C语言实现Base64编码

#include "base64.h"static char encodingTable[64] = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z','a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z','0','1','2','3','4','5','6','7','8','9','+','/'};int GetEncodeSize(const char *src){return (strlen(src) + 2) / 3 * 4 + 1;}int GetDecodeSize(const char *src){return strlen(src) / 4 * 3 + 1;}void Base64Encode(const char *src, char *dest){int i = 0, j = 0, k = 0;int len = strlen(src);int m = len / 3;int n = len % 3; for(; i < m; i++){k = i * 3;dest[j++] = encodingTable[(src[k] & 0xfc) >> 2];dest[j++] = encodingTable[((src[k] & 0x03) << 4) | ((src[k+1] & 0xf0) >> 4)];dest[j++] = encodingTable[((src[k+1] & 0x0f) << 2) | ((src[k+2] & 0xc0) >> 6)];dest[j++] = encodingTable[src[k+2] & 0x3f];}k = i * 3;if(n == 1){dest[j++] = encodingTable[(src[k] & 0xfc) >> 2];dest[j++] = encodingTable[(src[k] & 0x03) << 4];dest[j++] = '=';dest[j++] = '=';}else if(n == 2){dest[j++] = encodingTable[(src[k] & 0xfc) >> 2];dest[j++] = encodingTable[((src[k] & 0x03) << 4) | ((src[k+1] & 0xf0) >> 4)];dest[j++] = encodingTable[(src[k+1] & 0x0f) << 2];dest[j++] = '=';}dest[j] = '/0';}void Base64Decode(const char *src, char *dest){int i = 0, j = 0, k = 0;int size = strlen(src);int len = 0, m = 0, n = 0;char *ch = (char *)malloc(size + 1);memset(ch, 0, size + 1); for(; i < size; i++){if(src[i] >= 'A' && src[i] <= 'Z')ch[i] = src[i] - 'A';else if(src[i] >= 'a' && src[i] <= 'z')ch[i] = src[i] - 'a' + 26;else if(src[i] >= '0' && src[i] <= '9')ch[i] = src[i] - '0' + 52;else if(src[i] == '+')ch[i] = 62;else if(src[i] == '/')ch[i] = 63;}len = strlen(ch);m = len / 4;n = len % 4;for(i = 0; i < m; i++){k = i * 4;dest[j++] = (ch[k] << 2) | ((ch[k+1] & 0x30) >> 4);dest[j++] = ((ch[k+1] & 0x0f) << 4) | ((ch[k+2] & 0x3c) >> 2);dest[j++] = ((ch[k+2] & 0x03) << 6) | ch[k+3];}k = i * 4;if(n == 1){dest[j++] = ch[k] << 2;}else if(n == 2){dest[j++] = (ch[k] << 2) | ((ch[k+1] & 0x30) >> 4);dest[j++] = (ch[k+1] & 0x0f) << 4;}else if(n == 3){dest[j++] = (ch[k] << 2) | ((ch[k+1] & 0x30) >> 4);dest[j++] = ((ch[k+1] & 0x0f) << 4) | ((ch[k+2] & 0x3c) >> 2);dest[j++] = (ch[k+2] & 0x03) << 6;}dest[j] = '/0';free(ch);}