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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Every sufficiently competent C programmer knows
Date: Tue, 11 Mar 2025 08:37:49 -0500
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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.

> The information given is clearly insufficient to determine whether HHH
> reports at all or what it reports. That HHH returns an int is given but
> not how that int (or anything else) relates to "HHH cannot possibly reach
> the return instruction".
> 

HHH simply sees that same non-terminating pattern that
every competent C programmer sees.

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer