Deciphering File Formats in .NET

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Deciphering File Formats
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index, but not before it multiplies it by 8. This line essentially takes ESI, which was an index to the current file entry, and multiplies it by 19 * 8 = 152. Sounds familiar doesn t it You re right: 152 is the file entry length. By computing [ECX+EAX*8+8], Cryptex is obtaining the value of offset +8 at the current file entry. We already know that offset +8 contains the file size in clusters, and this value is being sent back to the caller using a parameter that was passed in to receive this value. Cryptex needs the file size in order to extract the file. After loading the file size, Cryptex checks for what is apparently another output parameter that is supposed to receive additional output data from this function, this time at [ESP+28]. If it is nonzero, Cryptex copies the value from offset +C at the file entry into the pointer that was passed and proceeds to copy offset +10 into offset +4 in the pointer that was passed, and so on, until a total of four DWORDs, or 16 bytes are copied. As a reminder, those 16 bytes are the ones that looked like junk when you dumped the file list earlier. Before returning to the caller, the function loads offset +4 at the current file entry and sets that into EAX it is returning it to the caller. To summarize, this sequence scans the file list looking for a specific file name, and once that entry is found it returns three individual items to the caller. The file size in clusters, an unknown, seemingly random 16-byte sequence, and another unknown DWORD from offset +4 in the file entry. Let s proceed to see how this data is used by the file extraction routine.
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Decrypting the File
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After returning from 004017B0, Cryptex proceeds to scan the supplied file name for backslashes and loops until the last backslash is encountered. The actual scanning is performed using the C runtime library function strchr, which simply returns the address of the first instance of the character, if one is found. The address that points to the last backslash is stored in [ESP+20]; this is essentially the clean version of the file name without any path information. One instruction that draws attention in this otherwise trivial sequence is the one at 00401C9E.
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You might recall that we ve already seen a similar instruction in the previous chapter. In that case, it was used as an infrastructure to allow people to trap system APIs in Windows. This case is not relevant here, so why would the compiler insert an instruction that does nothing into the middle of a function The answer is simple. The address in which this instruction begins is unaligned, which means that it doesn t start on a 32-bit boundary. Executing unaligned instructions (or accessing unaligned memory addresses in general)
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takes longer for 32-bit processors. By placing this instruction before the loop starts the compiler ensured that the loop won t begin on an unaligned instruction. Also, notice that again the compiler could have used NOPs, but instead used this instruction which does nothing, yet accurately fills the 2-byte gap that was present. After obtaining a backslash-free version of the file name, the function goes to create the new file that will contain the extracted data. After creating the file the function checks that 004017B0 actually found a file by testing EBP, which is where the function s return value was stored. If it is zero, Cryptex displays a file not found error message and quits. If EBP is nonzero, Cryptex calls the familiar 00401030, which reads and decrypts a sector, while using EBP (the return value from 004017B0) as the second parameter, which is treated as the cluster number to read and decrypt. So, you now know that 004017B0 returns a cluster index, but you re not sure what this cluster index is. It doesn t take much guesswork to figure out that this is the cluster index of the file you re trying to extract, or at least the first cluster for the file you re trying to extract (most files are probably going to occupy more than one cluster). If you go back to our discussion of the file lookup function, you see that its return value came from offset +4 in the file entry (see instruction at 004018BC). The bottom line is that you now know that offset +4 in the file entry contains the index of the first data cluster. If you look in the debugger, you will see that the third parameter is a pointer into which the data was decrypted, and that after the function returns this buffer contains the lovely asterisks! It is important to note that the asterisks are preceded by a 4-byte value: 0000046E. A quick conversion reveals that this number equals 1134, which is the exact file size of the original asterisks.txt file you encrypted earlier.
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