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I also tried to crack exactly this program a while ago. The company (I believe it is MilSoft, one of the most reputable software companies in Turkey) sent this
by ckaygusu 12y ago
I also tried to crack exactly this program a while ago. The company (I believe it is MilSoft, one of the most reputable software companies in Turkey) sent this challenge to university students to hire a part-time CS student. Nevertheless, this was the first time I've ever attempted to crack something and while I had little to no idea what was going on, it was a very thrilling experience. I think I went on 14 hours without taking a break.
I began by trying to run the program in GDB, got SIGSEGV'd. Afterwards I inspected the faulty address and tried to avoid it by changing its value, instead it crashed at somewhere else. After trying this hopeless catch-and-run for several hours, I decided I needed a better disassembly tool and went on to IDA Pro.
This particular program contains a trick that intrigued me very much, and it is the reason why I was getting SIGSEGV'd at different locations when altering the program code.
The main payload of this program is simply XOR-encrypted by some key. The whole thing begins by decrypting the payload and then begins its execution as normal. The gist is, the particular key that encrypted the main payload is the decryption code itself (for the unacquainted, assembly code is also just a byte stream). Here, this exact part:
0x804762d: mov $0xaa,%dl
0x804762f: mov $0x8048480,%edi
0x8047634: mov $0x8048cbc,%ecx
0x8047639: mov %edi,0x80476f3
0x804763f: mov %ecx,0x80476f7
0x8047645: sub %edi,%ecx
0x8047647: mov $0x804762f,%esi
0x804764c: push $0x80476c1
0x8047651: pusha
0x8047652: mov $0x55,%al
0x8047654: xor $0x99,%al
0x8047656: mov $0x8047656,%edi
0x804765b: mov $0x80476e5,%ecx
0x8047660: sub $0x8047656,%ecx
0x8047666: repnz scas %es:(%edi),%al
0x8047668: je 0x804770a
0x804766e: mov %edi,0x80476eb
0x8047674: popa
0x8047675: add 0x80476eb,%edx
0x804767b: ret
As far as I can remember, the key was a bit more than that, but I'm sure it was including this part.
At the end of every iteration (of something involving this loop which I can't precisely recall now) the program checks whether it is running under debug mode (essentially makes a PTRACE call and reads its output, the OP also talks about it) If this is the case, it makes a jump to random address, so even if you are just neatly watching the program run under debug mode, you weren't going to achieve anything.
The next thing that occured to me is to manipulate how PTRACE returns its value, but I thought it would involve some kernel code fiddling and running the program under the modified kernel, which is WAY beyond my ability for now. I didn't know how to do it, but later by some very stupid trick I managed to pass this decryption part and the program made a jump to something like "__glibc_start". I needed to save the altered program and run it under gdb again (I don't remember why), but I was using the trial version of IDA Pro which prohibits me of such a thing. After making a few more desperate attempts I gave up.
But this "using the code as the key".. I think spending 14 hours to see this done was well worth it.
- sillysaurus3 12y agoAt the end of every iteration (of something involving this loop which I can't precisely recall now) the program checks whether it is running under debug mode (essentially makes a PTRACE call and reads its output, the OP also talks about it) If this is the case, it makes a jump to random address, so even if you are just neatly watching the program run under debug mode, you weren't going to achieve anything. Could you change the jmp into a nop? That should let you attach in debug mode.
- ckaygusu 12y agoI believe it was also the part of the key.
- sillysaurus3 12y agoCould you change the jmp into a nop, then xor every nth byte of the program with jmp xor nop, where n = whatever offset the jmp was at in the key? The result should be a valid decryption.
- ckaygusu 12y agoI see. That would work I suppose, but at that time I couldn't think of this.
- annnnd 12y agoWouldn't that also change the xor-ed instructions? That is, wouldn't the program behave differently?
- sillysaurus3 12y agoNah, if instruction "foo" is xor'd with "jmp", then xoring foo with "jmp xor nop" will remove the jmp and add the nop. Then the nop is removed during decryption, because "foo xor nop" xor nop = foo. The reason this is clean and convenient is because nop is a single byte (0x0f or 0x90), which lets you replace any instruction of any size with nops. But if you had to transform jmp into an instruction of a different size, things could get hairy if the byte sizes don't line up. But you could still replace several instructions with other code.