Question

已知数据和CRC16的校验码,求该校验码的算法.

有十行数据,图片中的16进制数据从300H到390H,每行最后两列是CRC16校验码,如黄条所在行的数据:00 00 29 80 00 00 24 80 00 00 89 38 00 44 F8 07,其中F8 07就是该行CRC16校验码,求该CRC16校验码的算法.
黄行中,有用的数据就是29 80,24 80,89 38,另外的00 44只是数据循环覆盖时产生的,如:再写一条数据 的话,就会覆盖下面的350H行,后面的00 35就会变成00 45,就象序号一样.那现在的校验码就不清楚是全部校验还是局部校验,本案只想求一个针对性的算法,通用算法和原理一类的文章请不要发了,谢谢.在网上下载过几个CRC计算器,反复试算都对不上.所以在此想请大家帮忙看看.

现在是已知数据,已知校验码,不知多项式.也可以不用具体算法,只要告之用何计算器能 演算出这个数据结果.当然是您已经验证过的.如果没试过就不用说了,多谢.单贴最高悬赏只能200分,可以另外开贴给分.有特殊要求可以加Q:398493599,可以RMB报价.

Answer 1

将存有数据的字节数组进行逐位计算，求得字节形式的CRC

typedef unsigned __int16 INT16U;
#define CRC_SEED 0xFFFF // 该位称为预置值，使用人工算法（长除法）时 需要将除数多项式先与该与职位 异或 ，才能得到最后的除数多项式
#define POLY16 0x1021 // 该位为简式书写 实际为0x11021
INT16U crc16(unsigned char *buf,unsigned short length)
{
INT16U shift,data,val;
int i;

shift = CRC_SEED;

for(i=0;i<length;i++) {
if((i % 8) == 0)
data = (*buf++)<<8;
val = shift ^ data;
shift = shift<<1;
data = data <<1;
if(val&0x8000)
shift = shift ^ POLY16;
}
return shift;
}

2、查表法

static unsigned short ccitt_table[256] = {
0x0000, 0x1021, 0x2042, 0x3063, 0x4084, 0x50A5, 0x60C6, 0x70E7,
0x8108, 0x9129, 0xA14A, 0xB16B, 0xC18C, 0xD1AD, 0xE1CE, 0xF1EF,
0x1231, 0x0210, 0x3273, 0x2252, 0x52B5, 0x4294, 0x72F7, 0x62D6,
0x9339, 0x8318, 0xB37B, 0xA35A, 0xD3BD, 0xC39C, 0xF3FF, 0xE3DE,
0x2462, 0x3443, 0x0420, 0x1401, 0x64E6, 0x74C7, 0x44A4, 0x5485,
0xA56A, 0xB54B, 0x8528, 0x9509, 0xE5EE, 0xF5CF, 0xC5AC, 0xD58D,
0x3653, 0x2672, 0x1611, 0x0630, 0x76D7, 0x66F6, 0x5695, 0x46B4,
0xB75B, 0xA77A, 0x9719, 0x8738, 0xF7DF, 0xE7FE, 0xD79D, 0xC7BC,
0x48C4, 0x58E5, 0x6886, 0x78A7, 0x0840, 0x1861, 0x2802, 0x3823,
0xC9CC, 0xD9ED, 0xE98E, 0xF9AF, 0x8948, 0x9969, 0xA90A, 0xB92B,
0x5AF5, 0x4AD4, 0x7AB7, 0x6A96, 0x1A71, 0x0A50, 0x3A33, 0x2A12,
0xDBFD, 0xCBDC, 0xFBBF, 0xEB9E, 0x9B79, 0x8B58, 0xBB3B, 0xAB1A,
0x6CA6, 0x7C87, 0x4CE4, 0x5CC5, 0x2C22, 0x3C03, 0x0C60, 0x1C41,
0xEDAE, 0xFD8F, 0xCDEC, 0xDDCD, 0xAD2A, 0xBD0B, 0x8D68, 0x9D49,
0x7E97, 0x6EB6, 0x5ED5, 0x4EF4, 0x3E13, 0x2E32, 0x1E51, 0x0E70,
0xFF9F, 0xEFBE, 0xDFDD, 0xCFFC, 0xBF1B, 0xAF3A, 0x9F59, 0x8F78,
0x9188, 0x81A9, 0xB1CA, 0xA1EB, 0xD10C, 0xC12D, 0xF14E, 0xE16F,
0x1080, 0x00A1, 0x30C2, 0x20E3, 0x5004, 0x4025, 0x7046, 0x6067,
0x83B9, 0x9398, 0xA3FB, 0xB3DA, 0xC33D, 0xD31C, 0xE37F, 0xF35E,
0x02B1, 0x1290, 0x22F3, 0x32D2, 0x4235, 0x5214, 0x6277, 0x7256,
0xB5EA, 0xA5CB, 0x95A8, 0x8589, 0xF56E, 0xE54F, 0xD52C, 0xC50D,
0x34E2, 0x24C3, 0x14A0, 0x0481, 0x7466, 0x6447, 0x5424, 0x4405,
0xA7DB, 0xB7FA, 0x8799, 0x97B8, 0xE75F, 0xF77E, 0xC71D, 0xD73C,
0x26D3, 0x36F2, 0x0691, 0x16B0, 0x6657, 0x7676, 0x4615, 0x5634,
0xD94C, 0xC96D, 0xF90E, 0xE92F, 0x99C8, 0x89E9, 0xB98A, 0xA9AB,
0x5844, 0x4865, 0x7806, 0x6827, 0x18C0, 0x08E1, 0x3882, 0x28A3,
0xCB7D, 0xDB5C, 0xEB3F, 0xFB1E, 0x8BF9, 0x9BD8, 0xABBB, 0xBB9A,
0x4A75, 0x5A54, 0x6A37, 0x7A16, 0x0AF1, 0x1AD0, 0x2AB3, 0x3A92,
0xFD2E, 0xED0F, 0xDD6C, 0xCD4D, 0xBDAA, 0xAD8B, 0x9DE8, 0x8DC9,
0x7C26, 0x6C07, 0x5C64, 0x4C45, 0x3CA2, 0x2C83, 0x1CE0, 0x0CC1,
0xEF1F, 0xFF3E, 0xCF5D, 0xDF7C, 0xAF9B, 0xBFBA, 0x8FD9, 0x9FF8,
0x6E17, 0x7E36, 0x4E55, 0x5E74, 0x2E93, 0x3EB2, 0x0ED1, 0x1EF0
};

unsigned short crc_ccitt(unsigned char *q, int len)
{
unsigned short crc = 0;

while (len-- > 0)
crc = ccitt_table[(crc >> 8 ^ *q++) & 0xff] ^ (crc << 8);
return ~crc

}

追问

你好,你这个是CCITT,计算后的结果和我的数据是不一样的.

Answer 2

这个介绍的很详细， 你参考一下：
http://www.cnblogs.com/Msisiterc/archive/2011/12/20/2294910.html