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On 6/13/2025 4:11 AM, Fred. Zwarts wrote:You can't, because every competent programmer knows that it is correct.Op 12.jun.2025 om 17:51 schreef olcott:I m not going to correct you on this again.On 6/12/2025 4:37 AM, Fred. Zwarts wrote:>Op 12.jun.2025 om 02:52 schreef olcott:void DDD()>The analyser should not decide about an instance, but about what is specified by the code.
If you don't understand the difference between object
instances of OOP classes and the classes themselves
then you might not understand.
>
int main()
{
D(); // calls H(D) and the parameter to this H(D)
} // is not the caller of this instance of H(D)
>
{
HHH(DDD);
return;
}
>
The input to HHH(DDD) specifies a sequence of configurations
that cannot possibly reach their own "return" statement
final halt state.
The input is a pointer to code that includes the code to abort and halt.
I will simply stop looking at anything that you say.Which exactly proves my point, because HHH does abort, not only the simulating HHH, but also the simulated HHH is programmed to abort. So, it would stop running.
That HHH is not able to reach that part of the code is a failure of HHH, not a property of the program specified in the input.counter-factual and over-your-head.
It is a verified fact that unless the outermost HHH(DDD)
aborts its simulation of DDD then this simulated DDD, the
directly executed DDD() and the directly executed HHH()
would never stop running.
That you don't have enough technical skill to verifyNot understanding what simulation means is not stupid, but the resistance against learning is.
this fact should cause you to have the position that
you don't know if this fact is verified. That you
take the position that I am wrong knowing that you
don't understand these things seems dishonest.
In a simulator that simulates only one instruction after which aborts, all other instructions are unreachable.That HHH cannot reach it, does not change the specified code.Within a correct execution trace unreachable code is
the same as non-existent code. That you don't know that
is merely a lack of sufficient technical competence on
your part.
No, the input DDD is a pointer to a function (as a competent C programmer should know). This function has addresses to other functions, which includes the code to abort and halt.The program aborts as world-class simulators en direct execution of exactly the same input prove.int main()
>
{
DDD(); // calls HHH(DDD) with a different instance of DDD
}
You say that I am wrong and cannot provide all of the details>>
That you are insufficiently competent to see this is proven
by the fact that you have no rebuttal anchored in correct
reasoning.
>
Ad hominem attacks only prove your lack of counter-arguments.
showing exactly how and why I am wrong does not count as a
rebuttal. It only counts as incorrect intuition.
You prove your ignorance by the lack of details that
you provide in your purely dogmatic assertions that I
am wrong.
void DDD()
{
HHH(DDD);
return;
}
The input to HHH(DDD) specifies non-halting behavior
in that DDD correctly simulated by HHH cannot possibly
reach its own "return" statement final halt state.
Only a correct simulation of the input to HHH(DDD) byAnd the actual behaviour specified is a program with finite recursion that halts.
HHH derives the actual behavior specified by this input.
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