Question

我想为字符串/数据生成校验和

1. The same data should produce the same Checksum
2. Two different data strings can't product same checksum. Random collision of 0.1% can be negligible
3. No encryption/decryption of data 
4. Checksum length need not be too huge and contains letters and characters.
5. Must be too fast and efficient. Imagine generating checksum(s) for 100 Mb of text data should be in less than 5mins. Generating 1000 checksums for less than 1 KB of each segment data should be in less than 10 seconds.

非常感谢任何算法或实现参考和建议。

Answer 1

您可以编写自定义哈希函数：（c ++）

long long int hash(String s){
    long long k = 7;
    for(int i = 0; i < s.length(); i++){
        k *= 23;
        k += s[i];
        k *= 13;
        k %= 1000000009;
    }
    return k;
}

这应该可以为您提供（大多数样本无冲突）哈希值。

Answer 2

一个非常常见的快速校验和是CRC-32，一种32位多项式循环冗余校验。以下是C中的三种实现，它们的速度与复杂性有所不同，CRC-32 :(来自http://www.hackersdelight.org/hdcodetxt/crc.c.txt）

#include <stdio.h>
#include <stdlib.h>

// ---------------------------- reverse --------------------------------

// Reverses (reflects) bits in a 32-bit word.
unsigned reverse(unsigned x) {
   x = ((x & 0x55555555) <<  1) | ((x >>  1) & 0x55555555);
   x = ((x & 0x33333333) <<  2) | ((x >>  2) & 0x33333333);
   x = ((x & 0x0F0F0F0F) <<  4) | ((x >>  4) & 0x0F0F0F0F);
   x = (x << 24) | ((x & 0xFF00) << 8) |
       ((x >> 8) & 0xFF00) | (x >> 24);
   return x;
}

// ----------------------------- crc32a --------------------------------

/* This is the basic CRC algorithm with no optimizations. It follows the
logic circuit as closely as possible. */

unsigned int crc32a(unsigned char *message) {
   int i, j;
   unsigned int byte, crc;

   i = 0;
   crc = 0xFFFFFFFF;
   while (message[i] != 0) {
      byte = message[i];            // Get next byte.
      byte = reverse(byte);         // 32-bit reversal.
      for (j = 0; j <= 7; j++) {    // Do eight times.
         if ((int)(crc ^ byte) < 0)
              crc = (crc << 1) ^ 0x04C11DB7;
         else crc = crc << 1;
         byte = byte << 1;          // Ready next msg bit.
      }
      i = i + 1;
   }
   return reverse(~crc);
}

// ----------------------------- crc32b --------------------------------

/* This is the basic CRC-32 calculation with some optimization but no
table lookup. The the byte reversal is avoided by shifting the crc reg
right instead of left and by using a reversed 32-bit word to represent
the polynomial.
   When compiled to Cyclops with GCC, this function executes in 8 + 72n
instructions, where n is the number of bytes in the input message. It
should be doable in 4 + 61n instructions.
   If the inner loop is strung out (approx. 5*8 = 40 instructions),
it would take about 6 + 46n instructions. */

unsigned int crc32b(unsigned char *message) {
   int i, j;
   unsigned int byte, crc, mask;

   i = 0;
   crc = 0xFFFFFFFF;
   while (message[i] != 0) {
      byte = message[i];            // Get next byte.
      crc = crc ^ byte;
      for (j = 7; j >= 0; j--) {    // Do eight times.
         mask = -(crc & 1);
         crc = (crc >> 1) ^ (0xEDB88320 & mask);
      }
      i = i + 1;
   }
   return ~crc;
}

// ----------------------------- crc32c --------------------------------

/* This is derived from crc32b but does table lookup. First the table
itself is calculated, if it has not yet been set up.
Not counting the table setup (which would probably be a separate
function), when compiled to Cyclops with GCC, this function executes in
7 + 13n instructions, where n is the number of bytes in the input
message. It should be doable in 4 + 9n instructions. In any case, two
of the 13 or 9 instrucions are load byte.
   This is Figure 14-7 in the text. */

unsigned int crc32c(unsigned char *message) {
   int i, j;
   unsigned int byte, crc, mask;
   static unsigned int table[256];

   /* Set up the table, if necessary. */

   if (table[1] == 0) {
      for (byte = 0; byte <= 255; byte++) {
         crc = byte;
         for (j = 7; j >= 0; j--) {    // Do eight times.
            mask = -(crc & 1);
            crc = (crc >> 1) ^ (0xEDB88320 & mask);
         }
         table[byte] = crc;
      }
   }

   /* Through with table setup, now calculate the CRC. */

   i = 0;
   crc = 0xFFFFFFFF;
   while ((byte = message[i]) != 0) {
      crc = (crc >> 8) ^ table[(crc ^ byte) & 0xFF];
      i = i + 1;
   }
   return ~crc;
}

如果您只是google "CRC32"，您将获得比您可能吸收的更多信息。

为字符串生成校验和

2 个答案: