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On 8/5/2024 5:59 PM, Richard Damon wrote:Not when the emulation is conditional.On 8/5/24 9:49 AM, olcott wrote:Do you really not understand that recursive emulation <is>On 8/5/2024 2:39 AM, Mikko wrote:>On 2024-08-04 18:59:03 +0000, olcott said:>
>On 8/4/2024 1:51 PM, Richard Damon wrote:>On 8/4/24 9:53 AM, olcott wrote:>On 8/4/2024 1:22 AM, Fred. Zwarts wrote:>Op 03.aug.2024 om 18:35 schreef olcott:>>>> ∞ instructions of DDD correctly emulated by HHH[∞] never>reach their own "return" instruction final state.>
>
So you are saying that the infinite one does?
>
Dreaming again of HHH that does not abort? Dreams are no substitute for facts.
The HHH that aborts and halts, halts. A tautology.
void DDD()
{
HHH(DDD);
return;
}
>
That is the right answer to the wrong question.
I am asking whether or not DDD emulated by HHH
reaches its "return" instruction.
But the "DDD emulated by HHH" is the program DDD above,
When I say DDD emulated by HHH I mean at any level of
emulation and not and direct execution.
If you mean anything other than what the words mean you wihout
a definition in the beginning of the same message then it is
not reasonable to expect anyone to understand what you mean.
Instead people may think that you mean what you say or that
you don't know what you are saying.
>
If you don't understand what the word "emulate" means look it up.
>
DDD (above) cannot possibly reach its own "return" instruction halt
state when its machine code is correctly emulated by HHH.
>
Only because an HHH that does so never returns to anybody.
>
isomorphic to infinite recursion?
void Infinite_Recursion()Nope, but DDD() does, as the correct x86 emulation of it shows (not the incorrect, and partial emulation done by the HHH that it calls)
{
Infinite_Recursion();
return;
}
Does infinite recursion ever reach its own "return"
instruction halt state?
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