Re: Who here understands that the last paragraph is Necessarily true?

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Sujet : Re: Who here understands that the last paragraph is Necessarily true?
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory
Date : 17. Jul 2024, 02:12:36
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <246e3d712c3ebce03096f17a624ff960e7651885@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
User-Agent : Mozilla Thunderbird
On 7/16/24 9:41 AM, olcott wrote:
On 7/16/2024 2:28 AM, Fred. Zwarts wrote:
Op 15.jul.2024 om 21:40 schreef olcott:
On 7/15/2024 2:30 PM, Fred. Zwarts wrote:
Op 15.jul.2024 om 04:33 schreef olcott:
On 7/14/2024 9:04 PM, Richard Damon wrote:
On 7/14/24 9:27 PM, olcott wrote:
>
Any input that must be aborted to prevent the non termination
of simulating termination analyzer HHH necessarily specifies
non-halting behavior or it would never need to be aborted.
>
Excpet, as I have shown, it doesn't.
>
Your problem is you keep on ILEGALLY changing the input in your argument because you have misdefined what the input is.
>
>
_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]
>
The input *is* the machine address of this finite
string of bytes: 558bec6863210000e853f4ffff83c4045dc3
>
>
It seems that you do not understand x86 language. The input is not a string of bytes, but an address (00002163). This points to the starting of the code of DDD. But a simulation needs a program, not a function calling undefined other functions. Therefore, all functions called by DDD (such as HHH) are included in the code to simulate.
>
*The input is the machine address of this finite*
*string of bytes: 558bec6863210000e853f4ffff83c4045dc3*
>
It seems that olcott does not understand the x86 language.
The input for HHH is an address, not a finite string.
 This is such a terribly incorrect reply that I am ignoring it. Try again
 
No, it was a VERY RIGHT reply.
You DON'T know the x86 language, particularly how a call instruction works.

Date Sujet#  Auteur
12 Jul 25 o 

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