在C中实现MD5 [英] Implementation of MD5 in C
本文介绍了在C中实现MD5的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我想在C项目中实现MD5哈希函数，并且想自己做，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，以供以后使用，而只是从一个空字符串的MD5哈希开始.正确填充后，(我认为)它应如下所示:
I want to implement an MD5 hashing function in a C project, and I want to make my own, because one thing I am afraid of is using someone else's code (mainly because I have trouble understanding such codes). So, I headed straight to the Wiki page for the pseudocode: http://en.wikipedia.org/wiki/MD5 I decided to leave the padding and breaking into 512 bit chunks stuff for later and just start with an MD5 hash of an empty string. Properly padded, it should (I think) look like this:
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语言中复制Wiki的主循环(仅处理一个512位块的循环)的方式:
And here is how I reproduced Wiki's main loop (the one that deals with just one 512-bit chunk) in C:
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:
The leftRotate function, also modified from the pseudocode on Wikipedia:
unsigned int leftRotate(unsigned int x, int n){
return ((x) << n) | ((x) >> (32 - n));
}
这将输出以下内容:
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:
Converting these decimal values to hex, I get the following:
a0 = 578518856 -> 227B7F48
b0 = 3524428790 -> D21283F6
c0 = 1003076545 -> 3BC9BBC1
d0 = 1531243034 -> 5B44EA1A
这甚至与空字符串的实际MD5摘要有点不同(它是d41d8cd98f00b204e9800998ecf8427e).所以，我的问题是，我要去哪里错了?
Which isn't even a bit similar to the actual MD5 digest of an empty string (which is d41d8cd98f00b204e9800998ecf8427e). So, my question is, where am I going wrong?
推荐答案
#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;
}
这篇关于在C中实现MD5的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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