C语言中MD5的实现答案

【问题标题】：Implementation of MD5 in CC语言中MD5的实现
【发布时间】：2014-08-20 05:06:24
【问题描述】：

我想在一个 C 项目中实现一个 MD5 散列函数，我想做自己的，因为我害怕的一件事是使用别人的代码（主要是因为我无法理解这些代码）。所以，我直接前往 Wiki 页面获取伪代码：http://en.wikipedia.org/wiki/MD5 我决定留下填充并将其分成 512 位块以备后用，然后从空字符串的 MD5 哈希开始。正确填充，它应该（我认为）看起来像这样：

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

这就是我如何在 C 中重现 Wiki 的主循环（只处理一个 512 位块的循环）：

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，也是从维基百科上的伪代码修改而来的：

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

这会输出以下内容：

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

这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，我哪里出错了？

【问题讨论】：

显示你的leftRotate函数。
如果你在 little-endian 系统上执行此操作，并且消息的第一位是 1，那么 message[0] 不应该是 0x00000080 而不是 0x80000000 吗？
@MarkPlotnick 是的，就是这样。如果将答案的字节作为 4 字节无符号整数访问，则它们也需要重新排序。最好将它们输出为一系列无符号字符。
@ooga：谢谢，已编辑！
不，他的意思是0x00000080。反转的是字节（不是 nybbles！）。

标签： c hash md5

【解决方案1】：

#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;
}

【讨论】：

@Mints97 怎么没有main？？？而且它不需要原型，因为函数是在main 之前定义的。
对不起，对不起，我晚上总是那么慢=）我编辑了我原来的评论=）
等等，我如何输入和获取输出？

【解决方案2】：

可能是目标架构的问题（字节序、类型的大小……）
我的回答来自http://bytes.com/topic/c/answers/855645-simple-md5-hash-different-output-different-os

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，您必须打印生成的 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);
}

编辑2：
啊，太晚了：/

【讨论】：

字节顺序只是字节顺序的问题，但是，但是我重新排列了十六进制字符串的字节，我仍然没有得到正确的摘要。我检查了 sizeof(unsigned int)，它在我的操作系统（Windows 8 x64）上应该是 4。
Windows 意味着 x86 意味着 little-endian，所以可能不是这样。