Sujet : Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH
De : mikko.levanto (at) *nospam* iki.fi (Mikko)
Groupes : comp.theoryDate : 10. Feb 2025, 09:35:21
Autres entêtes
Organisation : -
Message-ID : <vocdo9$14kc0$1@dont-email.me>
References : 1 2 3 4 5
User-Agent : Unison/2.2
On 2025-02-09 15:24:53 +0000, olcott said:
On 2/9/2025 4:08 AM, Mikko wrote:
On 2025-02-08 14:55:09 +0000, olcott said:
On 2/8/2025 4:25 AM, Mikko wrote:
On 2025-02-07 23:13:04 +0000, olcott said:
Experts in the C programming language will know that DD
correctly simulated by HHH cannot possibly reach its own
"if" statement.
Wrong, they understand that nothing below exludes the possibility that
HHH is a program that can correctly simulate DD to its "if" statement.
Show the execution trace of that.
Your request does not make sense. Non-existence of a exclusion does not
have an execution trace.
The code of HHH might exlude that but that is not sohwn below.
The finite string DD specifies non-terminating recursive
simulation to simulating termination analyzer HHH.
No, it does not. DD as quoted below pecifies nothing about the behaviour
of HHH, only its argument types and return type.
>> int Halt_Status = HHH(DD); // line 3 of DD
Requires HHH to simulate itself simulating DD recursively.
No, it does not. I only requires that the execution of HHH with a function
pointer to DD must be started. OP does not show what happens next.
Within the context that HHH <is> a simulating termination
analyzer line 3 of DD proves that DD cannot possibly reach
its own "if" statement.
That is not the context of OP.
-- Mikko
…