是否可以找到前綴的CRC32，只給出整個數據的CRC32？

Question

我必須在表中查找，並且我有一個字符串標識符和字符串的CRC32。如果發生錯誤，我必須減小尺寸並查找標識符的前綴。 因此，我需要計算前綴的校驗和並重復每個前綴的過程。

這個算法我的C代碼是這樣的：

find_prefix(char* string, uint16_t size, uint32_t crc, hash_table_t *hash_table){ 
    do{ 
    if(perform_lookup(string, size, crc, hash_table)==HIT){ 
     return size; 
    } 
    size--; 
    crc=calculate_crc(string, size, 0xFFFFFFFF); //Is there a better way? 
    } while(size); 
} 

我的問題是：我能避免CRC計算和推導的前綴CRC，給整個字符串，字符串的CRC本身？

我發現了一些相關的問題here和here，但我有一個限制：被previosly填充表時，標識的CRC在硬件計算的，所以我不能修改算法，並且兩個鏈接提供作爲計算校驗和的一種不同方式（基本上使用所有組件的XOR）。

非常感謝。

Answer 1

是的，這是可能的。 （除了一個簡單的觀察，如果你有問題中陳述的字符串，那麼你可以簡單地計算前綴的CRC。）

爲了更改你的問題，我有兩個字符串A和B，它們的連接是AB。如果我只有AB的CRC並且有字符串B，我可以計算A的CRC嗎？

您可以使用CRC的高位字節來反轉計算CRC的過程，以確定使用哪個表項。它幾乎與計算B的CRC的速度一樣快。用於zip，gzip等標準CRC-32的示例代碼：

/* Given the CRC of the concatenated string AB and the string B, calculate the 
    CRC of A. Placed in the public domain by Mark Adler. */ 

#include <stdio.h> 

#define local static 

/* Byte-wise CRC table for CRC-32 */ 
local unsigned long crc_table[] = { 
    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 
}; 

local unsigned char rev[256]; 

local void revgen(void) 
{ 
    unsigned k; 

    for (k = 0; k < 256; k++) 
     rev[crc_table[k] >> 24] = k; 
} 

#define ONES 0xffffffff 

local unsigned long revcrc(unsigned long crc, 
          const unsigned char *buf, size_t len) 
{ 
    unsigned k; 

    crc = crc^ONES; 
    while (len--) { 
     k = rev[crc >> 24]; 
     crc = ((crc^crc_table[k]) << 8) | (k^buf[len]); 
    } 
    return crc^ONES; 
} 

int main(void) 
{ 
    unsigned long crc = 0x9ef61f95;  /* CRC-32 of "foobar" */ 
             /* CRC-32 of "foo" is 0x8c736521 */ 

    revgen(); 
    printf("0x%08lx (should be 0x8c736521)\n", 
      revcrc(crc, (unsigned char *)"bar", 3)); 
    return 0; 
} 

Answer 2

不，您需要在最後一個字符集成到CRC時移出的位。

您可以展開循環一小段，並在一次迭代中計算出(string, size - 8),(string, size - 7) ......的CRC。