c语言crc16
A. 关于c语言中16位数据的处理
显然buffer是字节类型的数组,将高字节乘以256(左移8位)与低字节相加,得到一个16位的整数,这个就是CRC的长度。
B. crc16校验的c语言程序
unsigned short crc_dsp(unsigned short reg, unsigned char data_crc)
//reg为crc寄存器, data_crc为将要处理的8bit数据流
{
unsigned short msb; //crc寄存器将移出的最高1bit
unsigned short data;
unsigned short gx = 0x8005, i = 0; //i为左移次数, gx为生成多项式
data = (unsigned short)data_crc;
data = data << 8;
reg = reg ^ data;
do
{
msb = reg & 0x8000;
reg = reg << 1;
if(msb == 0x8000)
{
reg = reg ^ gx;
}
i++;
}
while(i < 8);
return (reg);
}
C. 高分求计算CRC校验码的C语言程序
你就是想要CRC8-CCITT的代码,这个到处都是。
http://www.rajivchakravorty.com/source-code/uncertainty/multimedia-sim/html/crc8_8c-source.html
我一直有CRC16,没试过这个,但应该差不多。
参考文献:http://blog.sina.com.cn/s/blog_5e330a280100fcp9.html
D. 求一个C语言实现的CRC16/CCITT-FALSE校验码函数
//函数功能:计算CRC16
//参数1:*pData 16位CRC校验数据,
//参数2:nLength 数据流长度
//参数3:init 初始化值
//参数4:ptable 16位CRC查找表
//逆序CRC计算
unsigned short GetRevCrc_16(unsigned char * pData, int nLength,
unsigned short init, const
unsigned short *ptable)
{
unsigned short cRc_16 = init;
unsigned char temp;
while(nLength-- > 0)
{
temp = cRc_16 & 0xFF;
cRc_16 = (cRc_16 >> 8) ^ ptable[(temp ^ *pData++) & 0xFF];
}
return cRc_16;
}
//正序CRC计算
unsigned short GetCrc_16(unsigned char * pData, int nLength,
unsigned short init, const
unsigned short *ptable)
{
unsigned short cRc_16 = init;
unsigned char temp;
while(nLength-- > 0)
{
temp = cRc_16 >> 8;
cRc_16 = (cRc_16 << 8) ^ ptable[(temp ^ *pData++) & 0xFF];
}
return cRc_16;
}
//Demo -- modbus-crc16测试
unsigned short CRC_GetModbus16(unsigned char *pdata, int len)
{
//MODBUS CRC-16表 8005 逆序
const unsigned short g_McRctable_16[256] =
{
0x0000, 0xC0C1, 0xC181, 0x0140, 0xC301, 0x03C0, 0x0280, 0xC241,
0xC601, 0x06C0, 0x0780, 0xC741, 0x0500, 0xC5C1, 0xC481, 0x0440,
0xCC01, 0x0CC0, 0x0D80, 0xCD41, 0x0F00, 0xCFC1, 0xCE81, 0x0E40,
0x0A00, 0xCAC1, 0xCB81, 0x0B40, 0xC901, 0x09C0, 0x0880, 0xC841,
0xD801, 0x18C0, 0x1980, 0xD941, 0x1B00, 0xDBC1, 0xDA81, 0x1A40,
0x1E00, 0xDEC1, 0xDF81, 0x1F40, 0xDD01, 0x1DC0, 0x1C80, 0xDC41,
0x1400, 0xD4C1, 0xD581, 0x1540, 0xD701, 0x17C0, 0x1680, 0xD641,
0xD201, 0x12C0, 0x1380, 0xD341, 0x1100, 0xD1C1, 0xD081, 0x1040,
0xF001, 0x30C0, 0x3180, 0xF141, 0x3300, 0xF3C1, 0xF281, 0x3240,
0x3600, 0xF6C1, 0xF781, 0x3740, 0xF501, 0x35C0, 0x3480, 0xF441,
0x3C00, 0xFCC1, 0xFD81, 0x3D40, 0xFF01, 0x3FC0, 0x3E80, 0xFE41,
0xFA01, 0x3AC0, 0x3B80, 0xFB41, 0x3900, 0xF9C1, 0xF881, 0x3840,
0x2800, 0xE8C1, 0xE981, 0x2940, 0xEB01, 0x2BC0, 0x2A80, 0xEA41,
0xEE01, 0x2EC0, 0x2F80, 0xEF41, 0x2D00, 0xEDC1, 0xEC81, 0x2C40,
0xE401, 0x24C0, 0x2580, 0xE541, 0x2700, 0xE7C1, 0xE681, 0x2640,
0x2200, 0xE2C1, 0xE381, 0x2340, 0xE101, 0x21C0, 0x2080, 0xE041,
0xA001, 0x60C0, 0x6180, 0xA141, 0x6300, 0xA3C1, 0xA281, 0x6240,
0x6600, 0xA6C1, 0xA781, 0x6740, 0xA501, 0x65C0, 0x6480, 0xA441,
0x6C00, 0xACC1, 0xAD81, 0x6D40, 0xAF01, 0x6FC0, 0x6E80, 0xAE41,
0xAA01, 0x6AC0, 0x6B80, 0xAB41, 0x6900, 0xA9C1, 0xA881, 0x6840,
0x7800, 0xB8C1, 0xB981, 0x7940, 0xBB01, 0x7BC0, 0x7A80, 0xBA41,
0xBE01, 0x7EC0, 0x7F80, 0xBF41, 0x7D00, 0xBDC1, 0xBC81, 0x7C40,
0xB401, 0x74C0, 0x7580, 0xB541, 0x7700, 0xB7C1, 0xB681, 0x7640,
0x7200, 0xB2C1, 0xB381, 0x7340, 0xB101, 0x71C0, 0x7080, 0xB041,
0x5000, 0x90C1, 0x9181, 0x5140, 0x9301, 0x53C0, 0x5280, 0x9241,
0x9601, 0x56C0, 0x5780, 0x9741, 0x5500, 0x95C1, 0x9481, 0x5440,
0x9C01, 0x5CC0, 0x5D80, 0x9D41, 0x5F00, 0x9FC1, 0x9E81, 0x5E40,
0x5A00, 0x9AC1, 0x9B81, 0x5B40, 0x9901, 0x59C0, 0x5880, 0x9841,
0x8801, 0x48C0, 0x4980, 0x8941, 0x4B00, 0x8BC1, 0x8A81, 0x4A40,
0x4E00, 0x8EC1, 0x8F81, 0x4F40, 0x8D01, 0x4DC0, 0x4C80, 0x8C41,
0x4400, 0x84C1, 0x8581, 0x4540, 0x8701, 0x47C0, 0x4680, 0x8641,
0x8201, 0x42C0, 0x4380, 0x8341, 0x4100, 0x81C1, 0x8081, 0x4040
};
return GetRevCrc_16(pdata, len, 0xFFFF, g_McRctable_16);
}
E. 谁能帮我看看 这段VB求CRC-16的代码为什么和下面的C语言所求的CRC不一致。
把c的做成dll直接调用不就行了
Function CRC16_0(data() As Byte) As Long
Dim p1 As Long
Dim datalen As Long
datalen=UBound(data)
p1=VarPtr(data(1))
CRC16_0= cal_crc(p1,datalen)
End Function
F. CRC的C语言的程序
按位计算CRC采用CRC-CCITT多项式,多项式为0x11021,C语言编程时,参与计算为0x1021。当按位计算CRC时,例如计算二进制序列为1001 1010 1010 1111时,将二进制序列数左移16位,即为1001 1010 1010 1111 (0000 0000 0000 0000),实际上该二进制序列可拆分为1000 0000 0000 0000 (0000 0000 0000 0000) + 000 0000 0000 0000 (0000 0000 0000 0000) + 00 0000 0000 0000 (0000 0000 0000 0000) + 1 0000 0000 0000 (0000 0000 0000 0000) + ……
现在开始分析运算:
<1>对第一个二进制分序列求余数,竖式除法即为0x10000 ^ 0x11021运算,后面的0位保留;
<2>接着对第二个二进制分序列求余数,将第一步运算的余数*2后再和第二个二进制分序列一起对0x11021求余,这一步理解应该没什么问题。如果该分序列为0,无需计算。
<3>对其余的二进制序列求余与上面两步相同。
<4>计算到最后一位时即为整个二进制序列的余数,即为CRC校验码。
该计算方法相当于对每一位计算,运算过程很容易理解,所占内存少,缺点是一位一位计算比较耗时。
下面给出C语言实现方法:
代码如下:
unsigned char test[16] = {0x00,0x11,0x22,0x33,0x44,0x55,0x66,0x77,0x88,0x99,0xaa,0xbb,0xcc,0xdd,0xee,0xff};
unsigned char len = 16;
void main( void )
{
unsigned long temp = 0;
unsigned int crc;
unsigned char i;
unsigned char *ptr = test;
while( len-- ) {
for(i = 0x80; i != 0; i = i >> 1) {
temp = temp * 2;
if((temp & 0x10000) != 0)
temp = temp ^ 0x11021;
if((*ptr & i) != 0)
temp = temp ^ (0x10000 ^ 0x11021);
}
ptr++;
}
crc = temp;
printf("0x%x ",crc);
}
G. 将下面C语言版的CRC校验改为C#代码版
以下是我的分析,不知是否正确,你参考下1、首先来看你打java代码:crc=(byte)((crc>>1)^0x8c);和 crc=(byte)(crc>>1); 导致这个问题是因为byte的最高位符号位,转换的时候就出错了2、示例代码:package com.test;public class test {public static void main(String[] args) {byte[] ptr = { 1, 1, 1, 1, 1, 1 };byte res = getCrc(ptr);System.out.println();System.out.println((byte)( (1 >> 1) ^ 0x8c ) + ":" +( (1 >> 1) ^ 0x8c ) );}public static byte getCrc(byte[] ptr) {int crc = 0;for (int i = 0; i > 1) ^ 0x8c;} else {crc = crc >> 1;}}}return (byte) crc;}}
H. 把下面这段c语言的crc校验 转换成java的,麻烦了, 我是实在不会
unsigned short 都替换为int
unsigned int 也替换为int
unsigned char const *buf替换为byte[] buf
*buf++替换为buf[i]