Base64编码说明
Base64编码要求把3个8位字节(3*8=24)转化为4个6位的字节(4*6=24),之后在6位的前面补两个0,形成8位一个字节的形式。 如果剩下的字符不足3个字节,则用0填充,输出字符使用'=',因此编码后输出的文本末尾可能会出现1或2个'='。
为了保证所输出的编码位可读字符,Base64制定了一个编码表,以便进行统一转换。编码表的大小为2^6=64,这也是Base64名称的由来。
转码过程举例:
3*8=4*6
内存1个字符占8位
转前: s 1 3
先转成ascii:对应 115 49 51
2进制: 01110011 00110001 00110011
6个一组(4组) 011100110011000100110011
然后才有后面的 011100 110011 000100 110011
然后计算机是8位8位的存数 6不够,自动就补两个高位0了
所有有了 高位补0
科学计算器输入 00011100 00110011 00000100 00110011
得到 28 51 4 51
查对下照表 c z E z
Base64编码表
码值 | 字符 | 码值 | 字符 | 码值 | 字符 | 码值 | 字符 | |||
---|---|---|---|---|---|---|---|---|---|---|
0 | A | 16 | Q | 32 | g | 48 | w | |||
1 | B | 17 | R | 33 | h | 49 | x | |||
2 | C | 18 | S | 34 | i | 50 | y | |||
3 | D | 19 | T | 35 | j | 51 | z | |||
4 | E | 20 | U | 36 | k | 52 | 0 | |||
5 | F | 21 | V | 37 | l | 53 | 1 | |||
6 | G | 22 | W | 38 | m | 54 | 2 | |||
7 | H | 23 | X | 39 | n | 55 | 3 | |||
8 | I | 24 | Y | 40 | o | 56 | 4 | |||
9 | J | 25 | Z | 41 | p | 57 | 5 | |||
10 | K | 26 | a | 42 | q | 58 | 6 | |||
11 | L | 27 | b | 43 | r | 59 | 7 | |||
12 | M | 28 | c | 44 | s | 60 | 8 | |||
13 | N | 29 | d | 45 | t | 61 | 9 | |||
14 | O | 30 | e | 46 | u | 62 | + | |||
15 | P | 31 | f | 47 | v | 63 | / |
C++实现:
base64.h
1 #ifndef _BASE64_HH 2 #define _BASE64_HH 3 4 #ifdef __cplusplus 5 extern "C" { 6 #endif //__cplusplus 7 8 unsigned char* base64Decode(char* pszInput, unsigned int& iReturnSize, bool trimTrailingZeros = true); 9 10 char* base64Encode(char const* origSigned, unsigned iOrigLength); 11 12 #ifdef __cplusplus 13 } 14 #endif //__cplusplus 15 16 #endif //_BASE64_HH
base64.cpp
1 #include "base64.h" 2 #include <string.h> 3 4 static char base64DecodeTable[65]; 5 //初始化编码对照表 6 static const char base64Char[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"; 7 8 char* strDupSize(char const* str) 9 { 10 if (str == NULL) return NULL; 11 size_t len = strlen(str) + 1; 12 char* copy = new char[len]; 13 14 return copy; 15 } 16 17 //初始化解码对照表 18 static void initBase64DecodeTable() 19 { 20 for (int i = 0; i < 65; ++i) 21 base64DecodeTable[i] = (char)0x80; // default value: invalid 22 23 for (int i = 'A'; i <= 'Z'; ++i) 24 base64DecodeTable[i] = 0 + (i - 'A'); 25 for (int i = 'a'; i <= 'z'; ++i) 26 base64DecodeTable[i] = 26 + (i - 'a'); 27 for (int i = '0'; i <= '9'; ++i) 28 base64DecodeTable[i] = 52 + (i - '0'); 29 base64DecodeTable[(unsigned char)'+'] = 62; 30 base64DecodeTable[(unsigned char)'/'] = 63; 31 base64DecodeTable[(unsigned char)'='] = 64; //0x40 32 } 33 34 // 解码 35 unsigned char* base64Decode(char* pszInput, unsigned int& iReturnSize, bool trimTrailingZeros) 36 { 37 static bool haveInitedBase64DecodeTable = false; 38 if (!haveInitedBase64DecodeTable) 39 { 40 initBase64DecodeTable(); 41 haveInitedBase64DecodeTable = true; 42 } 43 44 unsigned char* pszOutput = new unsigned char[strlen(pszInput)]; 45 int k = 0; 46 /******************************************************************************************* 47 *技巧:先减去3 ---in case "pszInput" is not a multiple of 4 bytes (although it should be) * 48 * 个人觉得这里不用做处理,因为编码时候已经做过处理,密文肯定为4的整数倍 * 49 *******************************************************************************************/ 50 int const jMax = strlen(pszInput) - 3; 51 52 //将密文四个字节为一组,每组解码为三个字节 53 for (int j = 0; j < jMax; j += 4) 54 { 55 char inTmp[4], outTmp[4]; 56 for (int i = 0; i < 4; ++i) 57 { 58 inTmp[i] = pszInput[i+j]; 59 outTmp[i] = base64DecodeTable[(unsigned char)inTmp[i]]; 60 if ((outTmp[i]&0x40) != 0) //如果遇到'=',将其视为结束字符 61 outTmp[i] = 0x00; 62 } 63 64 pszOutput[k++] = (outTmp[0]<<2) | (outTmp[1]>>4); 65 pszOutput[k++] = (outTmp[1]<<4) | (outTmp[2]>>2); 66 pszOutput[k++] = (outTmp[2]<<6) | outTmp[3]; 67 } 68 69 if (trimTrailingZeros) 70 { 71 while (k > 0 && pszOutput[k-1] == '0') 72 --k; 73 } 74 iReturnSize = k; 75 unsigned char* result = new unsigned char[iReturnSize]; 76 memset(result, 0, iReturnSize); 77 memmove(result, pszOutput, iReturnSize); 78 delete[] pszOutput; 79 return result; 80 } 81 82 // 编码 83 char* base64Encode(char const* origSigned, unsigned iOrigLength) 84 { 85 unsigned char const* orig = (unsigned char const*)origSigned; 86 if (orig == NULL) 87 return NULL; 88 89 unsigned const iNumResultBytes = 4*(iOrigLength/3 + (iOrigLength%3 ? 1 : 0)); 90 char* result = new char[iNumResultBytes + 1]; 91 92 //源码三个字节为一组,编码后将三个字节变为四个字节 93 unsigned i; 94 for (i = 0; i < iOrigLength/3; ++i) 95 { 96 // 取源码的第一个字节的前六位,前面加两个零生成第一个字节 97 result[4*i+0] = base64Char[(orig[3*i]>>2) & 0x3F]; 98 // 取源码的第一个字节的后两位和第二个字节的前四位,前面补两个零生成第二个字节 99 result[4*i+1] = base64Char[((orig[3*i]<<4) | (orig[3*i+1]>>4)) & 0x3F]; 100 // 取源码的第二个字节的后四位和第三个字节的前两位,前面补两个零生成第三个字节 101 result[4*i+2] = base64Char[((orig[3*i+1]<<2) | (orig[3*i+2]>>6)) & 0x3F]; 102 // 取源码的第三个字节的后六位,前面补两个零生成第四个字节 103 result[4*i+3] = base64Char[orig[3*i+2]&0x3F]; 104 } 105 //不足三个字节的最后补'='字符,补足三个字节 106 if (iOrigLength%3) 107 { 108 result[4*i+0] = base64Char[(orig[3*i]>>2) & 0x3F]; 109 if (iOrigLength%3 == 2) 110 { 111 result[4*i+1] = base64Char[((orig[3*i]<<4) | (orig[3*i+1]>>4))&0x3F]; 112 result[4*i+2] = base64Char[(orig[3*i+1]<<2)&0x3C]; 113 } 114 else 115 { 116 result[4*i+1] = base64Char[((orig[3*i])<<4)&0x3F]; 117 result[4*i+2] = '='; 118 } 119 result[4*i+3] = '='; 120 } 121 122 result[iNumResultBytes] = '