Question

使用md5怎么给一串字符串加密

Answer 1

#include <stdio.h> #include <stdlib.h> #define ARR_LEN 100 #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;\ } typedef struct { unsigned int count[2]; unsigned int state[4]; unsigned char buffer[64]; } MD5_CTX; 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); 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 */ /* Round 2 */ GG(a, b, c, d, x[ 1], 5, 0xf61e2562); /* 17 */ GG(d, a, b, c, x[ 6], 9, 0xc040b340); /* 18 */ GG(c, d, a, b, x[11], 14, 0x265e5a51); /* 19 */ GG(b, c, d, a, x[ 0], 20, 0xe9b6c7aa); /* 20 */ GG(a, b, c, d, x[ 5], 5, 0xd62f105d); /* 21 */ GG(d, a, b, c, x[10], 9, 0x2441453); /* 22 */ GG(c, d, a, b, x[15], 14, 0xd8a1e681); /* 23 */ GG(b, c, d, a, x[ 4], 20, 0xe7d3fbc8); /* 24 */ GG(a, b, c, d, x[ 9], 5, 0x21e1cde6); /* 25 */ GG(d, a, b, c, x[14], 9, 0xc33707d6); /* 26 */ GG(c, d, a, b, x[ 3], 14, 0xf4d50d87); /* 27 */ GG(b, c, d, a, x[ 8], 20, 0x455a14ed); /* 28 */ GG(a, b, c, d, x[13], 5, 0xa9e3e905); /* 29 */ GG(d, a, b, c, x[ 2], 9, 0xfcefa3f8); /* 30 */ GG(c, d, a, b, x[ 7], 14, 0x676f02d9); /* 31 */ GG(b, c, d, a, x[12], 20, 0x8d2a4c8a); /* 32 */ /* Round 3 */ HH(a, b, c, d, x[ 5], 4, 0xfffa3942); /* 33 */ HH(d, a, b, c, x[ 8], 11, 0x8771f681); /* 34 */ HH(c, d, a, b, x[11], 16, 0x6d9d6122); /* 35 */ HH(b, c, d, a, x[14], 23, 0xfde5380c); /* 36 */ HH(a, b, c, d, x[ 1], 4, 0xa4beea44); /* 37 */ HH(d, a, b, c, x[ 4], 11, 0x4bdecfa9); /* 38 */ HH(c, d, a, b, x[ 7], 16, 0xf6bb4b60); /* 39 */ HH(b, c, d, a, x[10], 23, 0xbebfbc70); /* 40 */ HH(a, b, c, d, x[13], 4, 0x289b7ec6); /* 41 */ HH(d, a, b, c, x[ 0], 11, 0xeaa127fa); /* 42 */ HH(c, d, a, b, x[ 3], 16, 0xd4ef3085); /* 43 */ HH(b, c, d, a, x[ 6], 23, 0x4881d05); /* 44 */ HH(a, b, c, d, x[ 9], 4, 0xd9d4d039); /* 45 */ HH(d, a, b, c, x[12], 11, 0xe6db99e5); /* 46 */ HH(c, d, a, b, x[15], 16, 0x1fa27cf8); /* 47 */ HH(b, c, d, a, x[ 2], 23, 0xc4ac5665); /* 48 */ /* Round 4 */ II(a, b, c, d, x[ 0], 6, 0xf4292244); /* 49 */ II(d, a, b, c, x[ 7], 10, 0x432aff97); /* 50 */ II(c, d, a, b, x[14], 15, 0xab9423a7); /* 51 */ II(b, c, d, a, x[ 5], 21, 0xfc93a039); /* 52 */ II(a, b, c, d, x[12], 6, 0x655b59c3); /* 53 */ II(d, a, b, c, x[ 3], 10, 0x8f0ccc92); /* 54 */ II(c, d, a, b, x[10], 15, 0xffeff47d); /* 55 */ II(b, c, d, a, x[ 1], 21, 0x85845dd1); /* 56 */ II(a, b, c, d, x[ 8], 6, 0x6fa87e4f); /* 57 */ II(d, a, b, c, x[15], 10, 0xfe2ce6e0); /* 58 */ II(c, d, a, b, x[ 6], 15, 0xa3014314); /* 59 */ II(b, c, d, a, x[13], 21, 0x4e0811a1); /* 60 */ II(a, b, c, d, x[ 4], 6, 0xf7537e82); /* 61 */ II(d, a, b, c, x[11], 10, 0xbd3af235); /* 62 */ II(c, d, a, b, x[ 2], 15, 0x2ad7d2bb); /* 63 */ II(b, c, d, a, x[ 9], 21, 0xeb86d391); /* 64 */ state[0] += a; state[1] += b; state[2] += c; state[3] += d; } int main(int argc, char *argv[]) { int i; unsigned char encrypt[ARR_LEN]; unsigned char decrypt[16]; MD5_CTX md5; printf("【测试1】\n"); printf("加密前：Baidu\n"); printf("加密后：d36e219e608d6f9aa15b1e42081689fa\n"); printf("\n"); printf("【测试2】\n"); printf("加密前：Bihaifeng\n"); printf("加密后：48ef81c0058e136a7a210a3c99e48429\n"); printf("\n"); printf ("======================================================\n\n"); printf("请输入需MD5加密的字符串：\n"); gets(encrypt); printf("\n"); printf ("======================================================\n\n"); MD5Init(&md5); MD5Update(&md5,encrypt,strlen((char *)encrypt)); MD5Final(&md5,decrypt); printf("加密前：%s\n加密后：",encrypt); for(i=0;i<16;i++) { printf("%02x",decrypt[i]); } printf("\n"); getch(); return 0; } 运行结果 以上源代码非原创~~

Answer 2

md5 不是加密算法，是单向的，就是说经过这个算法计算得到的结果不可能反推回原始值