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From: joes <noreply@example.com>
Newsgroups: comp.theory
Subject: Re: D correctly simulated by H proved for THREE YEARS ---
Date: Mon, 10 Jun 2024 08:35:09 -0000 (UTC)
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Am Sun, 09 Jun 2024 22:54:52 -0500 schrieb olcott:
> On 5/29/2021 2:26 PM, olcott wrote:
> https://groups.google.com/g/comp.theory/c/dTvIY5NX6b4/m/cHR2ZPgPBAAJ
> 
> THE ONLY POSSIBLE WAY for D simulated by H to have the same behavior as
> the directly executed D(D) is for the instructions of D to be
> incorrectly simulated by H (details provided below).
> 
> _D()
> [00000cfc](01)  55                      push ebp
> [00000cfd](02)  8bec                    mov ebp,esp
> [00000cff](03)  8b4508                  mov eax,[ebp+08]
> [00000d02](01)  50                      push eax       ; push D
> [00000d03](03)  8b4d08                  mov ecx,[ebp+08]
> [00000d06](01)  51                      push ecx       ; push D
> [00000d07](05)  e800feffff              call 00000b0c  ; call H
> [00000d0c](03)  83c408                  add esp,+08
> [00000d0f](02)  85c0                    test eax,eax
> [00000d11](02)  7404                    jz 00000d17
> [00000d13](02)  33c0                    xor eax,eax
> [00000d15](02)  eb05                    jmp 00000d1c
> [00000d17](05)  b801000000              mov eax,00000001
> [00000d1c](01)  5d                      pop ebp
> [00000d1d](01)  c3                      ret Size in
> bytes:(0034) [00000d1d]
> 
> In order for D simulated by H to have the same behavior as the directly
> executed D(D) H must ignore the instruction at machine address
> [00000d07]. *That is an incorrect simulation of D*
I don't understand. Does D(D) ignore the call to H(D,D)?

> H does not ignore that instruction and simulates itself simulating D.
> The simulated H outputs its own execution trace of D.

-- 
joes