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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Every sufficiently competent C programmer knows
Date: Tue, 11 Mar 2025 16:06:41 +0200
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On 2025-03-11 13:37:49 +0000, olcott said:

> 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.

Irrelevant. Did they say that "HHH correctly reports that it cannot
possibly reach the 'return' instruction"?

-- 
Mikko