简单地将循环中的256个修改为65536,只会一次又一次地重复相同的256个值。如何生成65536个不同的值?
#define CRC64_ECMA182_POLY 0x42F0E1EBA9EA3693ULL
static uint64_t crc64_table[256] = {0};
static void generate_crc64_table(void)
{
uint64_t i, j, c, crc;
for (i = 0; i < 256; i++) {
crc = 0;
c = i << 56;
for (j = 0; j < 8; j++) {
if ((crc ^ c) & 0x8000000000000000ULL)
crc = (crc << 1) ^ CRC64_ECMA182_POLY;
else
crc <<= 1;
c <<= 1;
}
crc64_table[i] = crc;
}
}发布于 2019-07-20 11:33:24
如果您想要65536个值,那么大概您想要一个16位表,所以也要将位循环升级到16。
static void generate_crc64_table(void)
{
uint64_t i, j, c, crc;
for (i = 0; i < 65536 ; i++) { // 65536 was 256
crc = 0;
c = i << 32; // 32 was 56
for (j = 0; j < 16; j++) { // 16 was 8
if ((crc ^ c) & 0x8000000000000000ULL)
crc = (crc << 1) ^ CRC64_ECMA182_POLY;
else
crc <<= 1;
c <<= 1;
}
crc64_table[i] = crc;
}
}不能保证这将产生一个有用的表,但所有的值都应该是不同的-至少。
发布于 2019-07-20 21:49:51
如果生成的CRC应该匹配小端点处理器(如X86 )上的位或面向字节的CRC,则需要交换每个2字节== 16位对的上/下字节。示例代码,不确定是否可以清除。请注意,generate函数中的len是== #2字节元素== # 16位元素。
#define CRC64_ECMA182_POLY 0x42F0E1EBA9EA3693ULL
static uint64_t crc64_table[65536] = {0};
static void generate_crc64_table(void)
{
uint64_t i, j, crc;
for (i = 0; i < 65536; i++) {
crc = i << 48;
for (j = 0; j < 16; j++)
// assumes two's complement math
crc = (crc << 1) ^ ((0ull-(crc>>63))&CRC64_ECMA182_POLY);
// swap byte pairs on index and values for table lookup
crc64_table[((i & 0xff00) >> 8) | ((i & 0x00ff) << 8)] =
((crc & 0xff00ff00ff00ff00ull) >> 8) | ((crc & 0x00ff00ff00ff00ffull) << 8);
}
}
static uint64_t generate_crc64(uint16_t *bfr, int len)
{
uint64_t crc = 0;
int i;
for (i = 0; i < len; i++)
// generates crc with byte pairs swapped
crc = crc64_table[(crc>>48) ^ *bfr++] ^ (crc << 16);
// unswap byte pairs for return
return ((crc & 0xff00ff00ff00ff00) >> 8) | ((crc & 0x00ff00ff00ff00ff) << 8);
}https://stackoverflow.com/questions/57122893
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