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On 8/16/2024 1:42 AM, Fred. Zwarts wrote:But it does if the HHH that DDD calls is the HHH that you actually have that does abort its simulation.Op 16.aug.2024 om 04:24 schreef olcott:void DDD()On 8/15/2024 8:57 PM, Richard Damon wrote:>On 8/15/24 8:12 AM, olcott wrote:>On 8/15/2024 2:00 AM, joes wrote:Am Wed, 14 Aug 2024 16:07:43 +0100 schrieb Mike Terry:On 14/08/2024 08:43, joes wrote:>Am Tue, 13 Aug 2024 21:38:07 -0500 schrieb olcott:On 8/13/2024 9:29 PM, Richard Damon wrote:>On 8/13/24 8:52 PM, olcott wrote:You were trying to label an incomplete/partial/aborted simulation asThat is what I said dufuss.A simulation of N instructions of DDD by HHH according to theNope, it is just the correct PARTIAL emulation of the first N
semantics of the x86 language is necessarily correct.
instructions of DDD, and not of all of DDD,
correct.
>how *HHH* returns*Try to show exactly how DDD emulated by HHH returns to its caller*A correct simulation of N instructions of DDD by HHH is sufficientNope, if a HHH returns to its caller,
to correctly predict the behavior of an unlimited simulation.DDDHHH simulates DDD enter the matrix
DDD calls HHH(DDD) Fred: could be eliminated HHH simulatesvoilasecond level
DDD calls HHH(DDD) recursion detected
HHH aborts, returns outside interference DDD haltsHHH halts>
You're misunderstanding the scenario? If your simulated HHH aborts its
simulation [line 5 above],
>>then the outer level H would have aborted its identical simulation
earlier. You know that, right?
That is the part that Joes and Fred do not understand.
>
You do not understand what we say. I have repeated many times that the simulated HHH is aborted before it would halt by itself.
{
HHH(DDD);
return;
}
DDD emulated by HHH according to the semantics of the
x86 language would never halt by itself.
Right, and reachable code is reachable code, becuase HHH is the program that it is.The simulating HHH fails to reach that point, proving an incomplete simulation.void Infinite_Recursion()
{
Infinite_Recursion();
OutputString("I never make it here!\n");
}
Unreachable code is unreachable code.
Sure I do, but it seems you don't.That is what you do not understand, either because of incompetence, or unwillingness.You might not even understand the above simple example.
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