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On 7/4/2024 11:26 AM, joes wrote:Apparently you only read the bottom.Am Thu, 04 Jul 2024 11:03:09 -0500 schrieb olcott:On 7/4/2024 10:06 AM, joes wrote:Am Thu, 04 Jul 2024 08:41:22 -0500 schrieb olcott:On 7/4/2024 8:26 AM, joes wrote:Hello?Am Thu, 04 Jul 2024 07:46:15 -0500 schrieb olcott:On 7/4/2024 5:15 AM, joes wrote:What x86 semantics say that HHH can’t return?Am Wed, 03 Jul 2024 09:45:57 -0500 schrieb olcott:On 7/3/2024 9:39 AM, joes wrote:I repeat.Am Wed, 03 Jul 2024 08:21:40 -0500 schrieb olcott:On 7/3/2024 3:26 AM, Fred. Zwarts wrote:Which semantics?Op 02.jul.2024 om 21:48 schreef olcott:On 7/2/2024 2:22 PM, Fred. Zwarts wrote:Op 02.jul.2024 om 20:43 schreef olcott:On 7/2/2024 1:59 AM, Mikko wrote:On 2024-07-01 12:44:57 +0000, olcott said:On 7/1/2024 1:05 AM, Mikko wrote:On 2024-06-30 17:18:09 +0000, olcott said:>
Richard just said that he affirms that when DDD
correctly simulated by HHH calls HHH(DDD) that this
call returns even though the semantics of the x86
language disagrees.
The execution environment should not make any difference.The DDD correctly emulated by HHH in its own process context cannotWhat semantics am I disagreeing with? Doesn’t HHH halt?The semantics of the x86 language proves that DDD correctly emulatedWhy not? Clearly HHH halts. Does it not return or what?*Machine address 00002174 of DDD is never reached*Yes, and nothing else. So when HHH returns, so does DDD.By definition DDD calls its simulator.HHH halts by definition. Why can’t DDD?As long as it is impossible for DDD correctly emulated by HHH toDDD correctly emulated by HHH calls an emulated HHH(DDD) thatBut HHH aborts, so the cycle does end.
emulates DDD that calls an emulated HHH(DDD)
in a cycle that cannot end unless aborted.
reach its own ret instruction then DDD never halts even when its
stops running because its emulation was aborted.
by HHH cannot possibly reach its own machine address 00002183.
possibly halt even if another entirely different instance of DDD does
halt.
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