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that the string starts at offset +14 in the structure, you can assume that there aren t any additional data entries after it in the structure, which would mean that the maximum length of a file name in Cryptex is 152 20, or 132 bytes. Once this loop ends, an interesting thing takes place. The first member in the buffer you read and decrypted earlier is tested, and if it is nonzero, Cryptex calls the function at 00401030, the function from Listing 6.7 that reads and decrypts a data chunk that we analyzed earlier. The second parameter, which is used as a kind of cluster number (remember how the function multiplies that number by 4104 ), is taken directly from that first member. Clearly the idea here is to read and decrypt another chunk of data and scan it for files. It looks likes the file list can span an arbitrary number clusters and is essentially implemented using a sort of cluster linked list. This brings up one question: Is the first cluster hard-coded to number one Let s take a look at the code that made the initial call to read the first file-list cluster, from Listing 6.6.

00401A1E 00401A24 00401A25 00401A26 00401A27 MOV EDX,DS:[406064] PUSH ECX PUSH EDX PUSH ESI CALL cryptex.00401030

(19.18)

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The first-cluster index is taken from a global variable with a familiar address. It turns out that 00406064 is a part of the Cryptex header loaded into 00406058 just a few lines earlier. So, it looks like offset +0C in the Cryptex header contains the index of the first cluster in the file table. Going back to Listing 6.7, after 00401030 returns, ESI is tested for a nonzero value again (even though it has already been tested and its value couldn t have been changed), and if it is nonzero Cryptex loops back into the code that reads the file table. You now know that the first member in these file table clusters is the next cluster element that tells Cryptex which cluster contains the following file table entries, if any. Because the size of each file entry is fixed, there must also be a fixed number of entries in each cluster. Since a local variable at [ESP+14] is used for counting the remaining number of items in the current cluster, you easily find the instruction at 00401A67, which initializes this variable to 0x1A (26 in decimal), so you know that each cluster can contain up to 26 file entries. Finally, it is important to pay attention to three lines in Listing 6.6 that we ve so far ignored.

2 Sheet the main sail in hard and move the boat towards the wind with minimum tiller movement 1 Try and start the tack at full speed

challenge, where the protected program takes the volume serial number and the username and generates a challenge, which is just a long number. The user is then given that number and is supposed to call the software vendor and ask for a valid product key that will be generated based on the supplied number. In such cases, a keygen would simply convert the challenge to the product key. As its name implies, KeygenMe-3 was meant to be keygenned, so by patching it you were essentially cheating. Let s rectify the situation by creating a keygen for KeygenMe-3.

Solution of Equation (19.18) for a system with n displacement components and m elements gives an n m matrix: u1 u1 u1 M 1 M 2 M m u2 u2 u2 M 1 M 2 M m (19.19) . . . . . . . . . un un un M 1 M 2 M m

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