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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Fri, 9 May 2025 11:14:12 +0200
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Op 09.mei.2025 om 04:13 schreef olcott:
> On 5/8/2025 8:30 PM, Keith Thompson wrote:
>> olcott <polcott333@gmail.com> writes:
>>> On 5/8/2025 6:49 PM, Keith Thompson wrote:
>>>> olcott <polcott333@gmail.com> writes:
>>>> [...]
>>>>> void DDD()
>>>>> {
>>>>>     HHH(DDD);
>>>>>     return;
>>>>> }
>>>>>
>>>>> If you are a competent C programmer then you
>>>>> know that DDD correctly simulated by HHH cannot
>>>>> possibly each its own "return" instruction.
>>>> "cannot possibly each"?
>>>> I am a competent C programmer (and I don't believe you can make
>>>> the same claim).  I don't know what HHH is.  The name "HHH" tells
>>>> me nothing about what it's supposed to do.  Without knowing what
>>>> HHH is, I can't say much about your code (or is it pseudo-code?).
>>>>
>>>
>>> For the purpose of this discussion HHH is exactly
>>> what I said it is. It correctly simulates DDD.
>>
>> Does HHH correctly simulate DDD *and do nothing else*?
>>
>> Does HHH correctly simulate *every* function whose address is passed
>> to it?  Must the passed function be one that takes no arguments
>> and does not return a value?
>>
>> Can HHH just *call* the function whose address is passed to it?
>> If it's a correct simulation, there should be no difference between
>> calling the function and "correctly simulating" it.
>>
>> My knowledge of C tells me nothing about *how* HHH might simulate
>> DDD.
>>
> 
> HHH can only simulate a function that take no arguments
> and has no return value. HHH also simulates the entire
> chain of functions that this function calls. These can
> take arguments or not and have return values or not.
> 
> Thus HHH ends up simulating itself (and everything
> that HHH calls) simulating DDD in an infinite
> sequence of recursive emulation until OOM error.
> 
>>> We need not know anything else about HHH to
>>> know that DDD correctly simulated by HHH cannot
>>> possibly REACH its own "return" instruction.
>>
>> Assuming that HHH(DDD) "correctly simulates" DDD, and assuming it
>> does nothing else, your code would be equivalent to this:
>>
>>      void DDD(void) {
>>          DDD();
>>          return;
>>      }
>>
> 
> Exactly. None of these people on comp.theory could
> get that even after three years.
> 
Only if you forget that your proposed HHH aborts and returns.
The whole discussion is about that proposed HHH, not the hypothetical 
HHH that does not abort.
You try to get people to agree with you by using the same name for very 
different things.