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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: Everyone here seems to consistently lie about this
Date: Mon, 5 Aug 2024 10:23:52 +0300
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On 2024-08-04 12:30:26 +0000, olcott said:

> On 8/4/2024 1:11 AM, Fred. Zwarts wrote:
>> Op 03.aug.2024 om 17:01 schreef olcott:
>>> On 8/3/2024 9:54 AM, Fred. Zwarts wrote:>>>
>>>> Talking nonsense does not hide you problem. I don't disagree with that 
>>>> semantics.
>>>> It is HHH that deviates from the semantics of the x86 language by 
>>>> skipping the last few instructions of a halting program, changing its 
>>>> behaviour in this way.
>>> 
>>> There are no last few instructions of any halting program
>>> that DDD correctly emulated by HHH skips.
>> 
>> Why substituting facts by dreams?
>> DDD halts when HHH halts. HHH skips tte last cycle of the simulated 
>> HHH,after which it would return to DDD, which would then return too.
>> 
>>> 
>>> Within the semantics of C and the semantics of the x86
>>> language (thus specifying a correct simulation) the call
>>> to HHH(DDD) from the simulated DDD cannot possibly return.
>>> 
>> 
>> Indeed, that is why it is incorrect.
> 
> Would the call from DDDD to ExecuteInput(DDDD) return?
> 
> // This is ordinary C and I compiled and ran it.
> 
> typedef void (*ptr)();
> 
> void ExecuteInput(ptr x)
> {
>    x();
> }
> 
> void DDDD()
> {
>    ExecuteInput(DDDD);
>    return;
> }
> 
> int main()
> {
>    ExecuteInput(DDDD);
> }

Most likely it will abort or clrash for stack overflow. A compiler may
detect that the functions have no side effects and no return value and
traslate them as

void DDDD() {
  reurn;
}

or, if it also detects that there is a indirect infinite recursion,
as

void DDDD() {
  LOOP: goto LOOP;
}

but your compiler doesn't think so much.

-- 
Mikko