• Re: Any honest person that knows the x86 language can see... DDD will H

    From Richard Damon@21:1/5 to olcott on Mon Jul 29 19:48:56 2024
    XPost: comp.theory

    On 7/29/24 10:07 AM, olcott wrote:
    HHH(Infinite_Recursion) and HHH(DDD) show the same non-halting
    behavior pattern in their derived execution traces of their
    inputs.


    No they don't.

    Infinite_Recursion calls Infinite_Recursion without any conditional instructions in the full cycle.

    DDD calls HHH(DDD) which does a CONDITIONAL emul

    Correct emulation is defined as emulating the machine language
    input according to the x86 semantics specified by this input.

    Right, which HHH doesn't do, since that would require it correctly
    emulating the instructions within HHH


    For DDD correctly emulated by HHH this includes HHH emulating
    itself emulating DDD according to the x86 semantics of itself.

    Right, which it doesn't do. Note, that is NOT look at the emulation that
    HHH does, it is look at HHH doing the emulation.


    HHH(DDD) shows the exact same execution trace behavior pattern
    as HHH(Infinite_Recursion) where 3-4 instructions are repeated
    with no conditional branch instructions in this trace that could
    prevent them from endlessly repeating.

    Nope, because HHH(DDD) sees DDD call HHH, not DDD, and thus the CORRECT emulation of this needs to look at the ACTUAL BEHAVIOR of the code of
    HHH, which it doesn't do.

    You are just showing that you are totally ignorant of what you are
    trying to talk about,


    void Infinite_Recursion()
    {
      Infinite_Recursion();
    }

    _Infinite_Recursion()
    [0000215a] 55         push ebp      ; 1st line
    [0000215b] 8bec       mov ebp,esp   ; 2nd line
    [0000215d] e8f8ffffff call 0000215a ; 3rd line
    [00002162] 5d         pop ebp
    [00002163] c3         ret
    Size in bytes:(0010) [00002163]

    *THREE lines repeat with no conditional branch instructions*
    Begin Local Halt Decider Simulation   Execution Trace Stored at:113934 [0000215a][00113924][00113928] 55         push ebp      ; 1st line
    [0000215b][00113924][00113928] 8bec       mov ebp,esp   ; 2nd line [0000215d][00113920][00002162] e8f8ffffff call 0000215a ; 3rd line [0000215a][0011391c][00113924] 55         push ebp      ; 1st line
    [0000215b][0011391c][00113924] 8bec       mov ebp,esp   ; 2nd line [0000215d][00113918][00002162] e8f8ffffff call 0000215a ; 3rd line
    Local Halt Decider: Infinite Recursion Detected Simulation Stopped

    If you cannot see that the above x86 machine code proves that
    it will never halt then you can't possibly understand what I
    have been saying.

    The first three lines of _Infinite_Recursion() repeat and there
    are no conditional branch in that sequence that can possibly keep
    it from repeating forever.

    HHH(DDD) is the exact same pattern is shown below. The first
    four lines of DDD repeat and there are are no conditional branch
    in that sequence that can possibly keep it from repeating forever.

    Nope.


    void DDD()
    {
      HHH(DDD);
    }

    _DDD()
    [00002177] 55               push ebp      ; 1st line [00002178] 8bec             mov ebp,esp   ; 2nd line
    [0000217a] 6877210000       push 00002177 ; push DDD
    [0000217f] e853f4ffff       call 000015d7 ; call HHH
    [00002184] 83c404           add esp,+04
    [00002187] 5d               pop ebp
    [00002188] c3               ret
    Size in bytes:(0018) [00002188]

    *FOUR lines repeat with no conditional branch instructions*
    Begin Local Halt Decider Simulation   Execution Trace Stored at:113895 [00002177][00113885][00113889] 55         push ebp      ; 1st line
    [00002178][00113885][00113889] 8bec       mov ebp,esp   ; 2nd line [0000217a][00113881][00002177] 6877210000 push 00002177 ; push DDD [0000217f][0011387d][00002184] e853f4ffff call 000015d7 ; call HHH

    And the below is *NOT* the correct emulation of what a call HHH does.

    Proving you have inadequate knowledge of the x86 processor to be making
    your claims.


    [00002177][0015e2ad][0015e2b1] 55         push ebp      ; 1st line
    [00002178][0015e2ad][0015e2b1] 8bec       mov ebp,esp   ; 2nd line [0000217a][0015e2a9][00002177] 6877210000 push 00002177 ; push DDD [0000217f][0015e2a5][00002184] e853f4ffff call 000015d7 ; call HHH
    Local Halt Decider: Infinite Recursion Detected Simulation Stopped

    Because HHH has no idea that it is calling itself HHH only sees
    the same Infinite Recursion behavior pattern that it saw with Infinite_Recursion().


    If it has no idea it is calling itself, why does it think that a call to 000015d7 will cause an emulation of the program at the address on the
    top of the stack?

    If it has been "told" that at 000015d7 is an unconditional emulator, it
    has been programmed incorrectly, and thus is just flawed, and the
    programmer (which was you) is just a LIAR.

    So, HHH just doesn't do what you claim, so I guess you are just
    admitting to being that ignorant pathological lying idiot that just has
    a reckless disregard for the truth,

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