python - 密码学:将 C 函数转换为 python

Question

我正在尝试解决密码学问题( https://callicode.fr/pydefis/Reverse/txt )，该算法使用以下 C 函数(F1)。我不懂C，我尝试将其转换为python(F2)，但没有成功。预先感谢您让我知道我做错了什么。

F1:

void Encrypt(unsigned char key[20], unsigned char *pPlainBuffer, unsigned char *pCipherBuffer, unsigned int nLength) 
{
int nKeyPos = 0;
unsigned int n;
unsigned char KeyA = 0;

if ((pPlainBuffer != NULL) && (pCipherBuffer != NULL)) {
for (n = 0; n < 20; n++) 
  KeyA ^= key[n];
  nKeyPos = KeyA % 20;
for (n = 0; n < nLength; n++) {
  pCipherBuffer[n] = pPlainBuffer[n]^(key[nKeyPos]*KeyA);
  KeyA += pCipherBuffer[n];
  nKeyPos = pCipherBuffer[n] % 20;
}
 }
}

F2:

def Encrypt(plainText, key):  # plainText is a list of hexadecimals representing ascii 
                              # characters, key is a list of int size 20  each int 
                              # beetween 0 and 255
    keyA = 0

    for n in range(20):
        keyA ^= key[n]

    nKeyPos = KeyA % 20

    for n in range(len(plainText)):
         codedText[n] = plainText[n]^(key[nKeyPos]*KeyA)
         keyA += codedText[n]
         nKeyPos = codedText[n] % 20

Answer 1

你有很多问题......最明显的是Python整数默认情况下不受字节宽度的限制，所以你必须显式设置宽度

此外，您必须在 python 中将字母转换为数字，因为它们本质上是不同的东西(在 C/c++ 中字母实际上只是数字)

def Encrypt(plainText, key):  # plainText is a list of hexadecimals representing ascii 
                              # characters, key is a list of int size 20  each int                                   # beetween 0 and 255
    keyA = 0    
    for letter in key:
        keyA ^= ord(letter) # conver letter to number    
    keyA = keyA & 0xFF # fix width to one byte
    key = itertools.cycle(key) # repeate forever
    codedText = ""
    for letter,keyCode in zip(plainText,key):
         xorVal = (ord(keyCode) * keyA) & 0xFF # make one byte wide
         newLetter = chr((ord(letter) ^ xorVal )&0xFF) # make one byte wide

         codedText += newLetter
         keyA += ord(newLetter)
         keyA = keyA&0xFF # keep it one byte wide