HHH maps its input to the behavior specified by it --- never reaches its halt state

On 8/8/2024 2:00 AM, Mikko wrote:

On 2024-08-07 13:07:06 +0000, olcott said:
 

On 8/7/2024 1:48 AM, Mikko wrote:

On 2024-08-05 15:16:27 +0000, olcott said:
>

I have been working in the x86 language back when my work
computer at the US Army corps of engineers was an IBM PC
with an 8088 processor, 512K of RAM and dual floppy drives.
>
I was creating dBASE III systems on this computer. This was
before the 8086 processor even existed thus the name x86
language did not yet exist.

>
Intel 8088 is a variant of 8086 for less expensive computers.
Intel 8086 already exsted when the first 8088 computers were
sold. Later Intel develped 80188, 80186, and other processors
that cold run programs that were written or compiled for 8086,
so someone coined the term x86 for the family.
>

>
Can you write programs in this language?
I have written many interrupt intercept TSR
programs in the 8088 versions of the language.
I was doing my own time slicing back in 1987.

 I have done that tor 8088 and some other poocessors but not
recently so my skills may be rusty. Only rarely there is any
need for machine language programming.
 

typedef void (*ptr)();
int HHH(ptr P); // simulating termination analyzer
void DDD()
{
   HHH(DDD);
   return;
}
Each HHH of every HHH that can possibly exist definitely
emulates zero to infinity instructions of DDD correctly.
Every expert in the C language sees that this emulated DDD
cannot possibly reaches its own "return" instruction halt state.
It seems that every rebuttal that anyone can possibly make is
necessarily erroneous because the above paragraph is a tautology.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer