请教MD5算法 用C语言实现

Posted 2023-05-08

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#if defined(__APPLE__)
#  define COMMON_DIGEST_FOR_OPENSSL
#  include <CommonCrypto/CommonDigest.h>
#  define SHA1 CC_SHA1
#else
#  include <openssl/md5.h>
#endif

// 这是我自己写的函数，用于计算MD5
// 参数 str：要转换的字符串
// 参数 lengthL: 字符串的长度 可以用 strlen(str) 直接获取参数str的长度
// 返回值：MD5字符串
char *str2md5(const char * str, int length) 
    int n;
    MD5_CTX c;
    unsigned char digest[16];
    char *out = (char*)malloc(33);

    MD5_Init(&c);

    while (length > 0) 
        if (length > 512) 
            MD5_Update(&c, str, 512);
         else 
            MD5_Update(&c, str, length);
        
        length -= 512;
        str += 512;
    

    MD5_Final(digest, &c);

    for (n = 0; n < 16; ++n) 
        snprintf(&(out[n*2]), 16*2, "%02x", (unsigned int)digest[n]);
    

    return out;


    int main(int argc, char **argv) 
        char *output = str2md5("hello", strlen("hello"));
        
        printf("%s\\n", output);
        // 上面会输出 hello的MD5字符串：
        // 5d41402abc4b2a76b9719d911017c592
        free(output);
        return 0;
    

参考技术A 主要就是调用库函数
MD5加密说到底也是函数计算
没有什么思路的问题，了解md5的发明算法，这应该是一个数学问题

我这里有代码，你可以参考一下

#include
#include
#include
#include

typedef unsigned char *POINTER;
typedef unsigned short int UINT2;
typedef unsigned long int UINT4;

typedef struct

UINT4 state[4];
UINT4 count[2];
unsigned char buffer[64];
MD5_CTX;

void MD5Init(MD5_CTX *);
void MD5Update(MD5_CTX *, unsigned char *, unsigned int);
void MD5Final(unsigned char [16], MD5_CTX *);

#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

static unsigned char PADDING[64] =
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
;

#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) + (UINT4)(ac); (a) = ROTATE_LEFT ((a), (s)); (a) += (b);
#define GG(a, b, c, d, x, s, ac) (a) += G ((b), (c), (d)) + (x) + (UINT4)(ac); (a) = ROTATE_LEFT ((a), (s)); (a) += (b);
#define HH(a, b, c, d, x, s, ac) (a) += H ((b), (c), (d)) + (x) + (UINT4)(ac); (a) = ROTATE_LEFT ((a), (s)); (a) += (b);
#define II(a, b, c, d, x, s, ac) (a) += I ((b), (c), (d)) + (x) + (UINT4)(ac); (a) = ROTATE_LEFT ((a), (s)); (a) += (b);

inline void Encode(unsigned char *output, UINT4 *input, unsigned int len)

unsigned int i, j;

for (i = 0, j = 0; j < len; i++, j += 4)
output[j] = (unsigned char)(input[i] & 0xff);
output[j+1] = (unsigned char)((input[i] >> 8) & 0xff);
output[j+2] = (unsigned char)((input[i] >> 16) & 0xff);
output[j+3] = (unsigned char)((input[i] >> 24) & 0xff);



inline void Decode(UINT4 *output, unsigned char *input, unsigned int len)

unsigned int i, j;

for (i = 0, j = 0; j < len; i++, j += 4)
output[i] = ((UINT4)input[j]) | (((UINT4)input[j+1]) << 8) |
(((UINT4)input[j+2]) << 16) | (((UINT4)input[j+3]) << 24);


inline void MD5Transform (UINT4 state[4], unsigned char block[64])

UINT4 a = state[0], b = state[1], c = state[2], d = state[3], x[16];
Decode (x, block, 64);
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 */
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, 0x2441453); /* 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 */
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, 0x4881d05); /* 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 */
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 */
state[0] += a;
state[1] += b;
state[2] += c;
state[3] += d;
memset ((POINTER)x, 0, sizeof (x));


inline void MD5Init(MD5_CTX *context)

context->count[0] = context->count[1] = 0;
context->state[0] = 0x67452301;
context->state[1] = 0xefcdab89;
context->state[2] = 0x98badcfe;
context->state[3] = 0x10325476;


inline void MD5Update(MD5_CTX *context, unsigned char *input, unsigned int inputLen)

unsigned int i, index, partLen;

index = (unsigned int)((context->count[0] >> 3) & 0x3F);
if ((context->count[0] += ((UINT4)inputLen << 3))
< ((UINT4)inputLen << 3))
context->count[1]++;
context->count[1] += ((UINT4)inputLen >> 29);

partLen = 64 - index;

if (inputLen >= partLen)
memcpy((POINTER)&context->buffer[index], (POINTER)input, partLen);
MD5Transform(context->state, context->buffer);

for (i = partLen; i + 63 < inputLen; i += 64)
MD5Transform (context->state, &input[i]);
index = 0;

else
i = 0;

memcpy((POINTER)&context->buffer[index], (POINTER)&input[i], inputLen-i);


inline void MD5Final(unsigned char digest[16], MD5_CTX *context)

unsigned char bits[8];
unsigned int index, padLen;

Encode (bits, context->count, 8);
index = (unsigned int)((context->count[0] >> 3) & 0x3f);
padLen = (index < 56) ? (56 - index) : (120 - index);
MD5Update (context, PADDING, padLen);
MD5Update (context, bits, 8);
Encode (digest, context->state, 16);
memset ((POINTER)context, 0, sizeof (*context));


void MD5Digest(char *pszInput, unsigned long nInputSize, char *pszOutPut)

MD5_CTX context;
unsigned int len = strlen (pszInput);

MD5Init (&context);
MD5Update (&context, (unsigned char *)pszInput, len);
MD5Final ((unsigned char *)pszOutPut, &context);


main()
char szDigest[16];
char encrypt[200];
printf("请输入要计算MD5值的字符串:");
gets(encrypt);
printf("\n加密结果:");
MD5Digest(encrypt,strlen(encrypt),szDigest);
int i;
for (i=0;i<16;i++) printf ("%02X",(unsigned char)szDigest[i]);
getchar();
参考技术B 请直接百度吧，C语言 MD5

请教一个 C语言 字符串数组之间比较的算法，谢谢

我把10个ip地址，赋值给了10个字符串数组，现在想比较它们之间的异同，相同的删除，重重复的优先级+1，想请您帮助的是，者10个字符串数组应该用什么办法进行比较呢，如果一一比较的话得40多次，太麻烦，请问有好的算法吗，谢谢

这种时候当然是使用标准容器拉
std::map可以满足你的需要

10个ip 地址 复制给10个std::string. 然后构造一个 std::map<std::string, int> 再逐个使用insert方法插入， 如果插入成功（通过检查insert的返回值, 具体请搜索msdn,这里篇幅有限。）

如果插入成功， 继续； 不成功，就表示有重复，将返回的那个已经存在的ip对应的优先级++， 再继续。

map的特点就是不重复，你省去了自己去写比较，去优化的繁琐，而且一般stl实现的效率都是很高的绝对不是你这种40多次的O(N)的，应该至少都是o(ln N) 参考技术A 定义一个二维数组char str[10][20];
然后把str[0]~str[9]看成10个字符串数组名就可以用循环了,再用一个数组int a[10]来记录优先级（a[]不要忘了初始化）
如：
n=9//用来保存还剩几个字符串
for(i=0;i<=n;i++)

for(j=i+1;j<=n;j++)//为什么这么循环就不用说了吧

if(strcmp(str[i],str[j]==0))

for(k=j+1;k<=n;k++)

strcpy(str[k-1],str[k]);
a[k-1]=a[k];

n--;
a[i]++;
//整个if就是从与前面相同的那个开始整体往前移一位达到删除的效果当然a[]也要移来保持一一对应，然后a[i]自加一表示优第i个的优先级加一


输出时str[0][20]~str[n][20]就是剩下的ip地址a[0]~a[n]就是ip地址所对应的优先级
代码没编译过，可能有些小问题，主要是说说方法