CDkey2 - skulllsecurity.org

cdkey2

;Continuation from previous cdkey asm, this time I try to understand the next variation in asm code.

  lea     edi,  [esi + 0xb]       ; edi = address of cdkey[0 + 11], address of 12th digit 
  mov     ecx,  0xc2              ; ecx = 194

  top:
  mov     eax,  ecx               ; eax = 194 or 0xc2 
  mov     ebx,  0xc               ; ebx = 12
  cdq                             ; Convert eax from Double word to Quadruple word. so eax becomes edx:eax
                                  ; this also zero out any value previously in edx.
  idiv    ebx                     ; eax / ebx. 194/12. values now becomes eax = 16, edx = 2
  mov     al,   [edi]             ; al = 12th digit hex value
  sub     ecx,  0x11              ; 0xc2 - 0x11. 194 - 17. ecx = 177 or 0xb1
  dec     edi                     ; edi = cdkey[11 - 1]. edi = address of 11th digit cdkey
  cmp     ecx,  7                 ; is ecx = 7? 177 = 7?
                                  ; this next few instructions basically swap digits around.
  mov     bl,   [edx + esi]       ; bl = cdkey [0 + 2] = 3rd digit cdkey hex value
  mov     [edi + 1],  bl          ; copy 3rd digit hex value to cdkey[10+1] 12th digit.
  mov     [edx + esi],  al        ; copy 12th digit hex value to cdkey[0+2] 3rd digit.
  
  jge     top                     ; if ecx is greater or equal to 7, jump to top.
                                  ; else, continue to next instructions.
                                  
#Sample of asm code rewritten in python:

#!/usr/bin/env python

cdkey = raw_input('Enter 13 digit cd key: ')

ecx = 194
edi = 12
iter = 0

while ecx >= 7:
  iter += 1
  remainder = ecx % 12
  ecx = ecx-17
  cdkeylist=list(cdkey)
  cdkeylist[edi] = cdkey[remainder]
  cdkeylist[remainder] = cdkey[edi]
  
  print('#{} Swapping digit position {} with {}. \n{} now becomes {}.\n'.format(iter, edi+1, remainder+1, cdkey, ''.join(cdkeylist)))
  edi = edi-1
  cdkey = ''.join(cdkeylist)