Re: People are still trying to get away with disagreeing with the semantics of the x86 language

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:

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?

Hello?

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.

*Machine address 00002174 of DDD is never reached*

Why not? Clearly HHH halts. Does it not return or what?

The semantics of the x86 language proves that DDD correctly emulated by
HHH cannot possibly reach its own machine address 00002183.