Liste des Groupes | Revenir à theory |
On 2024-07-10 18:27:27 +0000, joes said:It is a hierarchy of prerequisites of knowledge.
Am Wed, 10 Jul 2024 08:37:30 -0500 schrieb olcott:On 7/10/2024 2:18 AM, Mikko wrote:On 2024-07-09 14:14:16 +0000, olcott said:A correct emulator can correctly any correct x86 instructions.On 7/9/2024 1:14 AM, Mikko wrote:>On 2024-07-08 17:36:58 +0000, olcott said:When we ask:On 7/8/2024 11:16 AM, Fred. Zwarts wrote:>Op 08.jul.2024 om 18:07 schreef olcott:>>You believe that two equals infinity.
Try to show how infinity is one cycle too soon.
>
void Infinite_Loop()
{
HERE: goto HERE;
}
void Infinite_Recursion()
{
Infinite_Recursion();
}
void DDD()
{
HHH(DDD);
}
Two cycles is enough to correctly determine that none of the above
functions correctly emulated by HHH can possibly halt.
That you don't see this is ignorance or deception.
There is an important detail that determines whether an infinite
execution can be inferred. That is best illustrated by the following
examples:
void Finite_Loop()
{
int x = 10000;
HERE:
if (x > 0) {
x--;
goto HERE;
}
}
void Finite_Recursion(int n)
{
if (n > 0) {
Finite_Recursion(n + 1);
}
}
void DDD()
{
HHH(DDD); // HHH detects recursive simulation and then simulates
no more }
The important difference is that in my examples there is a
conditional instruction that can (and does) prevent infinite
exectuion.
>
Does the call from DDD emulated by HHH to HHH(DDD) return?
Why would anyone ask that? A question should make clear its topic.
Instead one could ask whether HHH can fully emulate DDD if that is what
one wants to know. Or one may think that HHH and DDD are so
unimteresting that there is no point to ask anyting about them.
>
When it emulates non-halting code then itself does not halt.Oh? Maybe you should give your simulator and decider different namesA charlatan doesn't want clarity but confusion. A good charlatan just
so they don't get confused.
dont what them so much that they would be noticed for that might expose
the charlatan.
Les messages affichés proviennent d'usenet.