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On 2/27/2025 2:45 PM, joes wrote:Except that is just a LIE.Am Thu, 27 Feb 2025 14:13:13 -0600 schrieb olcott:On 2/27/2025 3:50 AM, Fred. Zwarts wrote:>Op 27.feb.2025 om 06:00 schreef olcott:On 2/26/2025 5:50 PM, Richard Damon wrote:On 2/26/25 9:49 AM, olcott wrote:On 2/26/2025 3:48 AM, Fred. Zwarts wrote:Op 26.feb.2025 om 05:50 schreef olcott:>On 2/25/2025 10:21 PM, Richard Damon wrote:You already know that you are stupidly wrong about the emulation
being incorrect or you would have provided the correct emulation
sequence long ago.If we assume that HHH emulates the above machine code then there isAnd you are missing the point that HHH can not correctly emulate 15You dodged the actual question.What are the first 15 lines of DD correctly emulated by HHH?The error in the simulation occurs already at the 5th instruction,
the 'call 000015c3'. Instead of simulating this instruction,
What are the correct first 15 lines of DD emulated by HHH.
line of this DD, as it is missing information and thus the task is
improperly defined.
NO MORE relevant information needed and we know that machine address
0000213c of DD is followed by machine address 00002133 of DD.No. Relevant information includes the code at address 0x213c and_DD()
everything it points to.
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[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]
THE ONLY FREAKING THING THAT NEED BE KNOWN
ABOUT THE CODE AT THAT ADDRESS
(TO DETERMINE THE SEQUENCE OF MACHINE INSTRUCTIONS OF DD)
IS THAT IT EMULATES THE MACHINE INSTRUCTIONS OF DD.
From this base assumption we know that the firstNope, that is just impossible, as even if HHH was an emulator, the emulator emulating this DD will NEVER get back to that address, it will only emulate HHH emulating those instruction
five instructions of DD are repeated three times
when the first 15 instructions of DD are emulated.
You won't know thus if you have no idea what the
term "emulate" means.
Easy. It calls a program that returns *by definition* .If assuming that following the call would lead to infinite recursionIf it was true that the above instance of DD reaches its "ret"
(which is not true, because DD has been proven to halt),
instruction then you could show how it does this.
>
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