Re: Every sufficiently competent C programmer knows

On 3/11/2025 12:35 PM, Richard Heathfield wrote:

On 11/03/2025 17:01, olcott wrote:

On 3/11/2025 8:42 AM, Richard Heathfield wrote:

On 11/03/2025 13:26, olcott wrote:

On 3/11/2025 5:01 AM, Richard Heathfield wrote:

>
<snip>
>

And yet I can't even /see/ HHH, let alone judge what it does or does not do correctly. All I see is a call to it.
>

>
It is stipulated that HHH correctly emulates N
steps of the x86 machine code of its input functions.
This may or may not include HHH emulating itself
emulating an input.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;
}

 Yeah, I got all that.
 

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.

 We can't know that without seeing the code.
 

when HHH correctly emulates N steps of the above
functions is stipulated then there is no need to
see the code. This thought experiment applies to
every HHH that can possibly exist such that the
above functions are correctly emulated by HHH.
Clearly playing the same head games as Richard Damon.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer