Re: Every sufficiently competent C programmer knows --- Semantic Property of Finite String

On 3/14/25 12:39 PM, olcott wrote:

On 3/14/2025 9:10 AM, Richard Damon wrote:

On 3/14/25 12:26 AM, olcott wrote:

On 3/13/2025 10:03 PM, Richard Damon wrote:

On 3/13/25 10:05 PM, olcott wrote:

On 3/13/2025 6:09 PM, Richard Damon wrote:

On 3/13/25 4:46 PM, olcott wrote:

On 3/13/2025 4:27 AM, joes wrote:

Am Wed, 12 Mar 2025 21:41:34 -0500 schrieb olcott:

On 3/12/2025 7:56 PM, dbush wrote:

On 3/12/2025 8:41 PM, olcott wrote:

>
NOT WHEN IT IS STIPULATED THAT THE BEHAVIOR BEING MEASURED IS

>
The direct execution of DDD

>
is proven to be different than the behavior of DDD emulated by HHH
according to the semantics of the x86 language.

>
Which is weird, considering that a simulator should produce the same
behaviour.
>
>

DECIDERS ARE REQUIRED TO REPORT ON THE SEMANTIC OR SYNTACTIC PROPERTY OF
THEIR INPUT FINITE STRINGS.

And not if the input called a different simulator that didn't abort.
>

>
DDD correctly emulated by HHH cannot possibly
reach its own final state no matter what HHH
does.
>
DDD correctly emulated by HHH1 does reach its
own final state.

>
Which shows that HHH doesn't correctly emulate its input, unless you just lied and gave the two programs different inputs.
>

>
void DDD()
{
   HHH(DDD);
   return;
}
>
Someone that is not a liar could explain exactly
how DDD emulated by HHH according to the semantics
of the C language must have the same behavior as
DDD emulated by HHH1 according to the semantics
of the C language.

>
WHy? The above is NOT a program, as to be a program it needs the full code of HHH included.
>

>
That would be too confusing for this simple thought experiment.
The behavior of HHH is already fully specified.
>

>
No, that is the method you use to hide your deception.
>

 HHH is specified to correctly simulate a finite number
of steps of DDD. THAT IS 100% OF EVERYTHING THAT HHH
DOES FOR THE PURPOSE OF THIS THOUGHT EXPERIMENT.

And thus, since DDD calls the always finite executing HHH(DDD), that HHH will always return to DDD and DDD will halt.

 void DDD()
{
   HHH(DDD);
   return;
}
 The semantics of the finite string input DDD to HHH specifies
that it will continue to call HHH(DDD) in recursive simulation.
 

Nope, DDD calls HHH once, and then HHH will simulate a finite number of steps of DDD. It may simulated DDD calling HHH again and then simulate HHH simulating HHH simulating DDD, but it never get "stuck" in the recursion, since all the simulations were, by your own definition, finite in length,
You can not get an "infinte recursion" from a program that has bounded behavior, since HHH was DEFINED above to only simulate a finite number of steps of DDD.
Sorry, but your logic is just based on your own lies, lies you have admitted to having told, as you admit that you are using the wrong definition of core terms of art.