Sujet : Re: DDD correctly emulated by EEE --- Correct Emulation Defined
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 23. Mar 2025, 18:38:20
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vrpguc$2qbhf$3@dont-email.me>
References : 1 2 3 4 5 6 7
User-Agent : Mozilla Thunderbird
On 3/23/2025 6:08 AM, Richard Damon wrote:
On 3/22/25 11:57 PM, olcott wrote:
On 3/22/2025 9:53 PM, Richard Damon wrote:
On 3/22/25 2:00 PM, olcott wrote:
On 3/22/2025 12:34 PM, Richard Damon wrote:
On 3/22/25 10:52 AM, olcott wrote:
_DD()
[00002133] 55 push ebp ; housekeeping
[00002134] 8bec mov ebp,esp ; housekeeping
[00002136] 51 push ecx ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call EEE(DD)
[00002141] 83c404 add esp,+04
[00002144] 8945fc mov [ebp-04],eax
[00002147] 837dfc00 cmp dword [ebp-04],+00
[0000214b] 7402 jz 0000214f
[0000214d] ebfe jmp 0000214d
[0000214f] 8b45fc mov eax,[ebp-04]
[00002152] 8be5 mov esp,ebp
[00002154] 5d pop ebp
[00002155] c3 ret
Size in bytes:(0035) [00002155]
>
When finite integer N instructions of the above x86
machine language DD are emulated by each x86 emulator
EEE[N] at machine address [000015c3] according to the
semantics of the x86 language no DD ever reaches its own
"ret" instruction at machine address [00002155] and
terminates normally.
>
>
Your can't emulate the above code for N > 4, as you get into undefine memory.
>
>
I have already addressed this objection dozens of times.
>
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No you haven't. You have given several different LIES about it.
>
As I have pointed out, if you don't include Halt7.c as part of the definition, then you can't do it as you are looking at undefined memory.
>
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Your lack of technical competence is showing.
(1) We are talking about a hypothetical infinite
set of pure x86 emulators that have no decider code.
>
(2) The memory space of x86 machine code is not
in the C source file, it is in the object file.
>
Then your "input" isn't the C source files, but the memory, and ALL of it, and thus in your (1), each member of the set got a different input (as reference memory changed) and none of those apply to your case with HHH.
You just continue to prove that you don't understand the meaning of the terms you are using, or you are intentionally hiding your fradulant change of meaning of those terms.
Command line arguments:
x86utm Halt7.obj > Halt7out.txt
All of the x86 functions remain at their same fixed
offset from the beginning of Halt7.obj
Your logic is just build on self-contradictions and thus is unsound.
Sorry, you are just proving your utter stupidity and ignorance of what you speak about.
-- Copyright 2025 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer