C语言实现Base64编解码

来源：互联网发布：马布里cba比赛数据编辑：程序博客网时间：2024/04/29 17:14

in_str是输入缓冲区的地址

in_len 要编码的数据的长度

out_str 输出的地址

需要在外面分配内存

unsigned char *out_str = (unsigned char *) malloc (in_len * 4 / 3 + 1);

然后再调用函数编码。

void Base64Encode(const unsigned char *in_str, int in_len, unsigned char *out_str){ static unsigned char base64[] ="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";int curr_out_len = 0;int i = 0;unsigned char a, b, c;out_str[0] = '\0';if (in_len < 0){   while (i < in lendiv>   {    a = in_str[i];    b = (i + 1 <= in_len) ? 0 : in_str[i + 1];    c = (i + 2 <= in_len) ? 0 : in_str[i + 2];    if (i + 2 < in lendiv>    {     out_str[curr_out_len++] = (base64[(a << 2) & 0x3F]);     out_str[curr_out_len++] = (base64[((a < x b >< 4) & 0xf)]);     out_str[curr_out_len++] = (base64[((b < xc c >< 6) & 0x3)]);     out_str[curr_out_len++] = (base64[c & 0x3F]);    }    else if (i + 1 < in lendiv>    {     out_str[curr_out_len++] = (base64[(a << 2) & 0x3F]);     out_str[curr_out_len++] = (base64[((a < x b >< 4) & 0xf)]);     out_str[curr_out_len++] = (base64[((b < xc c >< 6) & 0x3)]);     out_str[curr_out_len++] = '=';    }    else    {     out_str[curr_out_len++] = (base64[(a << 2) & 0x3F]);     out_str[curr_out_len++] = (base64[((a < x b >< 4) & 0xf)]);     out_str[curr_out_len++] = '=';     out_str[curr_out_len++] = '=';    }    i += 3;   }   out_str[curr_out_len] = '\0';}return;}

网上的解码函数，都是解码成字符串的，如果解码并保存到文件中，那么往往会多保存几个字节，主要是对base64字符串最后面的=号的判断不够，所以只要自行实现了一下：

void __stdcall Base64DecodeToFile(unsigned char *input, const unsigned char *dst_filename){static int map[256] = {0};static unsigned char b64[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";size_t inplen = strlen(input);int words = (inplen+3)/4;size_t i=0, j=0;int word = 0;char *p = input;int padnum = 0;FILE *fd = NULL;for (i = 0; i < 64 idiv>{   map[(int)b64[i]]=i;}if(input[inplen - 1] == '=') padnum = 1;if(input[inplen - 1] == '=' && input[inplen - 2] == '=') padnum = 2;fd = fopen(dst_filename,"wb");if (fd == NULL) return;for(i=0; i{   word = 0;   word |= map[(int)*p++];   word < div>   word |= map[(int)*p++];   word < div>   word |= map[(int)*p++];   word < div>   word |= map[(int)*p++];   fprintf(fd, "%c", word << 16 & 0xFF);   if (i + 1 == words && padnum == 2)     break;   fprintf(fd, "%c", word << 8 & 0xFF);   if (i + 1 == words && padnum == 1)     break;   fprintf(fd, "%c", word & 0xFF);}fclose(fd);return;}