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On 10/28/24 11:08 PM, olcott wrote:Counter-factual. This algorithm has no ability to KNOW ITS OWN CODE.On 10/28/2024 9:56 PM, Richard Damon wrote:Which is what it would do, get stuck and fail to be a decider. It might figure out that it is emulating an emulating decider, at which point it knows that the decider might choose to abort its conditional emulation to return, so it needs to emulate further.On 10/28/24 9:09 PM, olcott wrote:>On 10/28/2024 6:56 PM, Richard Damon wrote:>>>
It is IMPOSSIBLE to emulate DDD per the x86 semantics without the code for HHH, so it needs to be part of the input.
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*You seemed to be a totally Jackass here*
You are not that stupid
You are not that ignorant
and this is not your ADD
>
_DDD()
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
>
At machine address 0000217a HHH emulates itself emulating
DDD without knowing that it is emulating itself.
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Then how did it convert the call HHH into an emulation of DDD again?
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When HHH (unknowingly) emulates itself emulating DDD this
emulated HHH is going to freaking emulate DDD.
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Did you think it was going to play poker?
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Only by recognizing itself, does it have grounds to say that if I don't abort, it never will, and thus I am stuck, so I need to abort.
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