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
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
文章来源地址https://uudwc.com/A/XkgY5
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<stdio.h>
#include<string.h>
#include<stdlib.h>
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+/";//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 <iostream>
#include <cstdio>
#include<string>
#include <map>
using 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;
map<int, unsigned char> base64_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
文章来源：https://uudwc.com/A/XkgY5
