Question

我一直试图自己重新创建MD5算法。我似乎无法使算法正确。看来我在填充和字节序上有问题。

#include <stdio.h>
#include <stdint.h>

/* F, G, H and I are basic MD5 functions.
 */
#define F(x, y, z) (((x) & (y)) | ((~x) & (z)))
#define G(x, y, z) (((x) & (z)) | ((y) & (~z)))
#define H(x, y, z) ((x) ^ (y) ^ (z))
#define I(x, y, z) ((y) ^ ((x) | (~z)))

/* ROTATE_LEFT rotates x left n bits.
 */
#define ROTATE_LEFT(x, n) (((x) << (n)) | ((x) >> (32-(n))))

/* FF, GG, HH, and II transformations for rounds 1, 2, 3, and 4.
Rotation is separate from addition to prevent recomputation.
 */
#define FF(a, b, c, d, x, s, ac) { \
 (a) += F ((b), (c), (d)) + (x) + (uint32_t)(ac); \
 (a) = ROTATE_LEFT ((a), (s)); \
 (a) += (b); \
  }
#define GG(a, b, c, d, x, s, ac) { \
 (a) += G ((b), (c), (d)) + (x) + (uint32_t)(ac); \
 (a) = ROTATE_LEFT ((a), (s)); \
 (a) += (b); \
  }
#define HH(a, b, c, d, x, s, ac) { \
 (a) += H ((b), (c), (d)) + (x) + (uint32_t)(ac); \
 (a) = ROTATE_LEFT ((a), (s)); \
 (a) += (b); \
  }
#define II(a, b, c, d, x, s, ac) { \
 (a) += I ((b), (c), (d)) + (x) + (uint32_t)(ac); \
 (a) = ROTATE_LEFT ((a), (s)); \
 (a) += (b); \
  }

#define S11 7
#define S12 12
#define S13 17
#define S14 22
#define S21 5
#define S22 9
#define S23 14
#define S24 20
#define S31 4
#define S32 11
#define S33 16
#define S34 23
#define S41 6
#define S42 10
#define S43 15
#define S44 21

void MD5_hash(uint32_t *message, uint32_t *digest) {
 const uint32_t d0 = 0x67452301;
 const uint32_t d1 = 0xEFCDAB89;
 const uint32_t d2 = 0x98BADCFE;
 const uint32_t d3 = 0x10325476;

 uint32_t a, b, c, d, *x;

 a = d0;
 b = d1;
 c = d2;
 d = d3;

 x = message;

 /* Round 1 */
 FF (a, b, c, d, x[ 0], S11, 0xd76aa478); /* 1 */
 FF (d, a, b, c, x[ 1], S12, 0xe8c7b756); /* 2 */
 FF (c, d, a, b, x[ 2], S13, 0x242070db); /* 3 */
 FF (b, c, d, a, x[ 3], S14, 0xc1bdceee); /* 4 */
 FF (a, b, c, d, x[ 4], S11, 0xf57c0faf); /* 5 */
 FF (d, a, b, c, x[ 5], S12, 0x4787c62a); /* 6 */
 FF (c, d, a, b, x[ 6], S13, 0xa8304613); /* 7 */
 FF (b, c, d, a, x[ 7], S14, 0xfd469501); /* 8 */
 FF (a, b, c, d, x[ 8], S11, 0x698098d8); /* 9 */
 FF (d, a, b, c, x[ 9], S12, 0x8b44f7af); /* 10 */
 FF (c, d, a, b, x[10], S13, 0xffff5bb1); /* 11 */
 FF (b, c, d, a, x[11], S14, 0x895cd7be); /* 12 */
 FF (a, b, c, d, x[12], S11, 0x6b901122); /* 13 */
 FF (d, a, b, c, x[13], S12, 0xfd987193); /* 14 */
 FF (c, d, a, b, x[14], S13, 0xa679438e); /* 15 */
 FF (b, c, d, a, x[15], S14, 0x49b40821); /* 16 */

 /* Round 2 */
 GG (a, b, c, d, x[ 1], S21, 0xf61e2562); /* 17 */
 GG (d, a, b, c, x[ 6], S22, 0xc040b340); /* 18 */
 GG (c, d, a, b, x[11], S23, 0x265e5a51); /* 19 */
 GG (b, c, d, a, x[ 0], S24, 0xe9b6c7aa); /* 20 */
 GG (a, b, c, d, x[ 5], S21, 0xd62f105d); /* 21 */
 GG (d, a, b, c, x[10], S22, 0x02441453); /* 22 */
 GG (c, d, a, b, x[15], S23, 0xd8a1e681); /* 23 */
 GG (b, c, d, a, x[ 4], S24, 0xe7d3fbc8); /* 24 */
 GG (a, b, c, d, x[ 9], S21, 0x21e1cde6); /* 25 */
 GG (d, a, b, c, x[14], S22, 0xc33707d6); /* 26 */
 GG (c, d, a, b, x[ 3], S23, 0xf4d50d87); /* 27 */
 GG (b, c, d, a, x[ 8], S24, 0x455a14ed); /* 28 */
 GG (a, b, c, d, x[13], S21, 0xa9e3e905); /* 29 */
 GG (d, a, b, c, x[ 2], S22, 0xfcefa3f8); /* 30 */
 GG (c, d, a, b, x[ 7], S23, 0x676f02d9); /* 31 */
 GG (b, c, d, a, x[12], S24, 0x8d2a4c8a); /* 32 */

 /* Round 3 */
 HH (a, b, c, d, x[ 5], S31, 0xfffa3942); /* 33 */
 HH (d, a, b, c, x[ 8], S32, 0x8771f681); /* 34 */
 HH (c, d, a, b, x[11], S33, 0x6d9d6122); /* 35 */
 HH (b, c, d, a, x[14], S34, 0xfde5380c); /* 36 */
 HH (a, b, c, d, x[ 1], S31, 0xa4beea44); /* 37 */
 HH (d, a, b, c, x[ 4], S32, 0x4bdecfa9); /* 38 */
 HH (c, d, a, b, x[ 7], S33, 0xf6bb4b60); /* 39 */
 HH (b, c, d, a, x[10], S34, 0xbebfbc70); /* 40 */
 HH (a, b, c, d, x[13], S31, 0x289b7ec6); /* 41 */
 HH (d, a, b, c, x[ 0], S32, 0xeaa127fa); /* 42 */
 HH (c, d, a, b, x[ 3], S33, 0xd4ef3085); /* 43 */
 HH (b, c, d, a, x[ 6], S34, 0x04881d05); /* 44 */
 HH (a, b, c, d, x[ 9], S31, 0xd9d4d039); /* 45 */
 HH (d, a, b, c, x[12], S32, 0xe6db99e5); /* 46 */
 HH (c, d, a, b, x[15], S33, 0x1fa27cf8); /* 47 */
 HH (b, c, d, a, x[ 2], S34, 0xc4ac5665); /* 48 */

 /* Round 4 */
 II (a, b, c, d, x[ 0], S41, 0xf4292244); /* 49 */
 II (d, a, b, c, x[ 7], S42, 0x432aff97); /* 50 */
 II (c, d, a, b, x[14], S43, 0xab9423a7); /* 51 */
 II (b, c, d, a, x[ 5], S44, 0xfc93a039); /* 52 */
 II (a, b, c, d, x[12], S41, 0x655b59c3); /* 53 */
 II (d, a, b, c, x[ 3], S42, 0x8f0ccc92); /* 54 */
 II (c, d, a, b, x[10], S43, 0xffeff47d); /* 55 */
 II (b, c, d, a, x[ 1], S44, 0x85845dd1); /* 56 */
 II (a, b, c, d, x[ 8], S41, 0x6fa87e4f); /* 57 */
 II (d, a, b, c, x[15], S42, 0xfe2ce6e0); /* 58 */
 II (c, d, a, b, x[ 6], S43, 0xa3014314); /* 59 */
 II (b, c, d, a, x[13], S44, 0x4e0811a1); /* 60 */
 II (a, b, c, d, x[ 4], S41, 0xf7537e82); /* 61 */
 II (d, a, b, c, x[11], S42, 0xbd3af235); /* 62 */
 II (c, d, a, b, x[ 2], S43, 0x2ad7d2bb); /* 63 */
 II (b, c, d, a, x[ 9], S44, 0xeb86d391); /* 64 */

 a += d0;
 b += d1;
 c += d2;
 d += d3;

 digest[0] = a;
 digest[1] = b;
 digest[2] = c;
 digest[3] = d;
}

int main(void) {
 uint32_t message[16], digest[4];

 message[0] = 0x61800000;
 message[1] = 0x00000000;
 message[2] = 0x00000000;
 message[3] = 0x00000000;
 message[4] = 0x00000000;
 message[5] = 0x00000000;
 message[6] = 0x00000000;
 message[7] = 0x00000000;
 message[8] = 0x00000000;
 message[9] = 0x00000000;
 message[10] = 0x00000000;
 message[11] = 0x00000000;
 message[12] = 0x00000000;
 message[13] = 0x00000000;
 message[14] = 0x08000000;
 message[15] = 0x00000000;

 digest[0] = 0x00000000;
 digest[1] = 0x00000000;
 digest[2] = 0x00000000;
 digest[3] = 0x00000000;

 MD5_hash(message, digest);

 printf("%08X %08X %08X %08X\n", digest[0], digest[1], digest[2], digest[3]);
}

以上代码是我的实现。现在，我的问题是，我如何填写某条消息？例如，如果我的消息是'a'，则消息为：0x61800000 ... 08000000000000。它是否正确？（message [0] = 0x61800000 ... message [14] = 0x08000000，message [15] = 0x000000000）。我想我的字节顺序或我对填充指令的解释可能是错的。有谁能请赐教我？

（上述代码的输出为：5058CD0E 2476E559 CF86AEA4 8A173599）;

Answer 1

我做了自己的实施，在实施过程中遇到了同样的墙/障碍。

A + Padding是0x61800000它是正确的，但用于长度指示的64位是big-endianness格式。因此，在这64位中，最后4个字节为零，前面的4个字节包含消息字节的长度* 8位，即最高有效位之前的最低有效位。

您从MD5哈希的RFC标准中复制的代码的另一个问题。 MD5Transform函数从char数组格式（从左到右）复制数据，并要求它们以（小端）格式的32位整数打包。执行计算周期后，您需要将（little-endian）32位整数的结果转换为char数组格式，要求您反转在MD5Transform初始调用期间完成的编码步骤。

希望这有帮助。

Answer 2

我最近在实施MD5方面遇到了类似的麻烦。如果您使用的是Windows或其他使用little-endian的操作系统，请尝试以下操作：

message[0] = 0x00008061;
message[1] = 0x00000000;
message[2] = 0x00000000;
message[3] = 0x00000000;
message[4] = 0x00000000;
message[5] = 0x00000000;
message[6] = 0x00000000;
message[7] = 0x00000000;
message[8] = 0x00000000;
message[9] = 0x00000000;
message[10] = 0x00000000;
message[11] = 0x00000000;
message[12] = 0x00000000;
message[13] = 0x00000000;
message[14] = 0x00000008;
message[15] = 0x00000000;

这可能不是你唯一的问题（如果确实如此）。我尝试使用正确的MD5算法在我的小端操作系统上使用你的big-endian paddings（已经确认使用little-endian padding提供正确的输出），它给了我以下输出：0ecd585059e57624a4ae86cf9935178a这可能意味着你只需再次查看您的算法。

MD5 Padding和Little Endians

2 个答案: