C/C++ Base64加密解密算法

2018-01-27 10:27:18来源:网络收集作者:程序诗人人点击

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实用,我也忘了是哪儿抄的,亲测无问题,直接撸进项目去吧


const char base[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=";
/* Base64 编码 */
char* base64_encode(const char* data, int data_len)
{
//int data_len = strlen(data);
int prepare = 0;
int ret_len;
int temp = 0;
char *ret = NULL;
char *f = NULL;
int tmp = 0;
char changed[4];
int i = 0;
ret_len = data_len / 3;
temp = data_len % 3;
if (temp > 0)
{
ret_len += 1;
}
ret_len = ret_len*4 + 1;
ret = (char *)malloc(ret_len);
if ( ret == NULL)
{
printf("No enough memory./n");
return nullptr;
}
memset(ret, 0, ret_len);
f = ret;
while (tmp < data_len)
{
temp = 0;
prepare = 0;
memset(changed, '/0', 4);
while (temp < 3)
{
//printf("tmp = %d/n", tmp);
if (tmp >= data_len)
{
break;
}
prepare = ((prepare << 8) | (data[tmp] & 0xFF));
tmp++;
temp++;
}
prepare = (prepare<<((3-temp)*8));
//printf("before for : temp = %d, prepare = %d/n", temp, prepare);
for (i = 0; i < 4 ;i++ )
{
if (temp < i)
{
changed[i] = 0x40;
}
else
{
changed[i] = (prepare>>((3-i)*6)) & 0x3F;
}
*f = base[changed[i]];
//printf("%.2X", changed[i]);
f++;
}
}
*f = '/0';
return ret;
}
/* 转换算子 */
static char find_pos(char ch)
{
char *ptr = (char*)strrchr(base, ch);//the last position (the only) in base[]
return (ptr - base);
}
/* Base64 解码 */
char* base64_decode(const char *data, int data_len)
{
int ret_len = (data_len / 4) * 3;
int equal_count = 0;
char *ret = NULL;
char *f = NULL;
int tmp = 0;
int temp = 0;
int prepare = 0;
int i = 0;
if (*(data + data_len - 1) == '=')
{
equal_count += 1;
}
if (*(data + data_len - 2) == '=')
{
equal_count += 1;
}
if (*(data + data_len - 3) == '=')
{//seems impossible
equal_count += 1;
}
switch (equal_count)
{
case 0:
ret_len += 4;//3 + 1 [1 for NULL]
break;
case 1:
ret_len += 4;//Ceil((6*3)/8)+1
break;
case 2:
ret_len += 3;//Ceil((6*2)/8)+1
break;
case 3:
ret_len += 2;//Ceil((6*1)/8)+1
break;
}
ret = (char *)malloc(ret_len);
if (ret == NULL)
{
printf("No enough memory./n");
return nullptr;
}
memset(ret, 0, ret_len);
f = ret;
while (tmp < (data_len - equal_count))
{
temp = 0;
prepare = 0;
while (temp < 4)
{
if (tmp >= (data_len - equal_count))
{
break;
}
prepare = (prepare << 6) | (find_pos(data[tmp]));
temp++;
tmp++;
}
prepare = prepare << ((4-temp) * 6);
for (i=0; i<3 ;i++ )
{
if (i == temp)
{
break;
}
*f = (char)((prepare>>((2-i)*8)) & 0xFF);
f++;
}
}
*f = '/0';
return ret;
}

更多实用类:C++操作json
更多文章:http://blog.csdn.net/what951006
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