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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 08:31:21 -0500
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On 3/11/2025 5:28 AM, Mikko wrote:
> On 2025-03-10 23:41:13 +0000, olcott said:
> 
>> 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.
> 
> Every competent programmer knows that the information given is
> insufficient to determine whether HHH emulates at all, and whether
> it emulates correctly if it does.
> 
>> Since HHH does see that same pattern that competent
>> C programmers see it correctly aborts its emulation
>> and rejects these inputs as non terminating.
> 
> Whether HHH does see those patterns cannot be inferred from the information
> given. Only about DDD one can see that it halts if HHH returns. In 
> addition,
> the given information does not tell whether HHH can see patterns that are
> not there.
> 
> How many competent programmers you have asked?
> 

Two C programmers with masters degrees in computer science
agree that DDD correctly emulated by HHH cannot possibly
reach its own "return" instruction and terminate normally.

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