sh edx
xor eax, eax ; Zeroing working registers
xor edx, edx
RepeatString: ; Byte-by-byte string-processing loop

lodsb ; Reading the next byte into AL
test al, al ; Has the end of the string been reached?
jz short Exit

; The value of the counter that processes 1 byte of the string
; must be choosen so that all bits are intermixed, but parity
; (oddness) is provided for the result of transformations
; performed by the XOR operation.

mov ecx, 21h
RepeatChar:
xor edx, eax ; Repeatedly replacing XOR with ADC
ror eax, 3
rol edx, 5
call $+5 ; EBX = EIP
pop ebx ; /
xor byte ptr [ebx-0Dh], 26h;
; This instruction provides for the loop.
; The XOR instruction is replaced with ADC.
loop RepeatChar
jmp short RepeatString

Exit:

xchg eax, edx ; The result of work (ser.num) in EAX
pop edx ; Restoring the registers
pop ecx
pop ebx
pop esi
retn ; Returning from the function

这个算法有点怪异--

因为不断的调用一个函数,并且给它传递同样的参数有可能产生或者一样,或者完全不同的结果!这在于用户名的长度,当函数结束的时候,如果他是odd,xor被ADC替换,如果是偶数,看起来什么都没发生.

好的,这些是全部了,我希望你最少了解一件事情,这花了我整整两小时,欢迎反馈.