C语言求文件MD5的函数用法?

Question

byte[] data = new byte[DATA_SIZE]; MD5 md5 = new MD5CryptoServiceProvider(); byte[] result = md5.ComputeHash(data); 这个函数怎么用？。。

Answer 1

#ifndef MD5_H
#define MD5_H

typedef struct
{
unsigned int count[2];
unsigned int state[4];
unsigned char buffer[64];
}MD5_CTX;

#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))
#define ROTATE_LEFT(x,n) ((x << n) | (x >> (32-n)))
#define FF(a,b,c,d,x,s,ac) \
{ \
a += F(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
#define GG(a,b,c,d,x,s,ac) \
{ \
a += G(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
#define HH(a,b,c,d,x,s,ac) \
{ \
a += H(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
#define II(a,b,c,d,x,s,ac) \
{ \
a += I(b,c,d) + x + ac; \
a = ROTATE_LEFT(a,s); \
a += b; \
}
void MD5Init(MD5_CTX *context);
void MD5Update(MD5_CTX *context,unsigned char *input,unsigned int inputlen);
void MD5Final(MD5_CTX *context,unsigned char digest[16]);
void MD5Transform(unsigned int state[4],unsigned char block[64]);
void MD5Encode(unsigned char *output,unsigned int *input,unsigned int len);
void MD5Decode(unsigned int *output,unsigned char *input,unsigned int len);

#endif

源文件md5.c

#include <memory.h>
#include "md5.h"

unsigned char PADDING[]={0x80,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

void MD5Init(MD5_CTX *context)
{
context->count[0] = 0;
context->count[1] = 0;
context->state[0] = 0x67452301;
context->state[1] = 0xEFCDAB89;
context->state[2] = 0x98BADCFE;
context->state[3] = 0x10325476;
}
void MD5Update(MD5_CTX *context,unsigned char *input,unsigned int inputlen)
{
unsigned int i = 0,index = 0,partlen = 0;
index = (context->count[0] >> 3) & 0x3F;
partlen = 64 - index;
context->count[0] += inputlen << 3;
if(context->count[0] < (inputlen << 3))
context->count[1]++;
context->count[1] += inputlen >> 29;

if(inputlen >= partlen)
{
memcpy(&context->buffer[index],input,partlen);
MD5Transform(context->state,context->buffer);
for(i = partlen;i+64 <= inputlen;i+=64)
MD5Transform(context->state,&input[i]);
index = 0;
}
else
{
i = 0;
}
memcpy(&context->buffer[index],&input[i],inputlen-i);
}
void MD5Final(MD5_CTX *context,unsigned char digest[16])
{
unsigned int index = 0,padlen = 0;
unsigned char bits[8];
index = (context->count[0] >> 3) & 0x3F;
padlen = (index < 56)?(56-index):(120-index);
MD5Encode(bits,context->count,8);
MD5Update(context,PADDING,padlen);
MD5Update(context,bits,8);
MD5Encode(digest,context->state,16);
}
void MD5Encode(unsigned char *output,unsigned int *input,unsigned int len)
{
unsigned int i = 0,j = 0;
while(j < len)
{
output[j] = input[i] & 0xFF;
output[j+1] = (input[i] >> 8) & 0xFF;
output[j+2] = (input[i] >> 16) & 0xFF;
output[j+3] = (input[i] >> 24) & 0xFF;
i++;
j+=4;
}
}
void MD5Decode(unsigned int *output,unsigned char *input,unsigned int len)
{
unsigned int i = 0,j = 0;
while(j < len)
{
output[i] = (input[j]) |
(input[j+1] << 8) |
(input[j+2] << 16) |
(input[j+3] << 24);
i++;
j+=4;
}
}
void MD5Transform(unsigned int state[4],unsigned char block[64])
{
unsigned int a = state[0];
unsigned int b = state[1];
unsigned int c = state[2];
unsigned int d = state[3];
unsigned int x[64];
MD5Decode(x,block,64);
FF(a, b, c, d, x[ 0], 7, 0xd76aa478); /* 1 */
FF(d, a, b, c, x[ 1], 12, 0xe8c7b756); /* 2 */
FF(c, d, a, b, x[ 2], 17, 0x242070db); /* 3 */
FF(b, c, d, a, x[ 3], 22, 0xc1bdceee); /* 4 */
FF(a, b, c, d, x[ 4], 7, 0xf57c0faf); /* 5 */
FF(d, a, b, c, x[ 5], 12, 0x4787c62a); /* 6 */
FF(c, d, a, b, x[ 6], 17, 0xa8304613); /* 7 */
FF(b, c, d, a, x[ 7], 22, 0xfd469501); /* 8 */
FF(a, b, c, d, x[ 8], 7, 0x698098d8); /* 9 */
FF(d, a, b, c, x[ 9], 12, 0x8b44f7af); /* 10 */
FF(c, d, a, b, x[10], 17, 0xffff5bb1); /* 11 */
FF(b, c, d, a, x[11], 22, 0x895cd7be); /* 12 */
FF(a, b, c, d, x[12], 7, 0x6b901122); /* 13 */
FF(d, a, b, c, x[13], 12, 0xfd987193); /* 14 */
FF(c, d, a, b, x[14], 17, 0xa679438e); /* 15 */
FF(b, c, d, a, x[15], 22, 0x49b40821); /* 16 */

Answer 2

0x2000 0000在这里是16进制数 转换成十进制数是536870912bit = 67108864 BYTE = 65536 KB = 64MB
/* flen[1]=len/0x20000000; //flen单位是bit */
这行指令做的是 看那个被打开的文件的长度是64兆的多少倍 并且把计算出的结果（倍数）放到array flen[1]里（file length）并且存进去的是个整数 文件长度是个很长的很奇怪的数 除个536870912这么个怪数肯定不会是个整数（至少大多情况下）

所以有了下面这一行指令
/* flen[0]=(len%0x20000000)*8; */
这行指令把file length除64MB这个数的余数提出 并且乘了8（即除以64MB 余数* 8 让BYTE单位变成bit单位） 为程序的后面做运算做铺垫或者准备
并且存在了flen[0]里

这个程序不全 我想提问题的人应该有全的版本 我能给的帮助就这么多 希望能帮到你:)

Answer 3

冒号后面的叫初始化列表，你也可以在后面的大括号里写
weight=0，效果是一样的，就是你这个默认构造函数调用是，对象里的weight被初始化为0了，不过有些东西只能在初始化表里初始化，比如：引用什么的，具体你看《primer c++》里有详细介绍