Re: Every sufficiently competent C programmer knows --- Very Stupid Mistake or Liars

On 3/14/2025 10:55 AM, joes wrote:

Am Fri, 14 Mar 2025 10:13:41 -0500 schrieb olcott:

On 3/14/2025 9:10 AM, Richard Damon wrote:

On 3/13/25 11:53 PM, olcott wrote:

On 3/13/2025 10:03 PM, Richard Damon wrote:

On 3/13/25 10:07 PM, olcott wrote:

On 3/13/2025 6:09 PM, Richard Damon wrote:

On 3/13/25 9:41 AM, olcott wrote:

On 3/13/2025 6:18 AM, Dan Cross wrote:

In article <vqud4e$36e14$3@dont-email.me>,
Fred. Zwarts <F.Zwarts@HetNet.nl> wrote:

Op 12.mrt.2025 om 16:31 schreef olcott:

 

But a complete emulation can?
>

Yes, but an HHH that gives an answer doesn't do one, due to the
pathological design of the template used to build DD to the HHH it
calls (which is the only HHH that can exist, or you have violated the
basic rules of programing and logic).
We have two basic cases,
1) if HHH does the partial emulation you describe, then the complete
emulation of DD will see that DD call HHH, and it will emulate its
input for a while, then abort and theu return 0 to DD which will then
halt.

int main()
{
    HHH(DDD); // No DDD can possibly ever return.
}

Since HHH doesn;t call DDD, the statement is vacuous and shows a
fundamental ignorance of what is being talked about.

Important distinction.
 

Yes, No HHH can emulated DDD to the end, but since halting is DEFINED
by the behavior of the program, and for every HHH that aborts and
returns, the program of DDD, as tested with:
int main()
{
     DDD()
}
will return to main, that shows that every HHH that returns 0 fails to
be a Halt Decider or Termination Analyzer. PERIOD..

>
The only difference between HHH and HHH1 is that they are at different
locations in memory. DDD simulated by HHH1 has identical behavior to
DDD() executed in main().

Oh, I thought it was an unconditional simulator.
The same code cannot produce different behaviour.
 

The semantics of the finite string input DDD to HHH specifies that it
will continue to call HHH(DDD) in recursive simulation.
The semantics of the finite string input DDD to HHH1 specifies to
simulate to DDD exactly once.

No. DDD always specifies the one same thing.
 

counter-factual
The semantics of the finite string input DDD to HHH specifies
that it will continue to call HHH(DDD) in recursive simulation.
The semantics of the finite string input DDD to HHH1 specifies
to simulate to DDD exactly once.
void DDD()
{
   HHH(DDD);
   return;
}
The only difference between HHH and HHH1 is that they are
at different locations in memory. DDD simulated by HHH1
has identical behavior to DDD() directly executed in main().
The semantics of the finite string input DDD to HHH specifies
that it will continue to call HHH(DDD) in recursive simulation.
The semantics of the finite string input DDD to HHH1 specifies
to simulate to DDD exactly once.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer