CRC4 INTERLAKEN和ITU查找表的生成

Question

我试图在C ++中为已知的CRC算法CRC-4 / INTERLAKEN和CRC-4 / ITU生成CRC4查找表。 CRC定义如下：

width=4 poly=0x3 init=0xf refin=false refout=false xorout=0xf check=0xb residue=0x2 name="CRC-4/INTERLAKEN"
width=4 poly=0x3 init=0x0 refin=true  refout=true  xorout=0x0 check=0x7 residue=0x0 name="CRC-4/G-704"

我已经开始调整从answer here中找到的代码以生成查找表，以与在answer found here中找到的表进行比较。如果该代码的位顺序无关紧要，除了多项式相同，我还需要考虑什么？ INTERLAKEN和ITU算法的代码有何不同？]

代码：

#include <iomanip>
#include <iostream>

void make_crc_table(unsigned long crcTable[])
{
    unsigned long POLYNOMIAL = 0x3;
    unsigned long remainder;
    unsigned char b = 0;
    do
    {
        remainder = b;
        for (unsigned long bit = 8; bit > 0; --bit)
        {
            if (remainder & 1)
                remainder = (remainder >> 1) ^ POLYNOMIAL;
            else
                remainder = (remainder >> 1);
        }
        crcTable[(size_t)b] = remainder;
    } while (0 != ++b);
}

int main()
{
    unsigned long crcTable[256];
    make_crc_table(crcTable);
    // Print the CRC table
    for (size_t i = 0; i < 256; i++)
    {
        std::cout << "0x";
        std::cout << std::setfill('0') << std::setw(2) << std::hex << crcTable[i];
        if (i % 16 == 15)
            std::cout << "," << std::endl;
        else
            std::cout << ", ";
    }

    return 0;
}

输出：

0x00, 0x02, 0x03, 0x01, 0x01, 0x03, 0x02, 0x00, 0x02, 0x00, 0x01, 0x03, 0x03, 0x01, 0x00, 0x02,
0x03, 0x01, 0x00, 0x02, 0x02, 0x00, 0x01, 0x03, 0x01, 0x03, 0x02, 0x00, 0x00, 0x02, 0x03, 0x01,
0x01, 0x03, 0x02, 0x00, 0x00, 0x02, 0x03, 0x01, 0x03, 0x01, 0x00, 0x02, 0x02, 0x00, 0x01, 0x03,
0x02, 0x00, 0x01, 0x03, 0x03, 0x01, 0x00, 0x02, 0x00, 0x02, 0x03, 0x01, 0x01, 0x03, 0x02, 0x00,
0x02, 0x00, 0x01, 0x03, 0x03, 0x01, 0x00, 0x02, 0x00, 0x02, 0x03, 0x01, 0x01, 0x03, 0x02, 0x00,
0x01, 0x03, 0x02, 0x00, 0x00, 0x02, 0x03, 0x01, 0x03, 0x01, 0x00, 0x02, 0x02, 0x00, 0x01, 0x03,
0x03, 0x01, 0x00, 0x02, 0x02, 0x00, 0x01, 0x03, 0x01, 0x03, 0x02, 0x00, 0x00, 0x02, 0x03, 0x01,
0x00, 0x02, 0x03, 0x01, 0x01, 0x03, 0x02, 0x00, 0x02, 0x00, 0x01, 0x03, 0x03, 0x01, 0x00, 0x02,
0x03, 0x01, 0x00, 0x02, 0x02, 0x00, 0x01, 0x03, 0x01, 0x03, 0x02, 0x00, 0x00, 0x02, 0x03, 0x01,
0x00, 0x02, 0x03, 0x01, 0x01, 0x03, 0x02, 0x00, 0x02, 0x00, 0x01, 0x03, 0x03, 0x01, 0x00, 0x02,
0x02, 0x00, 0x01, 0x03, 0x03, 0x01, 0x00, 0x02, 0x00, 0x02, 0x03, 0x01, 0x01, 0x03, 0x02, 0x00,
0x01, 0x03, 0x02, 0x00, 0x00, 0x02, 0x03, 0x01, 0x03, 0x01, 0x00, 0x02, 0x02, 0x00, 0x01, 0x03,
0x01, 0x03, 0x02, 0x00, 0x00, 0x02, 0x03, 0x01, 0x03, 0x01, 0x00, 0x02, 0x02, 0x00, 0x01, 0x03,
0x02, 0x00, 0x01, 0x03, 0x03, 0x01, 0x00, 0x02, 0x00, 0x02, 0x03, 0x01, 0x01, 0x03, 0x02, 0x00,
0x00, 0x02, 0x03, 0x01, 0x01, 0x03, 0x02, 0x00, 0x02, 0x00, 0x01, 0x03, 0x03, 0x01, 0x00, 0x02,
0x03, 0x01, 0x00, 0x02, 0x02, 0x00, 0x01, 0x03, 0x01, 0x03, 0x02, 0x00, 0x00, 0x02, 0x03, 0x01,

Answer 1

更新-我更改了答案以查看您的代码，该代码使用了反射的输入和输出。这意味着多项式应从0x03反转为0x0c。为了确认这一点，表条目[0x80]应该为0x0c。