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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Who here is too stupid to know that DDD correctly simulated by HHH cannot possibly reach its own return instruction?
Date: Thu, 8 Aug 2024 10:44:39 +0300
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On 2024-08-07 13:19:33 +0000, olcott said:
> On 8/7/2024 3:18 AM, Fred. Zwarts wrote:
>> Op 04.aug.2024 om 21:00 schreef olcott:
>>> On 8/4/2024 1:54 PM, Richard Damon wrote:
>>>> On 8/4/24 9:11 AM, olcott wrote:
>>>>> On 8/4/2024 1:26 AM, Fred. Zwarts wrote:
>>>>>> Op 03.aug.2024 om 17:20 schreef olcott:>>
>>>>>>> When you try to show how DDD simulated by HHH does
>>>>>>> reach its "return" instruction you must necessarily
>>>>>>> must fail unless you cheat by disagreeing with the
>>>>>>> semantics of C. That you fail to have a sufficient
>>>>>>> understanding of the semantics of C is less than no
>>>>>>> rebuttal what-so-ever.
>>>>>>
>>>>>> Fortunately that is not what I try, because I understand that HHH
>>>>>> cannot possibly simulate itself correctly.
>>>>>>
>>>>>
>>>>> void DDD()
>>>>> {
>>>>> HHH(DDD);
>>>>> return;
>>>>> }
>>>>>
>>>>> In other words when HHH simulates itself simulating DDD it
>>>>> is supposed to do something other than simulating itself
>>>>> simulating DDD ??? Do you expect it to make a cup of coffee?
>>>>>
>>>>
>>>> No, but to be correct it need to complete that to the end.
>>>>
>>>
>>> Saying this and knowing there is no end cannot possibly
>>> be construed as anything but intentional deception.
>>>
>> And what is saying that there is no end for a program that aborts and
>
> void Infinite_Recursion()
> {
> Infinite_Recursion();
> return;
> }
>
> Unless we divide the behavior of the tester from the test
> subject Infinite_Recursion() would be determined to halt.
Is that false or non-sense? I can only determine that it is not true.
--
Mikko