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This example is the first step in Starcraft's CDKey Decode. This shuffles the characters in the key in a predictable way. I made this the second example because it's a little trickier, but it's not terribly difficult.
As in the previous example, try figuring this out on your own first!
lea edi, [esi+0Bh] mov ecx, 0C2h top: mov eax, ecx mov ebx, 0Ch cdq idiv ebx mov al, [edi] sub ecx, 11h dec edi cmp ecx, 7 mov bl, [edx+esi] mov [edi+1], bl mov [edx+esi], al jge top
Here, comments have been added explaining each line a little bit. These comments are added in an attempt to understand what the code's doing.
; This is actually continued from the last example, so esi contains the verified CDKey. lea edi, [esi+0Bh] ; edi is a pointer to the 12th character in the cdkey (0x0b = 11, and arrays start at 0). mov ecx, 0C2h ; Set ecx to 0xC2. Recall that ecx is often a loop counter. top: mov eax, ecx ; Move the loop counter to eax. mov ebx, 0Ch ; Set ebx to 0x0C (0x0C = 12, an arrays are indexed from 0, so the CDKey string goes from 0 to 12). cdq ; Get ready for a signed division. idiv ebx ; Divide the loop counter by 0x0C. It isn't clear yet whether this is division or modulus, but ; because an accumulator is being divided by the CDKey length, it's logical to assume that ; this is modular division. edx will likely be used, and eax will likely be overwritten. mov al, [edi] ; Move the value that edi points to (which is a character in the CDKey) to al. Recall that al is the ; right-most byte in eax. This confirms two things: that edi points to a character in the CDKey ; (since a character is 1 byte) and that the division above is modulus (because eax is overwritten). sub ecx, 11h ; Subtract 0x11 from ecx. Recall that ecx is often a loop counter, and likely is in this case. dec edi ; Decrement the pointer into the CDKey. edi started at the 12th character and is moving backwards. cmp ecx, 7 ; Compare ecx to 7, which confirms that ecx is the loop counter. The jump corresponding to this ; comparison is a few lines below. mov bl, [edx+esi] ; Recall that edx is the remainder from the above division, which is (accumulator % 12), and that ; esi points to the CDKey. So this takes the character corresponding to the accumulator and moves ; it into bl, which is the right-most byte of ebx. mov [edi+1], bl ; Overwrite the character that edi pointed to at the top of the loop. Recall that [edi] is moved ; into al, then decremented above, which is why this is +1 (to offset the decrement). mov [edx+esi], al ; Move al into the string corresponding to the modular division. jge top ; Jump as long as the ecx counter is greater than or equal to 7 ; ; Note that the loop here is simply a swap. edi decrements, starting from the 12th character and ; moving backwards. ecx is reduced by 0x11 each time, and the character at (ecx % 12) is swapped ; with the character pointed at by edi.