Convert this CRC32 algorithm to Python 3.3

Question

I need to convert this CRC32 algorithm to python (using 3.3), but I am a python noob. I tried the built in binascii.crc32(), but the CRC was incorrect. Apparently, STMicro does the CRC32 a bit different. I found an algorithm that works, now I just need it to be in python.

//****************************************************************************

DWORD Crc32Fast(DWORD Crc, DWORD Data)
{
  static const DWORD CrcTable[16] = { // Nibble lookup table for 0x04C11DB7 polynomial
    0x00000000,0x04C11DB7,0x09823B6E,0x0D4326D9,0x130476DC,0x17C56B6B,0x1A864DB2,0x1E475005,
    0x2608EDB8,0x22C9F00F,0x2F8AD6D6,0x2B4BCB61,0x350C9B64,0x31CD86D3,0x3C8EA00A,0x384FBDBD };

  Crc = Crc ^ Data; // Apply all 32-bits

  // Process 32-bits, 4 at a time, or 8 rounds

  Crc = (Crc << 4) ^ CrcTable[Crc >> 28]; // Assumes 32-bit reg, masking index to 4-bits
  Crc = (Crc << 4) ^ CrcTable[Crc >> 28]; //  0x04C11DB7 Polynomial used in STM32
  Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
  Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
  Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
  Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
  Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
  Crc = (Crc << 4) ^ CrcTable[Crc >> 28];

  return(Crc);
}

//****************************************************************************

DWORD Crc32FastBlock(DWORD Crc, DWORD Size, DWORD *Buffer) // 32-bit units
{
  while(Size--)
    Crc = Crc32Fast(Crc, *Buffer++);

  return(Crc);
}

I'm voting to close this question as off-topic because this isn't a code translation service — jonrsharpe, Oct 15 '15 at 15:03

Noctis Skytower · Accepted Answer · 2019-05-15T21:15:57.583

If my understanding is not mistaken, this should be the code that you want:

CRC_TABLE = (0x00000000, 0x04C11DB7, 0x09823B6E, 0x0D4326D9,
             0x130476DC, 0x17C56B6B, 0x1A864DB2, 0x1E475005,
             0x2608EDB8, 0x22C9F00F, 0x2F8AD6D6, 0x2B4BCB61,
             0x350C9B64, 0x31CD86D3, 0x3C8EA00A, 0x384FBDBD)


def dword(value):
    return value & 0xFFFFFFFF


def crc32_fast(crc, data):
    crc, data = dword(crc), dword(data)
    crc ^= data
    for _ in range(8):
        crc = dword(crc << 4) ^ CRC_TABLE[crc >> 28]
    return crc


def crc32_fast_block(crc, buffer):
    for data in buffer:
        crc = crc32_fast(crc, data)
    return crc


def crc32_fast_bytes(crc, bytes_data, byteorder='big'):
    if len(bytes_data) & 3:
        raise ValueError('bytes_data length must be multiple of four')
    for index in range(0, len(bytes_data), 4):
        data = int.from_bytes(bytes_data[index:index+4], byteorder)
        crc = crc32_fast(crc, data)
    return crc

The function crc32_fast_block expects an initial crc value and an iterable of numbers to run the algorithm on. crc32_fast_bytes is almost the same but expects a bytes value with a length being a multiple of four.

@DavidLechner An even better solution would be to provide an argument for specifying how the numbers are stored. Thanks for the suggestion! The code has been modified to support either option. — Noctis Skytower, May 15 '19 at 21:14

Convert this CRC32 algorithm to Python 3.3

1 Answers1