如何实现CRC-10算法
How to implement CRC-10 algorithm
我正在尝试使用多项式 0x633 实现 10 位 CRC。我翻阅了无数页，但关于 CRC-10 的内容并不多。
我写了一些示例代码，似乎计算得很好。我可以 运行 一些测试数据来获得 CRC，然后 运行 数据 + CRC 返回以获得 0，这是预期的。问题是我们正在用这个计算器测试它并得到不一致的结果：http://www.ghsi.de/pages/subpages/Online%20CRC%20Calculation/indexDetails.php?Polynom=11000110011&Message=1f+ff+30+04+05+34+a7。这包括多边形和测试数据。这个计算器的结果是“0x10e”，而我的是“0x3b1”。
这是代码，修改自https://cs.fit.edu/code/svn/cse2410f13team7/wireshark/wsutil/crc10.c:
#include <stdio.h>
#include <stdint.h>
uint16_t GetCRC10(uint16_t crc10, const uint8_t *data_blk_ptr, int data_blk_size);
static const uint16_t byte_crc10_table[256] = {
0x0000, 0x0233, 0x0255, 0x0066, 0x0299, 0x00aa, 0x00cc, 0x02ff,
0x0301, 0x0132, 0x0154, 0x0367, 0x0198, 0x03ab, 0x03cd, 0x01fe,
0x0031, 0x0202, 0x0264, 0x0057, 0x02a8, 0x009b, 0x00fd, 0x02ce,
0x0330, 0x0103, 0x0165, 0x0356, 0x01a9, 0x039a, 0x03fc, 0x01cf,
0x0062, 0x0251, 0x0237, 0x0004, 0x02fb, 0x00c8, 0x00ae, 0x029d,
0x0363, 0x0150, 0x0136, 0x0305, 0x01fa, 0x03c9, 0x03af, 0x019c,
0x0053, 0x0260, 0x0206, 0x0035, 0x02ca, 0x00f9, 0x009f, 0x02ac,
0x0352, 0x0161, 0x0107, 0x0334, 0x01cb, 0x03f8, 0x039e, 0x01ad,
0x00c4, 0x02f7, 0x0291, 0x00a2, 0x025d, 0x006e, 0x0008, 0x023b,
0x03c5, 0x01f6, 0x0190, 0x03a3, 0x015c, 0x036f, 0x0309, 0x013a,
0x00f5, 0x02c6, 0x02a0, 0x0093, 0x026c, 0x005f, 0x0039, 0x020a,
0x03f4, 0x01c7, 0x01a1, 0x0392, 0x016d, 0x035e, 0x0338, 0x010b,
0x00a6, 0x0295, 0x02f3, 0x00c0, 0x023f, 0x000c, 0x006a, 0x0259,
0x03a7, 0x0194, 0x01f2, 0x03c1, 0x013e, 0x030d, 0x036b, 0x0158,
0x0097, 0x02a4, 0x02c2, 0x00f1, 0x020e, 0x003d, 0x005b, 0x0268,
0x0396, 0x01a5, 0x01c3, 0x03f0, 0x010f, 0x033c, 0x035a, 0x0169,
0x0188, 0x03bb, 0x03dd, 0x01ee, 0x0311, 0x0122, 0x0144, 0x0377,
0x0289, 0x00ba, 0x00dc, 0x02ef, 0x0010, 0x0223, 0x0245, 0x0076,
0x01b9, 0x038a, 0x03ec, 0x01df, 0x0320, 0x0113, 0x0175, 0x0346,
0x02b8, 0x008b, 0x00ed, 0x02de, 0x0021, 0x0212, 0x0274, 0x0047,
0x01ea, 0x03d9, 0x03bf, 0x018c, 0x0373, 0x0140, 0x0126, 0x0315,
0x02eb, 0x00d8, 0x00be, 0x028d, 0x0072, 0x0241, 0x0227, 0x0014,
0x01db, 0x03e8, 0x038e, 0x01bd, 0x0342, 0x0171, 0x0117, 0x0324,
0x02da, 0x00e9, 0x008f, 0x02bc, 0x0043, 0x0270, 0x0216, 0x0025,
0x014c, 0x037f, 0x0319, 0x012a, 0x03d5, 0x01e6, 0x0180, 0x03b3,
0x024d, 0x007e, 0x0018, 0x022b, 0x00d4, 0x02e7, 0x0281, 0x00b2,
0x017d, 0x034e, 0x0328, 0x011b, 0x03e4, 0x01d7, 0x01b1, 0x0382,
0x027c, 0x004f, 0x0029, 0x021a, 0x00e5, 0x02d6, 0x02b0, 0x0083,
0x012e, 0x031d, 0x037b, 0x0148, 0x03b7, 0x0184, 0x01e2, 0x03d1,
0x022f, 0x001c, 0x007a, 0x0249, 0x00b6, 0x0285, 0x02e3, 0x00d0,
0x011f, 0x032c, 0x034a, 0x0179, 0x0386, 0x01b5, 0x01d3, 0x03e0,
0x021e, 0x002d, 0x004b, 0x0278, 0x0087, 0x02b4, 0x02d2, 0x00e1
};
/* update the data block's CRC-10 remainder one byte at a time */
uint16_t GetCRC10(uint16_t crc10, const uint8_t *data_blk_ptr, int data_blk_size)
{
register int i;
uint16_t crc10_accum = 0;
for (i = 0; i < data_blk_size; i++) {
crc10_accum = ((crc10_accum << 8) & 0x3ff)
^ byte_crc10_table[( crc10_accum >> 2) & 0xff]
^ *data_blk_ptr++;
}
crc10_accum = ((crc10_accum << 8) & 0x3ff)
^ byte_crc10_table[( crc10_accum >> 2) & 0xff]
^ (crc10>>2);
crc10_accum = ((crc10_accum << 8) & 0x3ff)
^ byte_crc10_table[( crc10_accum >> 2) & 0xff]
^ ((crc10<<6) & 0xFF);
return crc10_accum;
}
#define TEST_DATA_SIZE 9
void main()
{
uint8_t test_input_data[TEST_DATA_SIZE] = {0x1f, 0xff, 0x30, 0x4, 0x5, 0x34, 0xa7, 0x0, 0x0};
uint16_t crc;
uint16_t crc_final;
GenerateCRC10Table();
crc = GetCRC10(0, test_input_data, TEST_DATA_SIZE-2);
test_input_data[TEST_DATA_SIZE-2] ^= crc >> 8;
test_input_data[TEST_DATA_SIZE-1] ^= crc & 0xFF;
printf("%x\n", crc); // Error here. This crc doesn't match with the calculator.
crc_final = GetCRC10(0, test_input_data, TEST_DATA_SIZE);
if (crc_final == 0)
{
printf("Success");
}
}
这是用于生成查找的函数 table:
#define POLYNOMIAL 0x633
static uint16_t byte_crc10_table[256];
void gen_byte_crc10_table(void)
/* generate the table of CRC-10 remainders for all possible bytes */
{
register int i, j;
register unsigned short crc10_accum;
for ( i = 0; i < 256; i++ )
{
crc10_accum = ((unsigned short) i << 2);
for ( j = 0; j < 8; j++ )
{
if ((crc10_accum <<= 1) & 0x400) crc10_accum ^= POLYNOMIAL;
}
byte_crc10_table[i] = crc10_accum;
}
return;
}
我也用 0x233 正态多项式试过了。
这可能吗？本质上，我是在对一个完整的字节数组进行按位运算。我怀疑问题是计算器是逐位计算的，而我的计算器是逐字节计算的。由于 poly 是 10 位宽，因此它不能与 uint8 一起干净地工作。我真的很感激一些反馈。谢谢
因为10位多项式和8位数据的差异，计算器CRC和我的CRC不同是否有意义？
CRC 不仅仅由长度和多项式定义。还有传入字节的位排序、多项式应用的位排序、结果的位和字节排序、CRC 的初始化以及最终的异或运算。
该在线计算器在它计算的 CRC 中做出了一些任意选择，正如它自己在页面顶部指出的那样："Be careful: there are several ways to realize a CRC. They differ (at least) in the way which bit is shifted in first and also in the initialization of the flipflops."它选择无位反射、零初始化和零最终独占-或者.
生成的 CRC，使用您提供的多项式，0x233，实际上有一个名称和一个应用程序。它是 CRC-10/ATM（参见 link）。这是一个简单的按位实现：
unsigned crc10atm_bit(unsigned crc, void const *mem, size_t len) {
unsigned char const *data = mem;
if (data == NULL)
return 0;
while (len--) {
crc ^= (unsigned)(*data++) << 2;
for (unsigned k = 0; k < 8; k++)
crc = crc & 0x200 ? (crc << 1) ^ 0x233 : crc << 1;
}
crc &= 0x3ff;
return crc;
}
该例程使用的约定是，当传递 NULL 指针时返回 CRC 的初始值。
这是一个使用预先计算的 table:
的更快的字节方式实现
static unsigned short const table_byte[] = {
0x000, 0x233, 0x255, 0x066, 0x299, 0x0aa, 0x0cc, 0x2ff, 0x301, 0x132, 0x154,
0x367, 0x198, 0x3ab, 0x3cd, 0x1fe, 0x031, 0x202, 0x264, 0x057, 0x2a8, 0x09b,
0x0fd, 0x2ce, 0x330, 0x103, 0x165, 0x356, 0x1a9, 0x39a, 0x3fc, 0x1cf, 0x062,
0x251, 0x237, 0x004, 0x2fb, 0x0c8, 0x0ae, 0x29d, 0x363, 0x150, 0x136, 0x305,
0x1fa, 0x3c9, 0x3af, 0x19c, 0x053, 0x260, 0x206, 0x035, 0x2ca, 0x0f9, 0x09f,
0x2ac, 0x352, 0x161, 0x107, 0x334, 0x1cb, 0x3f8, 0x39e, 0x1ad, 0x0c4, 0x2f7,
0x291, 0x0a2, 0x25d, 0x06e, 0x008, 0x23b, 0x3c5, 0x1f6, 0x190, 0x3a3, 0x15c,
0x36f, 0x309, 0x13a, 0x0f5, 0x2c6, 0x2a0, 0x093, 0x26c, 0x05f, 0x039, 0x20a,
0x3f4, 0x1c7, 0x1a1, 0x392, 0x16d, 0x35e, 0x338, 0x10b, 0x0a6, 0x295, 0x2f3,
0x0c0, 0x23f, 0x00c, 0x06a, 0x259, 0x3a7, 0x194, 0x1f2, 0x3c1, 0x13e, 0x30d,
0x36b, 0x158, 0x097, 0x2a4, 0x2c2, 0x0f1, 0x20e, 0x03d, 0x05b, 0x268, 0x396,
0x1a5, 0x1c3, 0x3f0, 0x10f, 0x33c, 0x35a, 0x169, 0x188, 0x3bb, 0x3dd, 0x1ee,
0x311, 0x122, 0x144, 0x377, 0x289, 0x0ba, 0x0dc, 0x2ef, 0x010, 0x223, 0x245,
0x076, 0x1b9, 0x38a, 0x3ec, 0x1df, 0x320, 0x113, 0x175, 0x346, 0x2b8, 0x08b,
0x0ed, 0x2de, 0x021, 0x212, 0x274, 0x047, 0x1ea, 0x3d9, 0x3bf, 0x18c, 0x373,
0x140, 0x126, 0x315, 0x2eb, 0x0d8, 0x0be, 0x28d, 0x072, 0x241, 0x227, 0x014,
0x1db, 0x3e8, 0x38e, 0x1bd, 0x342, 0x171, 0x117, 0x324, 0x2da, 0x0e9, 0x08f,
0x2bc, 0x043, 0x270, 0x216, 0x025, 0x14c, 0x37f, 0x319, 0x12a, 0x3d5, 0x1e6,
0x180, 0x3b3, 0x24d, 0x07e, 0x018, 0x22b, 0x0d4, 0x2e7, 0x281, 0x0b2, 0x17d,
0x34e, 0x328, 0x11b, 0x3e4, 0x1d7, 0x1b1, 0x382, 0x27c, 0x04f, 0x029, 0x21a,
0x0e5, 0x2d6, 0x2b0, 0x083, 0x12e, 0x31d, 0x37b, 0x148, 0x3b7, 0x184, 0x1e2,
0x3d1, 0x22f, 0x01c, 0x07a, 0x249, 0x0b6, 0x285, 0x2e3, 0x0d0, 0x11f, 0x32c,
0x34a, 0x179, 0x386, 0x1b5, 0x1d3, 0x3e0, 0x21e, 0x02d, 0x04b, 0x278, 0x087,
0x2b4, 0x2d2, 0x0e1
};
unsigned crc10atm_byte(unsigned crc, void const *mem, size_t len) {
unsigned char const *data = mem;
if (data == NULL)
return 0;
while (len--)
crc = (crc << 8) ^
table_byte[((crc >> 2) ^ *data++) & 0xff];
crc &= 0x3ff;
return crc;
}
您可能更喜欢不同的选择，例如0x3ff 的初始值和相同值的最终异或。这将避免 CRC 为零的任意数量的零字符串。如果使用反射，也可以避免一些偏移。
问题中的代码正在实现CRC-10/ATM，其中包括在数据字节之后和10位CRC之前的6个填充0位。对于问题中的示例数据：
1f ff 30 04 05 34 a7
它应该得到 0x3b1，这是您发布的代码生成的。
替代实施，您发布的示例案例也会产生 0x3b1，并且问题代码和此替代实施都与 CRC-10/ATM 中的示例相匹配：
typedef unsigned char uint8_t;
typedef unsigned short uint16_t;
#define CRC10 (0x233<<6)
static uint16_t crctbl[256];
void gentbl(void)
{
uint16_t crc;
uint16_t c;
uint16_t i;
for(c = 0; c < 0x100; c++){
crc = c << 8;
for(i = 0; i < 8; i++)
/* assumes twos complement */
crc = (crc<<1)^((0-(crc>>15))&CRC10);
crctbl[c] = crc;
}
}
uint16_t crc10(uint8_t * bfr, int len)
{
uint16_t crc = 0x0000;
while(len--){
/* shifting 6 bits instead of 8 will leave crc cycled -2 bits when done */
crc ^= ((uint16_t)(*bfr++))<<6; /* xor byte<<6 into crc */
crc = (crc<<8)^crctbl[(crc>>8)]; /* cycle crc */
}
/* At this point, crc is cycled -2 bits, meaning that */
/* 2 bit of data have not been cycled yet, so cycle */
/* 8 bits to cycle the 2 bits of data and 6 padding bits of 0 */
crc = (crc<<8)^crctbl[(crc>>8)]; /* cycle crc +8 bits */
return(crc>>6); /* return 10 bit crc */
}
测试代码：
#include <stdio.h>
/* test messages including 10 bit CRC */
static uint8_t tst1[] =
{0x0A,0x0B,0x0C,0x0D,0x0E,0x0F,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0xf6};
static uint8_t tst2[] =
{0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x6B};
static uint8_t tst3[] =
{0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,
0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,
0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0x03,0x0F};
static uint8_t tst4[] =
{0x12,0x34,0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,
0x34,0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,0x34,
0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,0x02,0xED};
static uint8_t tst5[] =
{0x10,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,
0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,
0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x03,0xB9};
static uint8_t tst6[] =
{0x18,0x01,0x00,0x00,0x00,0x00,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,
0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x00,0x4A};
static uint8_t* ptst[6] = {tst1, tst2, tst3, tst4, tst5, tst6};
/* length of msgs including crc */
#define LEN (48)
int main(void)
{
uint16_t crc;
int i;
gentbl();
for(i = 0; i < (sizeof(ptst)/sizeof(ptst[0])); i++){
crc = crc10(ptst[i], LEN-2); /* generate a 10 bit CRC */
if((crc>>8) != ptst[i][LEN-2] || /* verify it matches test CRC */
(crc&0xff) != ptst[i][LEN-1])
break;
crc = crc10(ptst[i], LEN); /* verify CRC == 0 */
if(0 != crc)
break;
}
if(i == 6)
printf("passed\n");
else
printf("error\n");
return 0;
}
我正在尝试使用多项式 0x633 实现 10 位 CRC。我翻阅了无数页，但关于 CRC-10 的内容并不多。
我写了一些示例代码，似乎计算得很好。我可以 运行 一些测试数据来获得 CRC，然后 运行 数据 + CRC 返回以获得 0，这是预期的。问题是我们正在用这个计算器测试它并得到不一致的结果：http://www.ghsi.de/pages/subpages/Online%20CRC%20Calculation/indexDetails.php?Polynom=11000110011&Message=1f+ff+30+04+05+34+a7。这包括多边形和测试数据。这个计算器的结果是“0x10e”，而我的是“0x3b1”。
这是代码，修改自https://cs.fit.edu/code/svn/cse2410f13team7/wireshark/wsutil/crc10.c:
#include <stdio.h>
#include <stdint.h>
uint16_t GetCRC10(uint16_t crc10, const uint8_t *data_blk_ptr, int data_blk_size);
static const uint16_t byte_crc10_table[256] = {
0x0000, 0x0233, 0x0255, 0x0066, 0x0299, 0x00aa, 0x00cc, 0x02ff,
0x0301, 0x0132, 0x0154, 0x0367, 0x0198, 0x03ab, 0x03cd, 0x01fe,
0x0031, 0x0202, 0x0264, 0x0057, 0x02a8, 0x009b, 0x00fd, 0x02ce,
0x0330, 0x0103, 0x0165, 0x0356, 0x01a9, 0x039a, 0x03fc, 0x01cf,
0x0062, 0x0251, 0x0237, 0x0004, 0x02fb, 0x00c8, 0x00ae, 0x029d,
0x0363, 0x0150, 0x0136, 0x0305, 0x01fa, 0x03c9, 0x03af, 0x019c,
0x0053, 0x0260, 0x0206, 0x0035, 0x02ca, 0x00f9, 0x009f, 0x02ac,
0x0352, 0x0161, 0x0107, 0x0334, 0x01cb, 0x03f8, 0x039e, 0x01ad,
0x00c4, 0x02f7, 0x0291, 0x00a2, 0x025d, 0x006e, 0x0008, 0x023b,
0x03c5, 0x01f6, 0x0190, 0x03a3, 0x015c, 0x036f, 0x0309, 0x013a,
0x00f5, 0x02c6, 0x02a0, 0x0093, 0x026c, 0x005f, 0x0039, 0x020a,
0x03f4, 0x01c7, 0x01a1, 0x0392, 0x016d, 0x035e, 0x0338, 0x010b,
0x00a6, 0x0295, 0x02f3, 0x00c0, 0x023f, 0x000c, 0x006a, 0x0259,
0x03a7, 0x0194, 0x01f2, 0x03c1, 0x013e, 0x030d, 0x036b, 0x0158,
0x0097, 0x02a4, 0x02c2, 0x00f1, 0x020e, 0x003d, 0x005b, 0x0268,
0x0396, 0x01a5, 0x01c3, 0x03f0, 0x010f, 0x033c, 0x035a, 0x0169,
0x0188, 0x03bb, 0x03dd, 0x01ee, 0x0311, 0x0122, 0x0144, 0x0377,
0x0289, 0x00ba, 0x00dc, 0x02ef, 0x0010, 0x0223, 0x0245, 0x0076,
0x01b9, 0x038a, 0x03ec, 0x01df, 0x0320, 0x0113, 0x0175, 0x0346,
0x02b8, 0x008b, 0x00ed, 0x02de, 0x0021, 0x0212, 0x0274, 0x0047,
0x01ea, 0x03d9, 0x03bf, 0x018c, 0x0373, 0x0140, 0x0126, 0x0315,
0x02eb, 0x00d8, 0x00be, 0x028d, 0x0072, 0x0241, 0x0227, 0x0014,
0x01db, 0x03e8, 0x038e, 0x01bd, 0x0342, 0x0171, 0x0117, 0x0324,
0x02da, 0x00e9, 0x008f, 0x02bc, 0x0043, 0x0270, 0x0216, 0x0025,
0x014c, 0x037f, 0x0319, 0x012a, 0x03d5, 0x01e6, 0x0180, 0x03b3,
0x024d, 0x007e, 0x0018, 0x022b, 0x00d4, 0x02e7, 0x0281, 0x00b2,
0x017d, 0x034e, 0x0328, 0x011b, 0x03e4, 0x01d7, 0x01b1, 0x0382,
0x027c, 0x004f, 0x0029, 0x021a, 0x00e5, 0x02d6, 0x02b0, 0x0083,
0x012e, 0x031d, 0x037b, 0x0148, 0x03b7, 0x0184, 0x01e2, 0x03d1,
0x022f, 0x001c, 0x007a, 0x0249, 0x00b6, 0x0285, 0x02e3, 0x00d0,
0x011f, 0x032c, 0x034a, 0x0179, 0x0386, 0x01b5, 0x01d3, 0x03e0,
0x021e, 0x002d, 0x004b, 0x0278, 0x0087, 0x02b4, 0x02d2, 0x00e1
};
/* update the data block's CRC-10 remainder one byte at a time */
uint16_t GetCRC10(uint16_t crc10, const uint8_t *data_blk_ptr, int data_blk_size)
{
register int i;
uint16_t crc10_accum = 0;
for (i = 0; i < data_blk_size; i++) {
crc10_accum = ((crc10_accum << 8) & 0x3ff)
^ byte_crc10_table[( crc10_accum >> 2) & 0xff]
^ *data_blk_ptr++;
}
crc10_accum = ((crc10_accum << 8) & 0x3ff)
^ byte_crc10_table[( crc10_accum >> 2) & 0xff]
^ (crc10>>2);
crc10_accum = ((crc10_accum << 8) & 0x3ff)
^ byte_crc10_table[( crc10_accum >> 2) & 0xff]
^ ((crc10<<6) & 0xFF);
return crc10_accum;
}
#define TEST_DATA_SIZE 9
void main()
{
uint8_t test_input_data[TEST_DATA_SIZE] = {0x1f, 0xff, 0x30, 0x4, 0x5, 0x34, 0xa7, 0x0, 0x0};
uint16_t crc;
uint16_t crc_final;
GenerateCRC10Table();
crc = GetCRC10(0, test_input_data, TEST_DATA_SIZE-2);
test_input_data[TEST_DATA_SIZE-2] ^= crc >> 8;
test_input_data[TEST_DATA_SIZE-1] ^= crc & 0xFF;
printf("%x\n", crc); // Error here. This crc doesn't match with the calculator.
crc_final = GetCRC10(0, test_input_data, TEST_DATA_SIZE);
if (crc_final == 0)
{
printf("Success");
}
}
这是用于生成查找的函数 table:
#define POLYNOMIAL 0x633
static uint16_t byte_crc10_table[256];
void gen_byte_crc10_table(void)
/* generate the table of CRC-10 remainders for all possible bytes */
{
register int i, j;
register unsigned short crc10_accum;
for ( i = 0; i < 256; i++ )
{
crc10_accum = ((unsigned short) i << 2);
for ( j = 0; j < 8; j++ )
{
if ((crc10_accum <<= 1) & 0x400) crc10_accum ^= POLYNOMIAL;
}
byte_crc10_table[i] = crc10_accum;
}
return;
}
我也用 0x233 正态多项式试过了。
这可能吗？本质上，我是在对一个完整的字节数组进行按位运算。我怀疑问题是计算器是逐位计算的，而我的计算器是逐字节计算的。由于 poly 是 10 位宽，因此它不能与 uint8 一起干净地工作。我真的很感激一些反馈。谢谢
因为10位多项式和8位数据的差异，计算器CRC和我的CRC不同是否有意义？
CRC 不仅仅由长度和多项式定义。还有传入字节的位排序、多项式应用的位排序、结果的位和字节排序、CRC 的初始化以及最终的异或运算。
该在线计算器在它计算的 CRC 中做出了一些任意选择，正如它自己在页面顶部指出的那样："Be careful: there are several ways to realize a CRC. They differ (at least) in the way which bit is shifted in first and also in the initialization of the flipflops."它选择无位反射、零初始化和零最终独占-或者.
生成的 CRC，使用您提供的多项式，0x233，实际上有一个名称和一个应用程序。它是 CRC-10/ATM（参见 link）。这是一个简单的按位实现：
unsigned crc10atm_bit(unsigned crc, void const *mem, size_t len) {
unsigned char const *data = mem;
if (data == NULL)
return 0;
while (len--) {
crc ^= (unsigned)(*data++) << 2;
for (unsigned k = 0; k < 8; k++)
crc = crc & 0x200 ? (crc << 1) ^ 0x233 : crc << 1;
}
crc &= 0x3ff;
return crc;
}
该例程使用的约定是，当传递 NULL 指针时返回 CRC 的初始值。
这是一个使用预先计算的 table:
的更快的字节方式实现static unsigned short const table_byte[] = {
0x000, 0x233, 0x255, 0x066, 0x299, 0x0aa, 0x0cc, 0x2ff, 0x301, 0x132, 0x154,
0x367, 0x198, 0x3ab, 0x3cd, 0x1fe, 0x031, 0x202, 0x264, 0x057, 0x2a8, 0x09b,
0x0fd, 0x2ce, 0x330, 0x103, 0x165, 0x356, 0x1a9, 0x39a, 0x3fc, 0x1cf, 0x062,
0x251, 0x237, 0x004, 0x2fb, 0x0c8, 0x0ae, 0x29d, 0x363, 0x150, 0x136, 0x305,
0x1fa, 0x3c9, 0x3af, 0x19c, 0x053, 0x260, 0x206, 0x035, 0x2ca, 0x0f9, 0x09f,
0x2ac, 0x352, 0x161, 0x107, 0x334, 0x1cb, 0x3f8, 0x39e, 0x1ad, 0x0c4, 0x2f7,
0x291, 0x0a2, 0x25d, 0x06e, 0x008, 0x23b, 0x3c5, 0x1f6, 0x190, 0x3a3, 0x15c,
0x36f, 0x309, 0x13a, 0x0f5, 0x2c6, 0x2a0, 0x093, 0x26c, 0x05f, 0x039, 0x20a,
0x3f4, 0x1c7, 0x1a1, 0x392, 0x16d, 0x35e, 0x338, 0x10b, 0x0a6, 0x295, 0x2f3,
0x0c0, 0x23f, 0x00c, 0x06a, 0x259, 0x3a7, 0x194, 0x1f2, 0x3c1, 0x13e, 0x30d,
0x36b, 0x158, 0x097, 0x2a4, 0x2c2, 0x0f1, 0x20e, 0x03d, 0x05b, 0x268, 0x396,
0x1a5, 0x1c3, 0x3f0, 0x10f, 0x33c, 0x35a, 0x169, 0x188, 0x3bb, 0x3dd, 0x1ee,
0x311, 0x122, 0x144, 0x377, 0x289, 0x0ba, 0x0dc, 0x2ef, 0x010, 0x223, 0x245,
0x076, 0x1b9, 0x38a, 0x3ec, 0x1df, 0x320, 0x113, 0x175, 0x346, 0x2b8, 0x08b,
0x0ed, 0x2de, 0x021, 0x212, 0x274, 0x047, 0x1ea, 0x3d9, 0x3bf, 0x18c, 0x373,
0x140, 0x126, 0x315, 0x2eb, 0x0d8, 0x0be, 0x28d, 0x072, 0x241, 0x227, 0x014,
0x1db, 0x3e8, 0x38e, 0x1bd, 0x342, 0x171, 0x117, 0x324, 0x2da, 0x0e9, 0x08f,
0x2bc, 0x043, 0x270, 0x216, 0x025, 0x14c, 0x37f, 0x319, 0x12a, 0x3d5, 0x1e6,
0x180, 0x3b3, 0x24d, 0x07e, 0x018, 0x22b, 0x0d4, 0x2e7, 0x281, 0x0b2, 0x17d,
0x34e, 0x328, 0x11b, 0x3e4, 0x1d7, 0x1b1, 0x382, 0x27c, 0x04f, 0x029, 0x21a,
0x0e5, 0x2d6, 0x2b0, 0x083, 0x12e, 0x31d, 0x37b, 0x148, 0x3b7, 0x184, 0x1e2,
0x3d1, 0x22f, 0x01c, 0x07a, 0x249, 0x0b6, 0x285, 0x2e3, 0x0d0, 0x11f, 0x32c,
0x34a, 0x179, 0x386, 0x1b5, 0x1d3, 0x3e0, 0x21e, 0x02d, 0x04b, 0x278, 0x087,
0x2b4, 0x2d2, 0x0e1
};
unsigned crc10atm_byte(unsigned crc, void const *mem, size_t len) {
unsigned char const *data = mem;
if (data == NULL)
return 0;
while (len--)
crc = (crc << 8) ^
table_byte[((crc >> 2) ^ *data++) & 0xff];
crc &= 0x3ff;
return crc;
}
您可能更喜欢不同的选择，例如0x3ff 的初始值和相同值的最终异或。这将避免 CRC 为零的任意数量的零字符串。如果使用反射，也可以避免一些偏移。
问题中的代码正在实现CRC-10/ATM，其中包括在数据字节之后和10位CRC之前的6个填充0位。对于问题中的示例数据：
1f ff 30 04 05 34 a7
它应该得到 0x3b1，这是您发布的代码生成的。
替代实施，您发布的示例案例也会产生 0x3b1，并且问题代码和此替代实施都与 CRC-10/ATM 中的示例相匹配：
typedef unsigned char uint8_t;
typedef unsigned short uint16_t;
#define CRC10 (0x233<<6)
static uint16_t crctbl[256];
void gentbl(void)
{
uint16_t crc;
uint16_t c;
uint16_t i;
for(c = 0; c < 0x100; c++){
crc = c << 8;
for(i = 0; i < 8; i++)
/* assumes twos complement */
crc = (crc<<1)^((0-(crc>>15))&CRC10);
crctbl[c] = crc;
}
}
uint16_t crc10(uint8_t * bfr, int len)
{
uint16_t crc = 0x0000;
while(len--){
/* shifting 6 bits instead of 8 will leave crc cycled -2 bits when done */
crc ^= ((uint16_t)(*bfr++))<<6; /* xor byte<<6 into crc */
crc = (crc<<8)^crctbl[(crc>>8)]; /* cycle crc */
}
/* At this point, crc is cycled -2 bits, meaning that */
/* 2 bit of data have not been cycled yet, so cycle */
/* 8 bits to cycle the 2 bits of data and 6 padding bits of 0 */
crc = (crc<<8)^crctbl[(crc>>8)]; /* cycle crc +8 bits */
return(crc>>6); /* return 10 bit crc */
}
测试代码：
#include <stdio.h>
/* test messages including 10 bit CRC */
static uint8_t tst1[] =
{0x0A,0x0B,0x0C,0x0D,0x0E,0x0F,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0xf6};
static uint8_t tst2[] =
{0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x11,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x01,0x6B};
static uint8_t tst3[] =
{0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,
0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,
0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0x03,0x0F};
static uint8_t tst4[] =
{0x12,0x34,0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,
0x34,0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,0x34,
0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,0x34,0x56,0x78,0x90,0x12,0x02,0xED};
static uint8_t tst5[] =
{0x10,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,
0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,
0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x03,0xB9};
static uint8_t tst6[] =
{0x18,0x01,0x00,0x00,0x00,0x00,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,
0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x6A,0x00,0x4A};
static uint8_t* ptst[6] = {tst1, tst2, tst3, tst4, tst5, tst6};
/* length of msgs including crc */
#define LEN (48)
int main(void)
{
uint16_t crc;
int i;
gentbl();
for(i = 0; i < (sizeof(ptst)/sizeof(ptst[0])); i++){
crc = crc10(ptst[i], LEN-2); /* generate a 10 bit CRC */
if((crc>>8) != ptst[i][LEN-2] || /* verify it matches test CRC */
(crc&0xff) != ptst[i][LEN-1])
break;
crc = crc10(ptst[i], LEN); /* verify CRC == 0 */
if(0 != crc)
break;
}
if(i == 6)
printf("passed\n");
else
printf("error\n");
return 0;
}