openssl 基本算法小例

Posted 2022-03-18 qianbo_insist

openssl

openssl支持各种算法,在网络传输中我们需要用到各类算法，尤其是在安全协议里面，hmac算法是必须要用得，dtls协议里面也会用到各类对称算法，下面定义一个类，使用openssl得常用算法中得基本算法。

#pragma once

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
class c_hmac


	string v_key;
	string v_data;
	string v_name;

public:
	c_hmac(string alo_name,string key, string data)
	
		v_key = key;
		v_data = data;
		v_name = alo_name;
	
	uint8_t * hmac_algo(unsigned int &len);


;

实现

实现比较简单，版本使用openssl 1.1.1 版

#include "hmac.h"
#include <openssl/hmac.h>
#include <string.h>
#include <iostream>

#pragma comment(lib,"libcrypto.lib")
using namespace std;
#define strcasecmp _stricmp
uint8_t * c_hmac::hmac_algo(unsigned int &len) 
	const EVP_MD * engine = NULL;
	const char *name = v_name.c_str();
	if (strcasecmp("sha512", name) == 0) 
		engine = EVP_sha512();
	
	else if (strcasecmp("sha256", name) == 0) 
		engine = EVP_sha256();
	
	else if (strcasecmp("sha1", name) == 0) 
		engine = EVP_sha1();
	
	else if (strcasecmp("md5", name) == 0) 
		engine = EVP_md5();
	
	else if (strcasecmp("sha224", name) == 0) 
		engine = EVP_sha224();
	
	else if (strcasecmp("sha384", name) == 0) 
		engine = EVP_sha384();
	
	else 
		cout << "Algorithm " << name << " is not supported now!" << endl;
		return NULL;
	

	uint8_t *output = (uint8_t *)malloc(EVP_MAX_MD_SIZE);
	
	HMAC_CTX *hctx = HMAC_CTX_new();
	HMAC_CTX_reset(hctx);
	HMAC_Init_ex(hctx, v_key.c_str(), v_key.size(), engine, NULL);
	HMAC_Update(hctx, (uint8_t *)v_data.c_str(), v_data.size());        // input is OK; &input is WRONG !!!

	HMAC_Final(hctx, output, &len);
	//HMAC_CTX_cleanup(&ctx);
	HMAC_CTX_free(hctx);

	return output;

调用

调用分别是密码方法名称，我们使用sha256，关键密字和需要加密得数据，调用如下

#include "hmac.h"
using namespace std;

int main(int argc, char * argv[])



	c_hmac hmac("sha256","qianbo", "this is a test");
	unsigned int len = 0;
	uint8_t *ret = hmac.hmac_algo(len);

	if (NULL == ret) 
		cout << "Algorithm fail!" << endl;
	

	for (unsigned int i = 0; i < len; i++) 
		printf("%02x-", (unsigned int)ret[i]);
	
	cout << endl;

	if (ret) 
		free(ret);
	
	return 0;

结果

计算得结果如图所示，结果为32字节