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On 8/7/24 2:14 PM, olcott wrote:I really think that you must be a liar here becauseOn 8/7/2024 1:02 PM, joes wrote:But it doesn't, as a correct simulation of a DDD that calls an HHH that returns will stop running,Am Wed, 07 Aug 2024 08:54:41 -0500 schrieb olcott:>On 8/7/2024 2:29 AM, Mikko wrote:>On 2024-08-05 13:49:44 +0000, olcott said:It doesn't change anything about DDD. HHH was supposed to decide anythingI know what it means. But the inflected form "emulated" does not meanIn other words when DDD is defined to have a pathological relationship
what you apparently think it means. You seem to think that "DDD
emulated by HHH" means whatever HHH thinks DDD means but it does not.
DDD means what it means whether HHH emulates it or not.
>
to HHH we can just close our eyes and ignore it and pretend that it
doesn't exist?
and can't fulfill that promise. That doesn't mean that DDD is somehow
faulty, it's just a counterexample.
>
void DDD()
{
HHH(DDD);
return;
}
>
*HHH is required to report on the behavior of DDD*
Anyone that does not understand that HHH meets this criteria
has insufficient understanding.
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