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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Every sufficiently competent C programmer knows
Date: Mon, 10 Mar 2025 22:21:07 -0400
Organization: i2pn2 (i2pn.org)
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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.

That HHH can correct emulate N steps, that go past the "HHH(DD)" 
statement says that HHH is not correctly simulating the "program" given 
to it and inventing behavior.

Note, since you just defined the HHH DOES just emulate N steps and 
returns the value 0, says that when we pair this DD with that HHH to 
complete the program, we see that the actual behavior of DD will reach 
that final return, just not in the PARTIAL emulation it did.

Thus we see that Peter Olcott, who claims it is obvious that this 
doesn't happen is just an lying idiot.

Sorry, you are just showing that your logic is based on FRAUD and ERROR.