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用C/C++实现Base64编码和解码-创新互联

2022.12.19日,看了一下base64原理,然后想用代码实现一下,改了好久的bug终于完美成功了

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目录

一.收获

①移位运算符优先级高于与或非

②map容器可以方便查找,但使用时要注意find(keyvalue),是否解引用了空迭代器等

③unsigned char类型移位运算可以不考虑符号位,但是形参使用const char*更有通用性,所以需要进行一个强转

二.代码实现

1.纯c语言版

2.c++版

3.效果图

①编码

②解码


一.收获 ①移位运算符优先级高于与或非

②map容器可以方便查找,但使用时要注意find(keyvalue),是否解引用了空迭代器等

③unsigned char类型移位运算可以不考虑符号位,但是形参使用const char*更有通用性,所以需要进行一个强转 二.代码实现 1.纯c语言版
#include#include#includeunsigned char* base64_encode(const char* str0)
{
	unsigned char* str = (unsigned char*)str0;	//转为unsigned char无符号,移位操作时可以防止错误
	unsigned char base64_map[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";//数组形式,方便修改
	long len;				//base64处理后的字符串长度
	long str_len;			//源字符串长度
	long flag;				//用于标识模3后的余数
	unsigned char* res;		//返回的字符串
	str_len = strlen((const char*)str);
	switch (str_len % 3)	//判断模3的余数
	{
	case 0:flag = 0; len = str_len / 3 * 4; break;
	case 1:flag = 1; len = (str_len / 3 + 1) * 4; break;
	case 2:flag = 2; len = (str_len / 3 + 1) * 4; break;
	}
	res = (unsigned char*)malloc(sizeof(unsigned char) * len + 1);
	for (int i = 0, j = 0; j< str_len - flag; j += 3, i += 4)//先处理整除部分
	{
		//注意&运算和位移运算的优先级,是先位移后与或非
		res[i] = base64_map[str[j] >>2];
		res[i + 1] = base64_map[(str[j] & 0x3)<< 4 | str[j + 1] >>4];
		res[i + 2] = base64_map[(str[j + 1] & 0xf)<< 2 | (str[j + 2] >>6)];
		res[i + 3] = base64_map[str[j + 2] & 0x3f];
	}
	//不满足被三整除时,要矫正
	switch (flag)
	{
	case 0:break;	//满足时直接退出
	case 1:res[len - 4] = base64_map[str[str_len - 1] >>2];	//只剩一个字符时,右移两位得到高六位
		res[len - 3] = base64_map[(str[str_len - 1] & 0x3)<< 4];//获得低二位再右移四位,自动补0
		res[len - 2] = res[len - 1] = '='; break;				//最后两个补=
	case 2:
		res[len - 4] = base64_map[str[str_len - 2] >>2];				//剩两个字符时,右移两位得高六位
		res[len - 3] = base64_map[(str[str_len - 2] & 0x3)<< 4 | str[str_len - 1] >>4];	//第一个字符低二位和第二个字符高四位
		res[len - 2] = base64_map[(str[str_len - 1] & 0xf)<< 2];	//第二个字符低四位,左移两位自动补0
		res[len - 1] = '=';											//最后一个补=
		break;
	}
	res[len] = '\0';	//补上字符串结束标识
	return res;
}
unsigned char findPos(const unsigned char* base64_map, unsigned char c)//查找下标所在位置
{
	for (int i = 0; i< strlen((const char*)base64_map); i++)
	{
		if (base64_map[i] == c)
			return i;
	}
}
unsigned char* base64_decode(const char* code0)
{
	unsigned char* code = (unsigned char*)code0;
	unsigned char base64_map[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
	long len, str_len, flag = 0;
	unsigned char* res;
	len = strlen((const char*)code);
	if (code[len - 1] == '=')
	{
		if (code[len - 2] == '=')
		{
			flag = 1;
			str_len = len / 4 * 3 - 2;
		}

		else
		{
			flag = 2;
			str_len = len / 4 * 3 - 1;
		}

	}
	else
		str_len = len / 4 * 3;
	res = (unsigned char*)malloc(sizeof(unsigned char) * str_len + 1);
	for (int i = 0, j = 0; j< str_len - flag; j += 3, i += 4)
	{
		unsigned char a[4];
		a[0] = findPos(base64_map, code[i]);		//code[]每一个字符对应base64表中的位置,用位置值反推原始数据值
		a[1] = findPos(base64_map, code[i + 1]);
		a[2] = findPos(base64_map, code[i + 2]);
		a[3] = findPos(base64_map, code[i + 3]);
		res[j] = a[0]<< 2 | a[1] >>4;		//取出第一个字符对应base64表的十进制数的前6位与第二个字符对应base64表的十进制数的后2位进行组合  
		res[j + 1] = a[1]<< 4 | a[2] >>2;	//取出第二个字符对应base64表的十进制数的后4位与第三个字符对应bas464表的十进制数的后4位进行组合  
		res[j + 2] = a[2]<< 6 | a[3];	   //取出第三个字符对应base64表的十进制数的后2位与第4个字符进行组合 
	}

 switch (flag)
	{
	case 0:break;
	case 1:
	{
		a[0] = findPos(base64_map,code[len - 4]);
		a[1] = findPos(base64_map,code[len - 3]);
		res[str_len - 1] = a[0]<< 2 | a[1] >>4;
		break;
	}
	case 2: {
		a[0] = findPos(base64_map,code[len - 4]);
		a[1] = findPos(base64_map,code[len - 3]);
		a[2] = findPos(base64_map,code[len - 2]);
		res[str_len - 2] = a[0]<< 2 | a[1] >>4;
		res[str_len - 1] = a[1]<< 4 | a[2] >>2;
		break;
	}
	}
	res[str_len] = '\0';
	return res;
}

int main()
{
    //测试数据 hello
    //aGVsbG8=
	//aGVsbG82
    //aGVsbG==
	char str[100];
	int flag;
	printf("请选择功能:\n");
	printf("1.base64加密\n");
	printf("2.base64解密\n");
	scanf("%d", &flag);
	printf("请输入字符串:\n");
	scanf("%s", str);
	if (flag == 1)
		printf("加密后的字符串是:%s", base64_encode(str));
	else
		printf("解密后的字符串是:%s", base64_decode(str));
	return 0;
}

2.c++版

这里使用到了map

#include#include#include#includeusing namespace std;

unsigned char* base64_encode(const char* str0)
{
	unsigned char* str = (unsigned char*)str0;	//转为unsigned char无符号,移位操作时可以防止错误
	unsigned char base64_map[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";//也可以用map,这里用数组其实更方便
	long len;				//base64处理后的字符串长度
	long str_len;			//源字符串长度
	long flag;				//用于标识模3后的余数
	unsigned char* res;		//返回的字符串
	str_len = strlen((const char*)str);
	switch (str_len % 3)	//判断模3的余数
	{
	case 0:flag = 0; len = str_len / 3 * 4; break;
	case 1:flag = 1; len = (str_len / 3 + 1) * 4; break;
	case 2:flag = 2; len = (str_len / 3 + 1) * 4; break;
	}
	res = (unsigned char*)malloc(sizeof(unsigned char) * len + 1);
	for (int i = 0, j = 0; j< str_len - flag; j += 3, i += 4)//先处理整除部分
	{
		//注意&运算和位移运算的优先级,是先位移后与或非,括号不对有可能导致错误
		res[i] = base64_map[str[j] >>2];
		res[i + 1] = base64_map[(str[j] & 0x3)<< 4 | str[j + 1] >>4];
		res[i + 2] = base64_map[(str[j + 1] & 0xf)<< 2 | (str[j + 2] >>6)];
		res[i + 3] = base64_map[str[j + 2] & 0x3f];
	}
	//不满足被三整除时,要矫正
	switch (flag)
	{
	case 0:break;	//满足时直接退出
	case 1:res[len - 4] = base64_map[str[str_len - 1] >>2];	//只剩一个字符时,右移两位得到高六位
		res[len - 3] = base64_map[(str[str_len - 1] & 0x3)<< 4];//获得低二位再右移四位,自动补0
		res[len - 2] = res[len - 1] = '='; break;				//最后两个补=
	case 2:
		res[len - 4] = base64_map[str[str_len - 2] >>2];				//剩两个字符时,右移两位得高六位
		res[len - 3] = base64_map[(str[str_len - 2] & 0x3)<< 4 | str[str_len - 1] >>4];	//第一个字符低二位和第二个字符高四位
		res[len - 2] = base64_map[(str[str_len - 1] & 0xf)<< 2];	//第二个字符低四位,左移两位自动补0
		res[len - 1] = '=';											//最后一个补=
		break;
	}
	res[len] = '\0';	//补上字符串结束标识
	return res;
}

unsigned char* base64_decode(const char* code0)
{
	unsigned char* code = (unsigned char*)code0;
	mapbase64_map = {//map类型base64表
		{'A',0},{'B',1},{'C',2},{'D',3},{'E',4},{'F',5},{'G',6},{'H',7},{'I',8},{'J',9},{'K',10},
		{'L',11},{'M',12},{'N',13},{'O',14},{'P',15},{'Q',16},{'R',17},{'S',18},{'T',19},{'U',20},
		{'V',21},{'W',22},{'X',23},{'Y',24},{'Z',25},{'a',26},{'b',27},{'c',28},{'d',29},{'e',30},
		{'f',31},{'g',32},{'h',33},{'i',34},{'j',35},{'k',36},{'l',37},{'m',38},{'n',39},{'o',40},
		{'p',41},{'q',42},{'r',43},{'s',44},{'t',45},{'u',46},{'v',47},{'w',48},{'x',49},{'y',50},
		{'z',51},{'0',52},{'1',53},{'2',54},{'3',55},{'4',56},{'5',57},{'6',58},{'7',59},{'8',60},
		{'9',61},{'+',62},{'/',63}
	};
	long len, str_len,flag=0;
	unsigned char* res;
	len = strlen((const char*)code);
	if (code[len - 1] == '=')
	{
		if (code[len - 2] == '=')//两个等号,余一个字符
		{
			flag = 1;
			str_len = len / 4 * 3 - 2;
		}
	
		else//一个等号,余两个字符
		{
			flag = 2;
			str_len = len / 4 * 3 - 1;
		}
			
	}
	else
		str_len = len / 4 * 3;
	res = (unsigned char*)malloc(sizeof(unsigned char) * str_len + 1);
	unsigned char a[4];
	for (int i = 0, j = 0; j< str_len - flag; j += 3, i += 4)
	{
		
		a[0] = base64_map[code[i]];		//code[]每一个字符对应base64表中的位置,用位置值反推原始数据值
		a[1] = base64_map[code[i+1]];
		a[2] = base64_map[code[i+2]];
		a[3] = base64_map[code[i+3]];
		res[j] = a[0]<< 2 | a[1] >>4;		//取出第一个字符对应base64表的十进制数的前6位与第二个字符对应base64表的十进制数的后2位进行组合  
		res[j + 1] = a[1]<< 4 | a[2] >>2;	//取出第二个字符对应base64表的十进制数的后4位与第三个字符对应bas464表的十进制数的后4位进行组合  
		res[j + 2] = a[2]<< 6 | a[3];	   //取出第三个字符对应base64表的十进制数的后2位与第4个字符进行组合  
	}
	switch (flag)
	{
	case 0:break;
	case 1:
	{
		a[0] = base64_map[code[len - 4]];
		a[1] = base64_map[code[len - 3]];
		res[str_len - 1] = a[0]<< 2 | a[1] >>4;
		break;
	}
	case 2: {
		
		a[0] = base64_map[code[len - 4]];
		a[1] = base64_map[code[len - 3]];
		a[2] = base64_map[code[len - 2]];
		res[str_len - 2] = a[0]<< 2 | a[1] >>4;
		res[str_len - 1] = a[1]<< 4 | a[2] >>2;
		break;
	}
	}
	res[str_len] = '\0';
	return res;
}

int main()
{
	//测试数据
	//aGVsbG8=
	//aGVsbG82
	//aGVsbG==
	char str[100];
	int flag;
	printf("请选择功能:\n");
	printf("1.base64加密\n");
	printf("2.base64解密\n");
	scanf("%d", &flag);
	printf("请输入字符串:\n");
	scanf("%s", str);
	if (flag == 1)
		printf("加密后的字符串是:%s",base64_encode(str));
	else
		printf("解密后的字符串是:%s",base64_decode(str));
	return 0;
}

3.效果图 ①编码

②解码

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