Re: DDD correctly emulated by EEE --- Correct Emulation Defined --- CUT-AND-PASTE FAILED

On 3/23/2025 6:55 PM, Richard Damon wrote:

On 3/23/25 6:56 PM, olcott wrote:

On 3/23/2025 4:46 PM, Richard Damon wrote:

On 3/23/25 1:38 PM, olcott wrote:

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.
>

>
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.
>

>
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

>
So?
>
You still need to make the decision, is Halt7.c / Halt7.obj part of the INPUT to the decider, and thus either you can't change the code in it, or you need to consider each version a different input, or
>

>
_III()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov  ebp,esp  ; housekeeping
[00002175] 6872210000 push 00002172 ; push III
[0000217a] e853f4ffff call 000015d2 ; call EEE(III)
[0000217f] 83c404     add  esp,+04
[00002182] 5d         pop  ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
>
In other words an infinite set of pure x86 emulators
with each one stored at machine address 000015d2
that can be called from the above fixed finite string
of machine code IS UTTERLY BEYOND ANYTHING THAT YOU
CAN POSSIBLY IMAGINE.
>
I don't buy it. You are  neither that stupid nor
that ignorant.
>

 You can't have two different programs in one memory location at the same time.
 

CUT-AND-PASTE FAILED
I will dumb it down for you.
Try to come up with one x86 emulator EEE at machine
address 000015d2 that emulates III according to the
semantics of the x86 language and this emulated III
reaches its own machine address 00002183.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer