A have callback function with this prototype:
void serverAnswer(void *pUserData, int flag);
I need to pass a buffer const uint8_t *buf to this function using the void parameter.
How do i pass the buffer, and how can i access the elements of the buffer in the function?
Example (not working):
Call:
const uint8_t *buf;
int buf_len = 3;
serverAnswer(&buf, buf_len);
Access to buffer:
void serverAnswer(void *pUserData, int flag) {
uint8_t* p = (uint8_t *)pUserData;
uint8_t data = p[0];
}
&buf is a const uint8_t **. What you actually want to do is pass buf to serverAnswer, you'll need a const_cast to allow conversion from a const pointer to void*:
const uint8_t* buf;
serverAnswer(const_cast<uint8_t*>(buf), buf_len);
To avoid undefined behaviour you should put the const back in serverAnswer:
void serverAnswer(void *pUserData, int flag) {
const uint8_t* p = (const uint8_t *)pUserData;
uint8_t data = p[0];
}
You're actually passing a pointer to the array during your call. In your code, what you're casting back to an uint8_t*, is an uint8_t**.
What you need to do, is either passing the array directly, or casting it to uint8_t** instead of uint8_t* and dereferencing it.
IMHO, passing the array directly is the best way to go, because passing a pointer to a pointer is not needed when you don't want to modify the pointed pointer, which seems to be your case
