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

On 2/9/2025 10:39 AM, joes wrote:

Am Sun, 09 Feb 2025 09:24:53 -0600 schrieb olcott:

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:

 

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.

Yes, because HHH doesn’t halt.
 

No sense talking to people that don't have enough
technical skill to verify the actual facts.
The finite string x86 machine code of C function DD
that is input to HHH(DD) is correctly rejected by
simulating termination analyzer HHH as non-halting.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer