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在C中实现MD5

[英]Implementation of MD5 in C

 原文  2014-06-29 20:01:41  0   2    c/ hash/ md5

问题描述

我想在C项目中实现MD5哈希函数，而且我想自己做，因为我担心的一件事是使用别人的代码（主要是因为我在理解这样的代码时遇到麻烦）。 因此，我直接进入Wiki页面以获取伪代码： http : //en.wikipedia.org/wiki/MD5我决定离开填充并分成512位的块，以供日后使用，然后从MD5散列开始空字符串。 正确填充后，（我认为）它应如下所示： 

unsigned int message[16] = {0x80000000, 0x00000000, 0x00000000, 0x00000000,//binary: 100000000000000000000000...
                            0x00000000, 0x00000000, 0x00000000, 0x00000000,// ...0000000000000000000000000000...
                            0x00000000, 0x00000000, 0x00000000, 0x00000000,// ...0000000000000000000000000000...
                            0x00000000, 0x00000000, 0x00000000, 0x00000000};//...0000000000000000000000000000000

这是我在C语言中重现Wiki的主循环（仅处理一个512位块的循环）的方式： 

unsigned int a0, b0, c0, d0, 
             A, B, C, D,
             i, F, g, bufD;

//empty string appended with 1 bit and zeroes till it is 512 bytes long as 16 32-bit words
unsigned int message[16] = {0x80000000, 0x00000000, 0x00000000, 0x00000000,//binary: 100000000000000000000000...
                            0x00000000, 0x00000000, 0x00000000, 0x00000000,// ...0000000000000000000000000000...
                            0x00000000, 0x00000000, 0x00000000, 0x00000000,// ...0000000000000000000000000000...
                            0x00000000, 0x00000000, 0x00000000, 0x00000000};//...0000000000000000000000000000000

unsigned int shiftAmounts[64] = {7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22,
                                 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20,
                                 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23,
                                 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21};

unsigned int partsOfSines[64] = {0xd76aa478, 0xe8c7b756, 0x242070db, 0xc1bdceee,
                                 0xf57c0faf, 0x4787c62a, 0xa8304613, 0xfd469501,
                                 0x698098d8, 0x8b44f7af, 0xffff5bb1, 0x895cd7be,
                                 0x6b901122, 0xfd987193, 0xa679438e, 0x49b40821,
                                 0xf61e2562, 0xc040b340, 0x265e5a51, 0xe9b6c7aa,
                                 0xd62f105d, 0x02441453, 0xd8a1e681, 0xe7d3fbc8,
                                 0x21e1cde6, 0xc33707d6, 0xf4d50d87, 0x455a14ed,
                                 0xa9e3e905, 0xfcefa3f8, 0x676f02d9, 0x8d2a4c8a,
                                 0xfffa3942, 0x8771f681, 0x6d9d6122, 0xfde5380c,
                                 0xa4beea44, 0x4bdecfa9, 0xf6bb4b60, 0xbebfbc70,
                                 0x289b7ec6, 0xeaa127fa, 0xd4ef3085, 0x04881d05,
                                 0xd9d4d039, 0xe6db99e5, 0x1fa27cf8, 0xc4ac5665,
                                 0xf4292244, 0x432aff97, 0xab9423a7, 0xfc93a039,
                                 0x655b59c3, 0x8f0ccc92, 0xffeff47d, 0x85845dd1,
                                 0x6fa87e4f, 0xfe2ce6e0, 0xa3014314, 0x4e0811a1,
                                 0xf7537e82, 0xbd3af235, 0x2ad7d2bb, 0xeb86d391};

a0 = 0x67452301;
b0 = 0xefcdab89;
c0 = 0x98badcfe;
d0 = 0x10325476;

//gotta put this into a loop for each 512-bit chunk
A = a0;
B = b0;
C = c0;
D = d0;

for (i=0; i<64; i++){
    if (i < 16){
        F = (B & C) | (~B & D);
        g = i;
    } else if (i >= 16 && i < 32){
        F = (D & B) | (~D & C);
        g = (5*i + 1) % 16;
    } else if (i >= 32 && i < 48){
        F = B ^ C ^ D;
        g = (3*i + 5) % 16;
    } else if (i >= 48){
        F = C ^ (B | ~D);
        g = (7*i) % 16;
    }
    bufD = D;
    D = C;
    C = B;
    B += leftRotate((A + F + partsOfSines[i] + message[g]), shiftAmounts[i]);
    A = bufD;
}
a0 += A;
b0 += B;
c0 += C;
d0 += D;
//end future loop

printf ("a0=%u, b0=%u, c0=%u, d0=%u", a0, b0, c0, d0);

leftRotate函数，也从Wikipedia上的伪代码进行了修改： 

unsigned int leftRotate(unsigned int x, int n){
    return ((x) << n) | ((x) >> (32 - n));
}

输出以下内容： 

a0=578518856, b0=3524428790, c0=1003076545, d0=1531243034

将这些十进制值转换为十六进制，我得到以下信息： 

a0 = 578518856 ->  227B7F48
b0 = 3524428790 -> D21283F6
c0 = 1003076545 -> 3BC9BBC1
d0 = 1531243034 -> 5B44EA1A

甚至与空字符串的实际MD5摘要（即d41d8cd98f00b204e9800998ecf8427e）有点相似。 所以，我的问题是，我要去哪里错了？

2 个解决方案

解决方案1
 2  2014-06-29 20:09:56

可能是目标体系结构的问题（字节序，类型大小等）
我的答案来自http://bytes.com/topic/c/answers/855645-simple-md5-hash-different-output-different-os的位

编辑：
问题是因为考虑到内存中的小端存储，您添加的填充位不在正确的字节上。 此外，您必须打印结果md5哈希，因为它存储在内存中（在LE中）。 

#include <stdio.h>

unsigned int leftRotate(unsigned int x, unsigned int n){
    return (x << n) | (x >> (32 - n));
}

int main(){
    unsigned int a0, b0, c0, d0, 
                 A, B, C, D,
                 i, F, g, bufD;

    //empty string appended with 1 bit and zeroes till it is 512 bytes long as 16 32-bit words
    unsigned int message[16] = {0x00000080, 0x00000000, 0x00000000, 0x00000000,//binary: 100000000000000000000000...
                                0x00000000, 0x00000000, 0x00000000, 0x00000000,// ...0000000000000000000000000000...
                                0x00000000, 0x00000000, 0x00000000, 0x00000000,// ...0000000000000000000000000000...
                                0x00000000, 0x00000000, 0x00000000, 0x00000000};//...0000000000000000000000000000000

    unsigned int shiftAmounts[64] = {7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22,
                                     5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20,
                                     4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23,
                                     6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21};

    unsigned int partsOfSines[64] = {0xd76aa478, 0xe8c7b756, 0x242070db, 0xc1bdceee,
                                     0xf57c0faf, 0x4787c62a, 0xa8304613, 0xfd469501,
                                     0x698098d8, 0x8b44f7af, 0xffff5bb1, 0x895cd7be,
                                     0x6b901122, 0xfd987193, 0xa679438e, 0x49b40821,
                                     0xf61e2562, 0xc040b340, 0x265e5a51, 0xe9b6c7aa,
                                     0xd62f105d, 0x02441453, 0xd8a1e681, 0xe7d3fbc8,
                                     0x21e1cde6, 0xc33707d6, 0xf4d50d87, 0x455a14ed,
                                     0xa9e3e905, 0xfcefa3f8, 0x676f02d9, 0x8d2a4c8a,
                                     0xfffa3942, 0x8771f681, 0x6d9d6122, 0xfde5380c,
                                     0xa4beea44, 0x4bdecfa9, 0xf6bb4b60, 0xbebfbc70,
                                     0x289b7ec6, 0xeaa127fa, 0xd4ef3085, 0x04881d05,
                                     0xd9d4d039, 0xe6db99e5, 0x1fa27cf8, 0xc4ac5665,
                                     0xf4292244, 0x432aff97, 0xab9423a7, 0xfc93a039,
                                     0x655b59c3, 0x8f0ccc92, 0xffeff47d, 0x85845dd1,
                                     0x6fa87e4f, 0xfe2ce6e0, 0xa3014314, 0x4e0811a1,
                                     0xf7537e82, 0xbd3af235, 0x2ad7d2bb, 0xeb86d391};

    a0 = 0x67452301;
    b0 = 0xefcdab89;
    c0 = 0x98badcfe;
    d0 = 0x10325476;

    //gotta put this into a loop for each 512-bit chunk
    A = a0;
    B = b0;
    C = c0;
    D = d0;

    for (i=0; i<64; i++){
        if (i < 16){
            F = (B & C) | (~B & D);
            g = i;
        } else if (i >= 16 && i < 32){
            F = (D & B) | (~D & C);
            g = (5*i + 1) % 16;
        } else if (i >= 32 && i < 48){
            F = B ^ C ^ D;
            g = (3*i + 5) % 16;
        } else if (i >= 48){
            F = C ^ (B | ~D);
            g = (7*i) % 16;
        }
        bufD = D;
        D = C;
        C = B;
        B += leftRotate((A + F + partsOfSines[i] + message[g]), shiftAmounts[i]);
        A = bufD;
    }
    a0 += A;
    b0 += B;
    c0 += C;
    d0 += D;
    //end future loop

    printf("%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x\n", a0&0xff, (a0>>8)&0xff, (a0>>16)&0xff, (a0>>24)&0xff, b0&0xff, (b0>>8)&0xff, (b0>>16)&0xff, (b0>>24)&0xff, c0&0xff, (c0>>8)&0xff, (c0>>16)&0xff, (c0>>24)&0xff, d0&0xff, (d0>>8)&0xff, (d0>>16)&0xff, (d0>>24)&0xff);
}

EDIT2：
啊，为时已晚：/

解决方案2
 2 已采纳  2014-06-29 20:53:53

#include <stdio.h>

//empty string appended with 1 bit and zeroes till it is 512 bytes long as 16 32-bit words
unsigned int message[16] = {0x00000080, 0x00000000, 0x00000000, 0x00000000,
                            0x00000000, 0x00000000, 0x00000000, 0x00000000,
                            0x00000000, 0x00000000, 0x00000000, 0x00000000,
                            0x00000000, 0x00000000, 0x00000000, 0x00000000};

unsigned int shiftAmounts[64] = {7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22,
                                 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20,
                                 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23,
                                 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21};

unsigned int partsOfSines[64] = {0xd76aa478, 0xe8c7b756, 0x242070db, 0xc1bdceee,
                                 0xf57c0faf, 0x4787c62a, 0xa8304613, 0xfd469501,
                                 0x698098d8, 0x8b44f7af, 0xffff5bb1, 0x895cd7be,
                                 0x6b901122, 0xfd987193, 0xa679438e, 0x49b40821,
                                 0xf61e2562, 0xc040b340, 0x265e5a51, 0xe9b6c7aa,
                                 0xd62f105d, 0x02441453, 0xd8a1e681, 0xe7d3fbc8,
                                 0x21e1cde6, 0xc33707d6, 0xf4d50d87, 0x455a14ed,
                                 0xa9e3e905, 0xfcefa3f8, 0x676f02d9, 0x8d2a4c8a,
                                 0xfffa3942, 0x8771f681, 0x6d9d6122, 0xfde5380c,
                                 0xa4beea44, 0x4bdecfa9, 0xf6bb4b60, 0xbebfbc70,
                                 0x289b7ec6, 0xeaa127fa, 0xd4ef3085, 0x04881d05,
                                 0xd9d4d039, 0xe6db99e5, 0x1fa27cf8, 0xc4ac5665,
                                 0xf4292244, 0x432aff97, 0xab9423a7, 0xfc93a039,
                                 0x655b59c3, 0x8f0ccc92, 0xffeff47d, 0x85845dd1,
                                 0x6fa87e4f, 0xfe2ce6e0, 0xa3014314, 0x4e0811a1,
                                 0xf7537e82, 0xbd3af235, 0x2ad7d2bb, 0xeb86d391};

unsigned leftRotate(unsigned x, int n) {
    return (x << n) | (x >> (32 - n));
}

void printReverseEndian(unsigned n) {
  printf("%02x%02x%02x%02x", n & 0xff, (n >> 8) & 0xff, (n >> 16) & 0xff, n >> 24);
}

int main() {

  unsigned int a0, b0, c0, d0, 
               A, B, C, D,
               i, F, g, bufD;

  a0 = 0x67452301;
  b0 = 0xefcdab89;
  c0 = 0x98badcfe;
  d0 = 0x10325476;

  A = a0;
  B = b0;
  C = c0;
  D = d0;

  for (i=0; i<64; i++){
    if (i < 16){
        F = (B & C) | (~B & D);
        g = i;
    } else if (i < 32){
        F = (D & B) | (~D & C);
        g = (5*i + 1) % 16;
    } else if (i < 48){
        F = B ^ C ^ D;
        g = (3*i + 5) % 16;
    } else {
        F = C ^ (B | ~D);
        g = (7*i) % 16;
    }
    bufD = D;
    D = C;
    C = B;
    B += leftRotate((A + F + partsOfSines[i] + message[g]), shiftAmounts[i]);
    A = bufD;
  }
  a0 += A;
  b0 += B;
  c0 += C;
  d0 += D;


  printReverseEndian(a0);
  printReverseEndian(b0);
  printReverseEndian(c0);
  printReverseEndian(d0);

  return 0;
}

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