CRC4因特LAKEN和ITU查找表生成

Question

我试图为已知的CRC算法CRC-4/INTERLAKEN和CRC-4/ITU在C++中遗传CRC4查找表。“儿童权利公约”的定义如下：

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，以便与answer found here中的表进行比较。如果这段代码的位序不重要，那么除了多项式--这是相同的--我还需要考虑什么呢？因特拉肯算法和国际电联算法的代码会有什么不同？

守则：

#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;
}

产出：

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,

原始问题在这里结束。

在rcgldr答复后更新：

#include <iostream>
#include <bitset>
#include <iomanip>

void make_crc_table(unsigned int crcTable[])
{
    unsigned char POLYNOMIAL = 0xc;
    unsigned char remainder;
    unsigned char b = 0;
    do
    {
        remainder = b;
        for (int bit = 8; bit > 0; --bit)
        {
            if (remainder & 0x80)
                remainder = (remainder << 1) ^ POLYNOMIAL;
            else
                remainder = (remainder << 1);
        }
        crcTable[(size_t)b] = remainder;
    } while (0 != ++b);
}

int main()
{
    unsigned int crcTable[256];
    make_crc_table(crcTable);

    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;
}

守则的输出：

0x00, 0x0c, 0x18, 0x14, 0x30, 0x3c, 0x28, 0x24, 0x60, 0x6c, 0x78, 0x74, 0x50, 0x5c, 0x48, 0x44,
0xc0, 0xcc, 0xd8, 0xd4, 0xf0, 0xfc, 0xe8, 0xe4, 0xa0, 0xac, 0xb8, 0xb4, 0x90, 0x9c, 0x88, 0x84,
0x8c, 0x80, 0x94, 0x98, 0xbc, 0xb0, 0xa4, 0xa8, 0xec, 0xe0, 0xf4, 0xf8, 0xdc, 0xd0, 0xc4, 0xc8,
0x4c, 0x40, 0x54, 0x58, 0x7c, 0x70, 0x64, 0x68, 0x2c, 0x20, 0x34, 0x38, 0x1c, 0x10, 0x04, 0x08,
0x14, 0x18, 0x0c, 0x00, 0x24, 0x28, 0x3c, 0x30, 0x74, 0x78, 0x6c, 0x60, 0x44, 0x48, 0x5c, 0x50,
0xd4, 0xd8, 0xcc, 0xc0, 0xe4, 0xe8, 0xfc, 0xf0, 0xb4, 0xb8, 0xac, 0xa0, 0x84, 0x88, 0x9c, 0x90,
0x98, 0x94, 0x80, 0x8c, 0xa8, 0xa4, 0xb0, 0xbc, 0xf8, 0xf4, 0xe0, 0xec, 0xc8, 0xc4, 0xd0, 0xdc,
0x58, 0x54, 0x40, 0x4c, 0x68, 0x64, 0x70, 0x7c, 0x38, 0x34, 0x20, 0x2c, 0x08, 0x04, 0x10, 0x1c,
0x28, 0x24, 0x30, 0x3c, 0x18, 0x14, 0x00, 0x0c, 0x48, 0x44, 0x50, 0x5c, 0x78, 0x74, 0x60, 0x6c,
0xe8, 0xe4, 0xf0, 0xfc, 0xd8, 0xd4, 0xc0, 0xcc, 0x88, 0x84, 0x90, 0x9c, 0xb8, 0xb4, 0xa0, 0xac,
0xa4, 0xa8, 0xbc, 0xb0, 0x94, 0x98, 0x8c, 0x80, 0xc4, 0xc8, 0xdc, 0xd0, 0xf4, 0xf8, 0xec, 0xe0,
0x64, 0x68, 0x7c, 0x70, 0x54, 0x58, 0x4c, 0x40, 0x04, 0x08, 0x1c, 0x10, 0x34, 0x38, 0x2c, 0x20,
0x3c, 0x30, 0x24, 0x28, 0x0c, 0x00, 0x14, 0x18, 0x5c, 0x50, 0x44, 0x48, 0x6c, 0x60, 0x74, 0x78,
0xfc, 0xf0, 0xe4, 0xe8, 0xcc, 0xc0, 0xd4, 0xd8, 0x9c, 0x90, 0x84, 0x88, 0xac, 0xa0, 0xb4, 0xb8,
0xb0, 0xbc, 0xa8, 0xa4, 0x80, 0x8c, 0x98, 0x94, 0xd0, 0xdc, 0xc8, 0xc4, 0xe0, 0xec, 0xf8, 0xf4,
0x70, 0x7c, 0x68, 0x64, 0x40, 0x4c, 0x58, 0x54, 0x10, 0x1c, 0x08, 0x04, 0x20, 0x2c, 0x38, 0x34,

Answer 1

位序很重要。问题的代码是使用反射输入和输出。这意味着多项式应该从0x03到0x0c被反转。要确认这一点，表条目0x80应该是0x0c。

对于因特拉肯表，代码应该如下所示：

void make_crc_table(unsigned char crcTable[])
{
    unsigned char POLYNOMIAL = 0x30;
    unsigned char remainder;
    unsigned char b = 0;
    do
    {
        remainder = b;
        for (int bit = 8; bit > 0; --bit)
        {
            if (remainder & 0x80)
                remainder = (remainder << 1) ^ POLYNOMIAL;
            else
                remainder = (remainder << 1);
        }
        crcTable[(size_t)b] = remainder;
    } while (0 != ++b);
}

请注意，CRC将位于字节的上4位。完成后，代码将需要返回(crc>>4)^0x0f。