Re: DDD correctly emulated by HHH cannot possibly halt

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Sujet : Re: DDD correctly emulated by HHH cannot possibly halt
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory
Date : 11. Jul 2024, 03:21:06
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v6nc22$2501i$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
User-Agent : Mozilla Thunderbird
On 7/10/2024 8:11 PM, Richard Damon wrote:
On 7/10/24 9:01 PM, olcott wrote:
On 7/10/2024 7:37 PM, Richard Damon wrote:
On 7/10/24 8:24 PM, olcott wrote:
On 7/10/2024 7:01 PM, Richard Damon wrote:
On 7/10/24 9:41 AM, olcott wrote:
On 7/10/2024 8:27 AM, joes wrote:
Am Tue, 09 Jul 2024 23:19:25 -0500 schrieb olcott:
On 7/9/2024 11:01 PM, joes wrote:
  > That means that HHH doesn't return, in particular that it doesn't
  > abort.
DDD correctly emulated by any pure function HHH that correctly emulates
1 to ∞ steps of DDD can't make it past the above line of code no matter
what.
That line being the call to itself -> it can't simulate itself.
>
*DDD NEVER HALTS*
DDD ONLY calls HHH...
>
>
void DDD()
{
   HHH(DDD);
   return;
}
>
DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ lines of DDD can't make it
to the second line of DDD no matter what.
>
>
>
Nope, DDD does if HHH(DDD) returns.
>
>
You have a dead cat in your driveway does not mean that
you have a peanut butter sandwich on your front porch.
It has taken you at least 1000 messages to see that.
>
DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ lines of DDD can't make it
to the second line of DDD no matter what.
>
WRONG, you don't seem to understand the difference between DDD and HHH's emualtion of it.
>
>
>
Would you bet your immortal soul that DDD simulated
by HHH (as provided above) would terminate normally?
>
 That is a ambiguous statement, showing your attempt at deciet.
 
We stipulate that the only measure of a correct emulation
is the semantics of the x86 programming language. By this
measure when 1 to ∞ steps of DDD are correctly emulated
by each pure function x86 emulator HHH (of the infinite
set of every HHH that can possibly exist) then DDD cannot
possibly reach its own machine address of 00002174 and halt.
_DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]
Will you bet your immortal soul that any of the above
DDD of DDD/HHH pairs reach their own machine address
of 00002174 and halt?
Claiming ambiguity without specifically pointing this
out will be construed as deception thus swearing your
allegiance to the father of lies.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Date Sujet#  Auteur
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