How do I check if a TCP port is already being listened on?

Question

I have a third party library that acts as a HTTP server. I pass it an address and port, which it then uses to listen for incoming connections. This library listens in such a way that it doesn't receive exclusive usage of the port and address it's bound to. As a result, I can listen on the same port multiple times.

I need to run multiple instances of this HTTP server in the same process. Each instance has a default port, but if that port isn't available, it should use the next available port. This is where my problem is; I can end up with two HTTP servers listening on the same port.

I cannot change the HTTP server's code and the HTTP server will not alert me if it cannot listen on the port I give it, so I have to be able to check if a port is already in use before starting each HTTP server. I have tried checking if a port is already being listened on by binding my own socket with SO_REUSEADDR set to FALSE and SO_EXCLUSIVEADDRUSE set to TRUE, but the bind and listen calls both succeed when an existing HTTP server is already listening on that port.

How is this HTTP server achieving this effect, and how can I accurately check if a port is being listened on in this manner?

Answer 1

The quick and dirty method would be to try to connect() to the port on localhost. If the connect() call succeeds, then you know the port is currently being listened on (by whomever received the connection). If the connect call fails (in particular with ECONNREFUSED) then you can be pretty sure that nobody is listening on that port.

Of course, there's a race condition here: Nothing is really stopping another program from swooping in and grabbing the port immediately after you ran the above test, but before you get around to binding to the port yourself. So you should take the result of the test as more of a hint than an absolute rule, and (hopefully) have some way of handling it if you later find out that the port is in use after all.

Answer 2

Use a port number of 0. The OS will pick a free port.