[noPoll] Integration with existing IO loop

Dave Gilmour (FET) dave en fetcorp.com
Mie Abr 5 22:37:21 CEST 2017

I am attempting to integrate nopoll into an existing IO loop because I 
have other sockets that I must also look after.  I am having an issue 
with my first call to nopoll_conn_get_msg blocking.

I am initializing nopoll as follows:

nopoll_thread_handlers(mutex_create, mutex_destroy, mutex_lock, 
ctx = nopoll_ctx_new ();
noPollConn* listener = nopoll_listener_new (ctx, "", "8000");
nopoll_ctx_set_on_msg (ctx, on_message, NULL);
nopoll_ctx_set_on_ready(ctx, on_ready, NULL);
nopoll_log_enable(ctx, 1);
nopoll_log_set_handler(ctx, on_log, NULL);

I then determine which socket is being used for the master listener by 
sockWSMasterListener = nopoll_conn_socket(listener);

At the top of my io loop I do a FD_SET on the master listener socket (as 
well as for other sockets I'm managing) for the readfds that is passed 
to select, and call select.

I get an event on that the listener socket so I call
ws_client_conn = nopoll_conn_accept (ctx, listener);
in order to accept the connection and obtain the socket that is now 
associated with the new connection.

The next cycle through the io loop I do a FD_SET on the new socket (as 
well as for other sockets I'm managing and the master listener) for the 
readfds that is passed to select, and call select again

I get an event on the connected socket and call nopoll_conn_get_msg 

My problem is this call does not return. I have tracked it down to the 
following line of code in nopoll_conn.c:

/* get the first 2 bytes from the websocket header */
bytes = __nopoll_conn_receive (conn, buffer, 2); <<<<<<<<<<<<<<this call 

The problem is that when this happens I am not servicing any of my other 
sockets any more. Once a message is passed either from the client to the 
server or vice versa everything seems to go ahead without a problem. So 
my question is, what is nopoll waiting for at this point and how can I 
avoid the blocking.

My apologies for the long-windedness of this question.

Thank you.

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