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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 08:34:34 -0500
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On 3/11/2025 5:30 AM, Mikko wrote:
> On 2025-03-11 02:27:42 +0000, olcott said:
> 
>> On 3/10/2025 9:21 PM, Richard Damon wrote:
>>> 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.
>>
>> HHH need not be a program for this correct thought experiment.
>> The only detail required to know about HHH is that it correctly
>> emulates N steps of DD.
> 
> Wrong. One nneds also to know how a call to HHH is interpreted, in 
> particular
> if HHH is not a program.
> 

You are trying to get away with saying that
one C function cannot call another C function
according to the semantics of the C language?

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