Base64编解码详解-优快云博客

Base64编解码原理

Base64可以将ASCII字符串或者是二进制编码成只含A--Z,a--z,0--9,+,/ 这64个字符(这也是为啥叫Base64的原因).

1. 编码原理:将3个字符转换成4个字符( (3 X 8) = 24 字节= (4 X 6) )
   取输入流中的3个字节,每次取6位(共4次),以此6位值做索引去查表，输出相应的字符。如果输入流中字符不是3的倍数，最后结果缺的就以"="填充(输入流中最后只剩1个字符就填2个=;剩2个就填1个=).
base64digits[] =   "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"


2. 解码原理:将4个字符转换成3个字符.
            先读入输入流中的4个字符,然后根据上表中的值找索引(也可根据如下程序中的base64val[]索引查找)，生成4个6位的值，然后以字节(8位)截断,就形成3个字符的ASCII值,这样就还原了.

C语言中是通过移位>> <<和按位&｜实现的。
下面程序是网上的列子(有一个不正确的地方，偶稍微修改下了，并加了测试）

#include<stdio.h> #include<stdlib.h> #include<string.h> #include <ctype.h> static const char base64digits[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"; #define BAD -1 static const char base64val[] = { BAD,BAD,BAD,BAD, BAD,BAD,BAD,BAD, BAD,BAD,BAD,BAD, BAD,BAD,BAD,BAD, BAD,BAD,BAD,BAD, BAD,BAD,BAD,BAD, BAD,BAD,BAD,BAD, BAD,BAD,BAD,BAD, BAD,BAD,BAD,BAD, BAD,BAD,BAD,BAD, BAD,BAD,BAD, 62, BAD,BAD,BAD, 63, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61,BAD,BAD, BAD,BAD,BAD,BAD, BAD, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,BAD, BAD,BAD,BAD,BAD, BAD, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51,BAD, BAD,BAD,BAD,BAD }; #define DECODE64(c) (isascii(c) ? base64val[c] : BAD) void to64frombits(unsigned char *out, const unsigned char *in, int inlen) { for (; inlen >= 3; inlen -= 3) { *out++ = base64digits[in[0] >> 2]; *out++ = base64digits[((in[0] << 4) & 0x30) | (in[1] >> 4)]; *out++ = base64digits[((in[1] << 2) & 0x3c) | (in[2] >> 6)]; *out++ = base64digits[in[2] & 0x3f]; in += 3; } if (inlen > 0) { unsigned char fragment; *out++ = base64digits[in[0] >> 2]; fragment = (in[0] << 4) & 0x30; if (inlen > 1) fragment |= in[1] >> 4; *out++ = base64digits[fragment]; *out++ = (inlen < 2) ? '=' : base64digits[(in[1] << 2) & 0x3c]; *out++ = '='; } *out = '/0'; } int from64tobits(char *out, const char *in) { int len = 0; register unsigned char digit1, digit2, digit3, digit4; if (in[0] == '+' && in[1] == ' ') in += 2; if (*in == '/r') return(0); do { digit1 = in[0]; if (DECODE64(digit1) == BAD) return(-1); digit2 = in[1]; if (DECODE64(digit2) == BAD) return(-1); digit3 = in[2]; if (digit3 != '=' && DECODE64(digit3) == BAD) return(-1); digit4 = in[3]; if (digit4 != '=' && DECODE64(digit4) == BAD) return(-1); in += 4; *out++ = (DECODE64(digit1) << 2) | (DECODE64(digit2) >> 4); ++len; if (digit3 != '=') { *out++ = ((DECODE64(digit2) << 4) & 0xf0) | (DECODE64(digit3) >> 2); ++len; if (digit4 != '=') { *out++ = ((DECODE64(digit3) << 6) & 0xc0) | DECODE64(digit4); ++len; } } } while (*in && *in != '/r' && digit4 != '='); //don't forget this line! *out='/0'; return (len); } int main() { char in[]="linewer",*out,out2[8]; int inlen=strlen(in),n; n=(inlen % 3)?(inlen/3+1)*4 : (inlen/3*4); out=malloc(n+1);