Re: DD specifies non-terminating behavior to HHH --- COMPLETE PROOF

On 2/25/25 1:25 PM, olcott wrote:

On 2/25/2025 12:07 PM, joes wrote:

Am Tue, 25 Feb 2025 12:01:12 -0600 schrieb olcott:

On 2/25/2025 10:13 AM, Mikko wrote:

Althogh the subject line has the words "COMPLETE PROOF" there is no
proof or pointer to proof below.
>

typedef void (*ptr)();
int HHH(ptr P);
int DD()
{
    int Halt_Status = HHH(DD);
    if (Halt_Status)
      HERE: goto HERE;
    return Halt_Status;
}
int main()
{
    HHH(DD);
}
>
The above does specify that DD simulated by HHH cannot possibly
terminate normally by reaching its own "return" instruction.

 

This depends entirely on HHH not aborting *in every invocation*.
>

 Whether HHH aborts or never aborts DD simulated by
HHH never terminates normally.

No, becuawe "DD" is the program as it behaves when run or correctly simulated.
If HHH aborts, then if gave up the ability to define the behavior of DD, and thus we need to look at the ACTUAL behavior.

 Some of my reviewers have said that they have ADD thus
have great difficulty paying attention to ALL MY WORDS.
 Anyone that pays COMPLETE ATTENTON TO ALL OF THE ABOVE WORDS
and has sufficient technical skill understands that I
am correct. Most of the recent rebuttals have ignored
some of my words.

Maybe you need to pay attention to the actual problem, and the errors you make that have been pointed out, as opposed to continuely prove your stupidity by working on a strawman, and lying about what actually happens.

 

Ignoring the code in main() seemed dishonest.