Java C++实现相同MD5加密算法的方式
作者：冰 河 发布时间：2023-10-12 12:26:02
标签：Java,C++,MD5加密算法
Java与C++实现相同的MD5加密算法
1、Java版
package com.lyz.utils.common;
import java.io.UnsupportedEncodingException;
import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;
/**
* MD5加密
* @author liuyazhuang
*/
public class MD5Hash {
public static String md5Java(String message) {
String digest = null;
try {
MessageDigest md = MessageDigest.getInstance("MD5");
byte[] hash = md.digest(message.getBytes("UTF-8"));
//converting byte array to Hexadecimal String
StringBuilder sb = new StringBuilder(2 * hash.length);
for (byte b : hash) {
sb.append(String.format("％02x", b & 0xff));
}
digest = sb.toString();
} catch (UnsupportedEncodingException ex) {
//Logger.getLogger(StringReplace.class.getName()).log(Level.SEVERE, null, ex);
} catch (NoSuchAlgorithmException ex) {
//Logger.getLogger(StringReplace.class.getName()).log(Level.SEVERE, null, ex);
}
return digest;
}
public static void main(String[] args) {
System.out.println(md5Java("admin").toUpperCase());
}
}
2、C++代码
(1)md5.h
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>
void MD5Digest(char *pszInput, unsigned long nInputSize, char *pszOutPut);
(2)md5.cpp
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>
#include "md5.h"
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);
FF (d, a, b, c, x[ 1], S12, 0xe8c7b756);
FF (c, d, a, b, x[ 2], S13, 0x242070db);
FF (b, c, d, a, x[ 3], S14, 0xc1bdceee);
FF (a, b, c, d, x[ 4], S11, 0xf57c0faf);
FF (d, a, b, c, x[ 5], S12, 0x4787c62a);
FF (c, d, a, b, x[ 6], S13, 0xa8304613);
FF (b, c, d, a, x[ 7], S14, 0xfd469501);
FF (a, b, c, d, x[ 8], S11, 0x698098d8);
FF (d, a, b, c, x[ 9], S12, 0x8b44f7af);
FF (c, d, a, b, x[10], S13, 0xffff5bb1);
FF (b, c, d, a, x[11], S14, 0x895cd7be);
FF (a, b, c, d, x[12], S11, 0x6b901122);
FF (d, a, b, c, x[13], S12, 0xfd987193);
FF (c, d, a, b, x[14], S13, 0xa679438e);
FF (b, c, d, a, x[15], S14, 0x49b40821);
GG (a, b, c, d, x[ 1], S21, 0xf61e2562);
GG (d, a, b, c, x[ 6], S22, 0xc040b340);
GG (c, d, a, b, x[11], S23, 0x265e5a51);
GG (b, c, d, a, x[ 0], S24, 0xe9b6c7aa);
GG (a, b, c, d, x[ 5], S21, 0xd62f105d);
GG (d, a, b, c, x[10], S22, 0x2441453);
GG (c, d, a, b, x[15], S23, 0xd8a1e681);
GG (b, c, d, a, x[ 4], S24, 0xe7d3fbc8);
GG (a, b, c, d, x[ 9], S21, 0x21e1cde6);
GG (d, a, b, c, x[14], S22, 0xc33707d6);
GG (c, d, a, b, x[ 3], S23, 0xf4d50d87);
GG (b, c, d, a, x[ 8], S24, 0x455a14ed);
GG (a, b, c, d, x[13], S21, 0xa9e3e905);
GG (d, a, b, c, x[ 2], S22, 0xfcefa3f8);
GG (c, d, a, b, x[ 7], S23, 0x676f02d9);
GG (b, c, d, a, x[12], S24, 0x8d2a4c8a);
HH (a, b, c, d, x[ 5], S31, 0xfffa3942);
HH (d, a, b, c, x[ 8], S32, 0x8771f681);
HH (c, d, a, b, x[11], S33, 0x6d9d6122);
HH (b, c, d, a, x[14], S34, 0xfde5380c);
HH (a, b, c, d, x[ 1], S31, 0xa4beea44);
HH (d, a, b, c, x[ 4], S32, 0x4bdecfa9);
HH (c, d, a, b, x[ 7], S33, 0xf6bb4b60);
HH (b, c, d, a, x[10], S34, 0xbebfbc70);
HH (a, b, c, d, x[13], S31, 0x289b7ec6);
HH (d, a, b, c, x[ 0], S32, 0xeaa127fa);
HH (c, d, a, b, x[ 3], S33, 0xd4ef3085);
HH (b, c, d, a, x[ 6], S34, 0x4881d05);
HH (a, b, c, d, x[ 9], S31, 0xd9d4d039);
HH (d, a, b, c, x[12], S32, 0xe6db99e5);
HH (c, d, a, b, x[15], S33, 0x1fa27cf8);
HH (b, c, d, a, x[ 2], S34, 0xc4ac5665);
II (a, b, c, d, x[ 0], S41, 0xf4292244);
II (d, a, b, c, x[ 7], S42, 0x432aff97);
II (c, d, a, b, x[14], S43, 0xab9423a7);
II (b, c, d, a, x[ 5], S44, 0xfc93a039);
II (a, b, c, d, x[12], S41, 0x655b59c3);
II (d, a, b, c, x[ 3], S42, 0x8f0ccc92);
II (c, d, a, b, x[10], S43, 0xffeff47d);
II (b, c, d, a, x[ 1], S44, 0x85845dd1);
II (a, b, c, d, x[ 8], S41, 0x6fa87e4f);
II (d, a, b, c, x[15], S42, 0xfe2ce6e0);
II (c, d, a, b, x[ 6], S43, 0xa3014314);
II (b, c, d, a, x[13], S44, 0x4e0811a1);
II (a, b, c, d, x[ 4], S41, 0xf7537e82);
II (d, a, b, c, x[11], S42, 0xbd3af235);
II (c, d, a, b, x[ 2], S43, 0x2ad7d2bb);
II (b, c, d, a, x[ 9], S44, 0xeb86d391);
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();
}
3、运行效果
这里我们都以输入123456为例
(1)java输出结果如下:
(2)C++输出结果如下:
来源：https://blog.csdn.net/l1028386804/article/details/47025267
0
投稿猜你喜欢
一、项目要求实现一个通讯录通讯录可以用来存储100个人的信息，每个人的信息包括：姓名、性别、年龄、电话、住址提供方法：添加联系人信息删除指定- 官方文档：https://developers.weixin.qq.com/miniprogram/dev/framework/open-a
- 一、synchronized 有不足新事物的出现要不是替代老事物，要么就是对老事物的补充JUC 的 locks 就是对 synchroniz
- 使用限制JDBC未支持列表Sharding-JDBC暂时未支持不常用的JDBC方法。DataSource接口不支持timeout相关操作Co
- 前言近期一直在忙项目，我也是打工仔。不多说，我们开始玩一玩seata。什么都不说，我们按照惯例，先上一个图（图里不规范的使用请忽略）：简单一
- 前言Android提供了很多种保存应用程序数据的方法。其中一种就是用SharedPreferences对象来保存我们私有的键值（key-va
- 本文实例为大家分享了android TextView跑马灯效果的具体代码，供大家参考，具体内容如下一、要点设置四个属性android:sin
- strcpy函数详解如下1.函数介绍1.1.函数接口char * __cdecl strcpy(char * dst, const char
- 一、先看下项目结构CodeGenerator：生成器主类resources下的mapper.java.vm：一个模板类，用以在生成dao层时
- 本文实例讲述了Android基于SoftReference缓存图片的方法。分享给大家供大家参考，具体如下：Java中的SoftReferen
- 本文实例为大家分享了android自定义环形对比图的具体代码，供大家参考，具体内容如下1.首先在res/values里创建一个attr.xm
- 前言目前Flutter三大主流状态管理框架分别是provider、flutter_bloc、getx，三大状态管理框架各有优劣，本篇文章将介
- 如果有哪一个做程序员的小伙伴说自己没有遇到中文乱码问题，我是不愿意相信的。今天在做微信订阅号的智能回复时，又一时迷乱的跳进了中文乱码这个火坑
- 1、普通用户与系统管理员用户的权限要有严格的区分。如果一个普通用户在使用查询语句中嵌入另一个Drop Table语句，那么是否允许
- 1 前言任何一门语言都需要基本的流程控制语句，其思想也符合人类判断问题或做事的逻辑过程。什么是流程控制呢？流程就是做一件事情的顺序，或者说是
- 开发 Web 应用的思路实现一个简单的 JSP/Servlet。搭建创建 Web 应用工程的环境。创建 Web 应用工程。Web 应用工程的
- 仅供学习交流，禁止商业用途。如侵害利益，联系必删！前言最近一位小伙伴钟爱二次元文化，于是找到半次元这个app，但是很快他就遇到了问题。一、案
- mapper文件使用in("str1","str2")mybatis的xxxMapper.xml文件
- java 中遍历取值异常（Hashtable Enumerator）解决办法用迭代器取值时抛出的异常：java.util.NoSuchEle
- 可以静态绑定数据源，这样就自动为DataGridView控件添加 相应的行。假如需要动态为DataGridView控件添加新行，方法有很多种
