[英]Implementation of MD5 in C
我想在C項目中實現MD5哈希函數,而且我想自己做,因為我擔心的一件事是使用別人的代碼(主要是因為我在理解這樣的代碼時遇到麻煩)。 因此,我直接進入Wiki頁面以獲取偽代碼: http : //en.wikipedia.org/wiki/MD5我決定離開填充並分成512位的塊,以供日后使用,然后從MD5散列開始空字符串。 正確填充后,(我認為)它應如下所示:
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
這是我在C語言中重現Wiki的主循環(僅處理一個512位塊的循環)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函數,也從Wikipedia上的偽代碼進行了修改:
unsigned int leftRotate(unsigned int x, int n){
return ((x) << n) | ((x) >> (32 - n));
}
輸出以下內容:
a0=578518856, b0=3524428790, c0=1003076545, d0=1531243034
將這些十進制值轉換為十六進制,我得到以下信息:
a0 = 578518856 -> 227B7F48
b0 = 3524428790 -> D21283F6
c0 = 1003076545 -> 3BC9BBC1
d0 = 1531243034 -> 5B44EA1A
甚至與空字符串的實際MD5摘要(即d41d8cd98f00b204e9800998ecf8427e)有點相似。 所以,我的問題是,我要去哪里錯了?
可能是目標體系結構的問題(字節序,類型大小等)
我的答案來自http://bytes.com/topic/c/answers/855645-simple-md5-hash-different-output-different-os的位
編輯:
問題是因為考慮到內存中的小端存儲,您添加的填充位不在正確的字節上。 此外,您必須打印結果md5哈希,因為它存儲在內存中(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,為時已晚:/
#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;
}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.