我有一个类型为 unsigned char buffer[2850]
的大型缓冲区数组,我想将其转换为 Base64 字符串。我正在尝试使用可以找到的 libb64 库 on github
这是我尝试转换它的方式:
char* encode(const char* input)
{
/* set up a destination buffer large enough to hold the encoded data */
char* output = (char*)malloc(SIZE);
/* keep track of our encoded position */
char* c = output;
/* store the number of bytes encoded by a single call */
int cnt = 0;
/* we need an encoder state */
base64_encodestate s;
/*---------- START ENCODING ----------*/
/* initialise the encoder state */
base64_init_encodestate(&s);
/* gather data from the input and send it to the output */
cnt = base64_encode_block(input, strlen(input), c, &s);
c += cnt;
/* since we have encoded the entire input string, we know that
there is no more input data; finalise the encoding */
cnt = base64_encode_blockend(c, &s);
c += cnt;
/*---------- STOP ENCODING ----------*/
/* we want to print the encoded data, so null-terminate it: */
*c = 0;
return output;
}
char *encodedBuffer;
encodedBuffer = encode(buffer);
但是我收到了警告
Passing 'unsigned char [2850]' to parameter of type 'const char *' converts between pointers to integer types with different sign.
有没有更好的方法来转换它,或者因为我试图传入一个无符号字符数组,所以我需要改变什么。谢谢!
最佳答案
您的缓冲区是 char**
类型。您需要传递 char*
类型之一。
你可以使用
const char *buffer = "whatever";
encode(buffer);
关于c - 如何将 unsigned char 数组转换为 base64 字符串?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22256872/