Re: H(D,D) cannot even be asked about the behavior of D(D) V3 ---IGNORING ALL OTHER REPLIES

On 2024-06-17 13:03:56 +0000, olcott said:
 

On 6/17/2024 2:20 AM, Mikko wrote:

On 2024-06-16 12:47:09 +0000, olcott said:
>

On 6/16/2024 2:53 AM, Mikko wrote:

On 2024-06-15 16:22:09 +0000, olcott said:
>

On 6/13/2024 8:24 PM, Richard Damon wrote:
 > On 6/13/24 11:32 AM, olcott wrote:
 >>
 >> It is contingent upon you to show the exact steps of how H computes
 >> the mapping from the x86 machine language finite string input to
 >> H(D,D) using the finite string transformation rules specified by
 >> the semantics of the x86 programming language that reaches the
 >> behavior of the directly executed D(D)
 >>
 >
 > Why? I don't claim it can.
>
The first six steps of this mapping are when instructions
at the machine address range of [00000cfc] to [00000d06]
are simulated/executed.
>
After that the behavior of D correctly simulated by H diverges
from the behavior of D(D) because the call to H(D,D) by D
correctly simulated by H cannot possibly return to D.
>
_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]

>
When you put "V2" or "V3" or something similar on the subject line
you should tell what is different from the original version.
>

>
I ask what are the steps
I provide 6 steps and then ask what are the next steps.
I provide all of the steps.

>
In which version?
>

>
*This is the simplest possible version*
>
void DDD()
{
   H0(DDD);
}
>
After six steps of DDD are correctly emulated by H0
what machine address of DDD would it be at?
>
_DDD()
[00001fd2] 55               push ebp      ; housekeeping
[00001fd3] 8bec             mov ebp,esp   ; housekeeping
[00001fd5] 68d21f0000       push 00001fd2 ; push DDD
[00001fda] e8f3f9ffff       call 000019d2 ; call H0
[00001fdf] 83c404           add esp,+04   ; housekeeping
[00001fe2] 5d               pop ebp       ; housekeeping
[00001fe3] c3               ret           ; return to caller
Size in bytes:(0018) [00001fe3]

 So how is this a difference between the original version and V2 and V3?