" is based on making up lies about what you need to do to show what you want, and then just going with that fraud.
It seems you are trying to use every logical fallicy possible.