算法CRC-12 [英] Algorithm CRC-12

问题描述

我正在尝试为12位CRC和算法创建crc_table，但始终会得到错误的结果。

I am trying to do crc_table for 12-bit CRC and the algorithm, but always I got wrong results.

您能帮我吗？要创建crc表，我尝试：

Can you help me? To create crc table I try:

void crcInit(void)
{
    unsigned short  remainder;
    int    dividend;
    unsigned char  bit;

    for (dividend = 0; dividend < 256; ++dividend)
    {
        remainder = dividend << 4;

        for (bit = 8; bit > 0; --bit)
        {
            if (remainder & 0x800)
            {
                remainder = (remainder << 1) ^ 0x180D; //Polynomio of CRC-12
            }
            else
            {
                remainder = (remainder << 1);
            }
        }
       crcTable[dividend] = remainder;
    }

}

我对此进行了更新，CRC算法是：

I updated that and CRC algorithm is:

unsigned short crcFast(unsigned char const message[], int nBytes)
{
    unsigned short remainder = 0x0000;
    unsigned char  data;
    int  byte;


    /*
     * Divide the message by the polynomial, a byte at a time.
     */
    for (byte = 0; byte < nBytes; ++byte)
    {
        data = message[byte] ^ (remainder >> 4);
    remainder = crcTable[data] ^ (remainder << 8);
    }

    /*
     * The final remainder is the CRC.
     */
    return (remainder ^ 0);

}

但是它不起作用.....

But It isn't working.....

推荐答案

这似乎不正确：

if (remainder & 10000000)

您似乎打算将此数字用作是二进制，但实际上是十进制。您应该使用十六进制字面量（0x80）。

It looks like you are intending this number to be binary, but it is actually decimal. You should use a hex literal instead (0x80).

此数字以及您执行的移位大小似乎也存在问题：此测试应检查如果余数的高位被设置。由于您要执行12位CRC，因此掩码应为0x800（二进制100000000000）。高于此的偏移应该是：

There also seem to be problems with this number, and with the size of the shift you do: This test should check if the high-order bit of the remainder is set. Since you are doing a 12-bit CRC, the mask should be 0x800 (binary 100000000000). And the shift above that should probably be:

remainder = dividend << 4;

设置其余部分的最左8位。

to set the leftmost 8 bits of the remainder.

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