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On 8/16/2024 8:34 AM, Mikko wrote:Whaatever you "construe" does not change the fact that DDD specifiesOn 2024-08-16 12:02:00 +0000, olcott said:void DDD()
I must go one step at a time.That's reasonable in a discussion. The one thing you were discussing
above is what is the meaning of the output of HHH. Its OK to stay
at that step until we are sure it is understood.
{
HHH(DDD);
return;
}
Unless an unlimited emulation of DDD by HHH
can reach the "return" instruction of DDD it is
construed that this instance of DDD never halts.
For three years now at least most reviewers insistedIf you claim that HHH halts and DDD doesn't you disagree with
on disagreeing with the semantics of the x86 language.
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