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On 10/13/2024 3:06 AM, Mikko wrote:A phrase that you just used cannot be out of scope.On 2024-10-12 19:44:06 +0000, olcott said:void DDD()My point HERE AND NOW is that DDD emulated by everyThat does not mean anything as long as you don't define "every HHH"
HHH that can possibly exist cannot possibly reach
its own return instruction NO MATTER WHAT HHH DOES.
{
HHH(DDD);
return;
}
The only requirement for HHH is that it is an emulator that
emulates more than zero steps of DDD. HHH also must be able
to emulate itself emulating DDD.
so that one can determine whether a HHH that emulates whatever isThis is out-of-scope. The scope is already 100% fully
given as input except that instead of emulating its own code (it
it is called) as "return 1;" only is included in "every HHH".
specified above.
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