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On 2/22/2025 6:02 PM, Richard Damon wrote:Right, and since every instruction (except the final return) specifies to execute the next instruction, it is NEVER a "correct simulation" to abort the simulaition unless the instruction will cause your computer to actually lock up or self-distruct.On 2/22/25 11:52 AM, olcott wrote:*Correct simulation means emulates the machine code as specified*On 2/22/2025 5:05 AM, joes wrote:>Am Thu, 20 Feb 2025 18:25:27 -0600 schrieb olcott:>On 2/20/2025 4:38 PM, Alan Mackenzie wrote:Honestly, you're gonna die first, one way or the other.olcott <polcott333@gmail.com> wrote:As soon as people fully address rather than endlessly dodge my keyOn 2/20/2025 2:38 AM, Mikko wrote:>On 2025-02-20 00:31:33 +0000, olcott said:Yes. It would be a relief if you could move on to posting somethingEvery post that I have been talking about for two or more years hasI have given everyone here all of the complete source code for a fewTrue but irrelevant. OP did not specify that HHH means that
years
particular code.
referred to variations of that same code.
new and fresh.
points I will be done.
>Let's start with a root point.Since DD halts, that's dead in the water.
All of the other points validate this root point.
*Simulating termination analyzer HHH correctly determines*
*the non-halt status of DD*
>
Despicably intentionally dishonest attempts at the straw-man
deception aside:
>
DD correctly simulated by HHH cannot possibly terminate
normally by reaching its own "return" instruction.
>
Only because that statement is based on a false premise.
>
Since HHH doesn't correctly simulate its input, your statement is just a fabrication of your imagination.
It cannot mean imagining a different sequence than the one that the machine code specifies. That most people here are clueless aboutRight, so the simulation of the input below requires that the call at address 0000213c to be followed by the simulation of the contents of the memory location 000015c3.
x86 machine code is far less than no rebuttal at all.
_DD()Sure it can, if the emulator is a correct emulator, since the HHH that DD calls was defined to abort its emulation and return 0.
[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 HHH(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 DD emulated by HHH calls HHH(DD) this call cannot
possibly return to the emulator, conclusively proving
that
DD correctly simulated by HHH cannot possibly terminateThe problem is you base your logic on bald face lies, since your HHH doesn't do a correct simulation.
normally by reaching its own "return" instruction.
Assuming that it does return is simply stupid.
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