Re: Who here is too stupid to know that DDD correctly simulated by HHH cannot possibly reach its own return instruction?

On 8/7/2024 3:18 AM, Fred. Zwarts wrote:

Op 04.aug.2024 om 21:00 schreef olcott:

On 8/4/2024 1:54 PM, Richard Damon wrote:

On 8/4/24 9:11 AM, olcott wrote:

On 8/4/2024 1:26 AM, Fred. Zwarts wrote:

Op 03.aug.2024 om 17:20 schreef olcott:>>

When you try to show how DDD simulated by HHH does
reach its "return" instruction you must necessarily
must fail unless you cheat by disagreeing with the
semantics of C. That you fail to have a sufficient
understanding of the semantics of C is less than no
rebuttal what-so-ever.

>
Fortunately that is not what I try, because I understand that HHH cannot possibly simulate itself correctly.
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
In other words when HHH simulates itself simulating DDD it
is supposed to do something other than simulating itself
simulating DDD ???  Do you expect it to make a cup of coffee?
>

>
No, but to be correct it need to complete that to the end.
>

>
Saying this and knowing there is no end cannot possibly
be construed as anything but intentional deception.
>

And what is saying that there is no end for a program that aborts and

void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}
Unless we divide the behavior of the tester from the test
subject Infinite_Recursion() would be determined to halt.
Instead we define the "return" as the never reached halt
state of Infinite_Recursion().

halts after two cycles? Are you dreaming again of the non-aborting HHH that does not halt? Dreams are no substitute for facts.

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer