Re: Anyone with sufficient knowledge of C knows that DD specifies non-terminating behavior to HHH

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.

https://github.com/plolcott/x86utm/blob/master/Halt7.c
has fully operational HHH and DD

 No, it has not. There is no DD there.
 

 line 1354 through 1360

That page does not show so many lines.
Nothing above gives any reason to revise my comment that the subjec line
is false.
--
Mikko