Question

I have a question about the send method of C.

int send (int socket, void *buffer, size_t size, int flags);

My code:

char *buffer = (char *)malloc(100*sizeof(char));
send(s, buffer, 100*sizeof(char), MSG_NOSIGNAL);

Now I was wondering if I should free the buffer myself or does the send method free it?

My code where I free the buffer:

char *buffer = (char *)malloc(100*sizeof(char));
send(s, buffer, 100*sizeof(char), MSG_NOSIGNAL);
free(buffer);

When I saw the error I thought I freed the buffer to soon and the send method was still using the memory.

Let me know.

Answer 1

send does not free the buffer. You need to free it. Aside from the fact that it simply wasn’t designed that way (socket functions are independent of heap functions), send would not actually “know” if the memory should be freed since it could be a pointer to the middle of an allocated chunk of memory or it could be a pointer to stack data.

It should be “safe” to immediately free the buffer after the send call because the function will have copied the data to its own internal buffers. The code you show would not cause a corruption problem (although I suspect there is more code involved since the buffer is not initialized as shown). It is possible for the heap to be corrupted earlier in the process and not cause an exception until later (e.g., the free after the send).

The return value of send indicates the number of bytes successfully sent (although not necessarily successfully delivered). So while it would be “safe” to free it, you would lose the unsent data if it wasn’t all sent. If the entire buffer was not sent, you would need to call send again with the remaining amount.