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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: People are still trying to get away with disagreeing with the
 semantics of the x86 language
Date: Thu, 4 Jul 2024 13:26:39 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Thu, 04 Jul 2024 07:46:15 -0500 schrieb olcott:
> On 7/4/2024 5:15 AM, joes wrote:
>> Am Wed, 03 Jul 2024 09:45:57 -0500 schrieb olcott:
>>> On 7/3/2024 9:39 AM, joes wrote:
>>>> Am Wed, 03 Jul 2024 08:21:40 -0500 schrieb olcott:
>>>>> On 7/3/2024 3:26 AM, Fred. Zwarts wrote:
>>>>>> 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.
>>>> Which semantics?
>> I repeat.
What x86 semantics say that HHH can’t return?

>>>>> DDD correctly emulated by HHH calls an emulated HHH(DDD) that
>>>>> emulates DDD that calls an emulated HHH(DDD)
>>>>> in a cycle that cannot end unless aborted.
>>>> But HHH aborts, so the cycle does end.
>>> As long as it is impossible for DDD correctly emulated by HHH to reach
>>> its own ret instruction then DDD never halts even when its stops
>>> running because its emulation was aborted.
>> HHH halts by definition. Why can’t DDD?
> By definition DDD calls its simulator.
Yes, and nothing else. So when HHH returns, so does DDD.

-- 
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.