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tathanhdinh

tathanhdinh


code reformat

I'm cannot resolve the challenge yet, it's indeed very hard (at least for me). I would like just to know whether I've got the correct partial result or not.

I've managed to "dump" the key checking procedure, which locates on several non-contiguous pages (!?). The attached image is a part of it (I don't know how to capture all the function). I've found that there is a loop which reads each chararacter (input key is a wide string, each char is 2 bytes) by the instruction

movz ecx, [eax + ebx * 2]

the character is then checked with several values (e.g. "-", etc). But I still cannot go further.

panda_obfuscator.png

tathanhdinh

tathanhdinh


code reformat

I'm cannot resolve the challenge yet, it's indeed very hard (at least for me). I would like just to know whether I've got the correct partial result or not.

I've managed to "dump" the key checking procedure, the attached image is a part of it (I don't know how to capture all the function). I've found that there is a loop which reads each chararacter (input key is a wide string, each char is 2 bytes) by the instruction

movz ecx, [eax + ebx * 2]

the character is then checked with several values (e.g. "-", etc). But I still cannot go further.

panda_obfuscator.png

tathanhdinh

tathanhdinh

I'm cannot resolve the challenge yet, it's indeed very hard (at least for me). I would like just to know whether I've got the correct partial result or not.

I've managed to "dump" the key checking procedure, the attached image is a part of it (I don't know how to capture all the function). I've found that there is a loop which reads each chararacter (input key is a wide string, each char is 2 bytes) by the instruction

movz ecx, [eax + ebx * 2]

the character is then checked with several values (e.g. "-", etc). But I still cannot go further.

panda_obfuscator.png

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