python - 从c到python的crc24

有人能把这段代码翻译成python吗？我试了又试，但没有成功：

  #define CRC24_INIT 0xB704CEL
  #define CRC24_POLY 0x1864CFBL

  typedef long crc24;
  crc24 crc_octets(unsigned char *octets, size_t len)
  {
      crc24 crc = CRC24_INIT;
      int i;
      while (len--) {
          crc ^= (*octets++) << 16;
          for (i = 0; i < 8; i++) {
              crc <<= 1;
              if (crc & 0x1000000)
                  crc ^= CRC24_POLY;
          }
      }
      return crc & 0xFFFFFFL;
  }

我有一个rotate left函数（ROL24(value,bits_to_rotate_by)），我知道它是从一个著名程序员的源代码中得到的，但我没有得到八位字节上的*和++。我只知道++在c++中是如何工作的，我根本不知道*是什么
我的代码是：

def crc24(octets, length):# now octects is a binary string
 INIT = 0xB704CE
 POLY = 0x1864CFB
 crc = INIT
 index = 0
 while length:
  crc ^= (int(octets[index], 2) << 16)
  index += 1
  for i in xrange(8):
   crc = ROL(crc, 1)
   if crc & 0x1000000:
    crc ^= POLY
  length -= 1
 return crc & 0xFFFFFF

最佳答案

# Yes, there is no 'length' parameter here. We don't need it in Python.
def crc24(octets):
    INIT = 0xB704CE
    POLY = 0x1864CFB
    crc = INIT
    for octet in octets: # this is what the '*octets++' logic is effectively
    # accomplishing in the C code.
        crc ^= (octet << 16)
        # Throw that ROL function away, because the C code **doesn't** actually
        # rotate left; it shifts left. It happens to throw away any bits that are
        # shifted past the 32nd position, but that doesn't actulaly matter for
        # the correctness of the algorithm, because those bits can never "come back"
        # and we will mask off everything but the bottom 24 at the end anyway.
        for i in xrange(8):
            crc <<= 1
            if crc & 0x1000000: crc ^= POLY
    return crc & 0xFFFFFF

关于python - 从c到python的crc24，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/4544154/