讨论一下crc32的多项式问题？

[yemaozi2008]

本人因为要写一个帧，无奈的接触了CRC32，问了周围一些人，居然都是copy代码用的，根本解释不了，我自己琢磨了2天，算法原理基本清楚了，但对多项式的采用方面有写疑惑，如下：

在IEEE 802.3 帧中采用的是CRC-32校验码，多项式如下：
x32 + x26 + x23 + x22 + x16 + x12 + x11 + x10 + x8 + x7 + x5 + x4 + x2 + x + 1 多项式按正常写法是0x04C11DB7 ，翻转则为 0xEDB88320；

我的问题就是何时用正常，何时用翻转的多项式？

crc32的算法程序，按位或查表的，基本上搞懂了，比如说查表法，这里举个例子；

#define poly 0x04C11DB7
#define upoly 0xEDB88320

这是一个按位计算的程序
u_long g_crc(u_char *ptr,u_short len)
{
int i=0,j=0;
u_long iTemp=0,flag;
for (i=0; i<len; i++)
{
iTemp^=((*(ptr+i))<<24);
for(j=0; j<8; j++)
{
flag=iTemp & 0x80000000;
iTemp <<=1;
if(flag)iTemp^=poly;
}
}
return iTemp;
}

假如ptr指向一个数组SS[1]={0x01}；这里的出来的crc32码是0x04C11DB7

如果按查表法的话，则要生成一个crc32tbl[256];网上翻了很多资料，基本上都是
static const unsigned long crc32tbl[256] = {
0x00000000,0x77073096,0xEE0E612C,0x990951BA,0x076DC419,0x706AF48F,0xE963A535,
0x9E6495A3,0x0EDB8832,0x79DCB8A4,0xE0D5E91E,0x97D2D988,0x09B64C2B,0x7EB17CBD,
0xE7B82D07,0x90BF1D91,0x1DB71064,0x6AB020F2,0xF3B97148,0x84BE41DE,0x1ADAD47D,
0x6DDDE4EB,0xF4D4B551,0x83D385C7,0x136C9856,0x646BA8C0,0xFD62F97A,0x8A65C9EC,
0x14015C4F,0x63066CD9,0xFA0F3D63,0x8D080DF5,0x3B6E20C8,0x4C69105E,0xD56041E4,
0xA2677172,0x3C03E4D1,0x4B04D447,0xD20D85FD,0xA50AB56B,0x35B5A8FA,0x42B2986C,
0xDBBBC9D6,0xACBCF940,0x32D86CE3,0x45DF5C75,0xDCD60DCF,0xABD13D59,0x26D930AC,
0x51DE003A,0xC8D75180,0xBFD06116,0x21B4F4B5,0x56B3C423,0xCFBA9599,0xB8BDA50F,
0x2802B89E,0x5F058808,0xC60CD9B2,0xB10BE924,0x2F6F7C87,0x58684C11,0xC1611DAB,
0xB6662D3D,0x76DC4190,0x01DB7106,0x98D220BC,0xEFD5102A,0x71B18589,0x06B6B51F,
0x9FBFE4A5,0xE8B8D433,0x7807C9A2,0x0F00F934,0x9609A88E,0xE10E9818,0x7F6A0DBB,
0x086D3D2D,0x91646C97,0xE6635C01,0x6B6B51F4,0x1C6C6162,0x856530D8,0xF262004E,
0x6C0695ED,0x1B01A57B,0x8208F4C1,0xF50FC457,0x65B0D9C6,0x12B7E950,0x8BBEB8EA,
0xFCB9887C,0x62DD1DDF,0x15DA2D49,0x8CD37CF3,0xFBD44C65,0x4DB26158,0x3AB551CE,
0xA3BC0074,0xD4BB30E2,0x4ADFA541,0x3DD895D7,0xA4D1C46D,0xD3D6F4FB,0x4369E96A,
0x346ED9FC,0xAD678846,0xDA60B8D0,0x44042D73,0x33031DE5,0xAA0A4C5F,0xDD0D7CC9,
0x5005713C,0x270241AA,0xBE0B1010,0xC90C2086,0x5768B525,0x206F85B3,0xB966D409,
0xCE61E49F,0x5EDEF90E,0x29D9C998,0xB0D09822,0xC7D7A8B4,0x59B33D17,0x2EB40D81,
0xB7BD5C3B,0xC0BA6CAD,0xEDB88320,0x9ABFB3B6,0x03B6E20C,0x74B1D29A,0xEAD54739,
0x9DD277AF,0x04DB2615,0x73DC1683,0xE3630B12,0x94643B84,0x0D6D6A3E,0x7A6A5AA8,
0xE40ECF0B,0x9309FF9D,0x0A00AE27,0x7D079EB1,0xF00F9344,0x8708A3D2,0x1E01F268,
0x6906C2FE,0xF762575D,0x806567CB,0x196C3671,0x6E6B06E7,0xFED41B76,0x89D32BE0,
0x10DA7A5A,0x67DD4ACC,0xF9B9DF6F,0x8EBEEFF9,0x17B7BE43,0x60B08ED5,0xD6D6A3E8,
0xA1D1937E,0x38D8C2C4,0x4FDFF252,0xD1BB67F1,0xA6BC5767,0x3FB506DD,0x48B2364B,
0xD80D2BDA,0xAF0A1B4C,0x36034AF6,0x41047A60,0xDF60EFC3,0xA867DF55,0x316E8EEF,
0x4669BE79,0xCB61B38C,0xBC66831A,0x256FD2A0,0x5268E236,0xCC0C7795,0xBB0B4703,
0x220216B9,0x5505262F,0xC5BA3BBE,0xB2BD0B28,0x2BB45A92,0x5CB36A04,0xC2D7FFA7,
0xB5D0CF31,0x2CD99E8B,0x5BDEAE1D,0x9B64C2B0,0xEC63F226,0x756AA39C,0x026D930A,
0x9C0906A9,0xEB0E363F,0x72076785,0x05005713,0x95BF4A82,0xE2B87A14,0x7BB12BAE,
0x0CB61B38,0x92D28E9B,0xE5D5BE0D,0x7CDCEFB7,0x0BDBDF21,0x86D3D2D4,0xF1D4E242,
0x68DDB3F8,0x1FDA836E,0x81BE16CD,0xF6B9265B,0x6FB077E1,0x18B74777,0x88085AE6,
0xFF0F6A70,0x66063BCA,0x11010B5C,0x8F659EFF,0xF862AE69,0x616BFFD3,0x166CCF45,
0xA00AE278,0xD70DD2EE,0x4E048354,0x3903B3C2,0xA7672661,0xD06016F7,0x4969474D,
0x3E6E77DB,0xAED16A4A,0xD9D65ADC,0x40DF0B66,0x37D83BF0,0xA9BCAE53,0xDEBB9EC5,
0x47B2CF7F,0x30B5FFE9,0xBDBDF21C,0xCABAC28A,0x53B39330,0x24B4A3A6,0xBAD03605,
0xCDD70693,0x54DE5729,0x23D967BF,0xB3667A2E,0xC4614AB8,0x5D681B02,0x2A6F2B94,
0xB40BBE37,0xC30C8EA1,0x5A05DF1B,0x2D02EF8D };


这个表的生成很简单，但却一般是用的0xEDB88320翻转多项式，假如用0x04C11DB7这个正常多项式则必须还要交换位，下面分别给出两种的程序。

void gen_crc_table(void)
{
unsigned long crc;
int i, j;

for (i = 0; i < 256; i++)
{
crc = i;
for (j = 8; j > 0; j--)
{
if (crc & 1)

crc = (crc >> 1) ^ upoly;

else
crc >>= 1;

}
crc32tbl[i] = crc;
printf("%08x ",crc32tbl[i]);
}
}

用正常的就很麻烦了，先要搞个换位程序
u_long Reflect(u_long ref,char ch)
{ int i;
u_long value=0;
for(i=1;i<(ch+1);i++)
{
if(ref&1)
value|=1<<(ch-i);
ref>>=1;
}
return value;
}

void gen_crc_table(void)
{

int i,j;
unsigned long crc;
for(i=0;i<256;i++)
{
crc=Reflect(crc=i,32);
for(j=0;j<8;j++)
{

if(crc&0x80000000)
{crc<<=1;crc^=poly;}
else
crc<<=1;
}
crc32tbl[i]=Reflect(crc,32);
printf("%08x ",crc32tbl[i]);
}

}

可以看出，用正常多项式的话，数据最开始要翻转一次，做完crc又要翻转一次，也就是还原了，和上面的第一个程序得出的crc32码表是一样的；



查表的程序我就不给出了，假如数组SS[1]={0x01}，求CRC码，即使用手算也算得出来，crc32码就是0x04C11DB7，也就是多项式自身，而按查表法，因为只有1个字节，所以就是表中的crc32tbl[1]=0x77073096，这我就有点晕了，但我在网上看见过一个程序，算出的表不是和上面的一样，类似与crct32tbl[256]={0x00000000,0x04C11DB7,......},在做表的程序中多项式用的是PLOY（第一个做表程序），这样的表算出来的和我的按位计算的是对应的。

这个多项式的问题让我很头疼，请各位大侠给点指点。