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On 11/14/2024 3:09 AM, Mikko wrote:No, DDD emulated by HHH emulates HHH emulating DDD and then aborting that emulation and returning, and thus the COMPLETE emulation that it only partially did WILL reach the end state, so the correct answer is Halting.On 2024-11-13 23:11:30 +0000, olcott said:That is a ridiculously stupid thing to say because we
>On 11/13/2024 4:58 AM, Mikko wrote:>On 2024-11-12 13:58:03 +0000, olcott said:>
>On 11/12/2024 1:12 AM, joes wrote:>Am Mon, 11 Nov 2024 10:35:57 -0600 schrieb olcott:>On 11/11/2024 10:25 AM, joes wrote:>Am Mon, 11 Nov 2024 08:58:02 -0600 schrieb olcott:On 11/11/2024 4:54 AM, Mikko wrote:On 2024-11-09 14:36:07 +0000, olcott said:On 11/9/2024 7:53 AM, Mikko wrote:When DDD calls a simulator that aborts, that simulator returns to DDD,DDD emulated by HHH does not reach its "return" instruction final haltThe actual computation itself does involve HHH emulating itselfWhich is what you are doing: you pretend that DDD calls some other HHH
emulating DDD. To simply pretend that this does not occur seems
dishonest.
that doesn’t abort.
state whether HHH aborts its emulation or not.
which then halts.
>
It is not the same DDD as the DDD under test.
If the DDD under the test is not the same as DDD then the test
is performed incorrectly and the test result is not valid.
>
The DDD under test IS THE INPUT DDD
IT IS STUPIDLY WRONG-HEADED TO THINK OTHERWISE.
I agree that there is only one DDD but above you said otherwise.
>
already know that DDD emulated by HHH emulates itself
emulating DDD and DDD emulated by HHH1 *DOES NOT DO THAT*
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