Re: Every sufficiently competent C programmer knows

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Sujet : Re: Every sufficiently competent C programmer knows
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 11. Mar 2025, 14:31:21
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vqpdv9$202b2$2@dont-email.me>
References : 1 2
User-Agent : Mozilla Thunderbird
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

Date Sujet#  Auteur
24 May 25 o 

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