Is there a way to wait for a listening socket on win32?

Question

I have a server and client program on the same machine. The server is part of an application- it can start and stop arbitrarily. When the server is up, I want the client to connect to the server's listening socket. There are win32 functions to wait on file system changes (ReadDirectoryChangesW) and registry changes (RegNotifyChangeKeyValue)- is there anything similar for network changes? I'd rather not have the client constantly polling.

Answer 1

Consider using some form of Windows synchronization, such as semaphore. The client can wait on the synchronization primitive and the server can signal it when it starts up.

Answer 2

There is no such Win32 API, however this can be easily accomplished by using an event. The client would wait on that event to be signaled. The server would signal the event when it starts up.

The related API that you will need to use is CreateEvent, OpenEvent, SetEvent, ResetEvent and WaitForSingleObject.

If your server will run as a service, then for Vista and up it will run in session 0 isolation. That means you will need to use an event with a name prefixed with "Global\".

You probably do have a good reason for needing this, but before you implement this please consider:

• Is there some reason you need a connect right away? I see this as a non issue because if you perform an action in the client, you can at that point make a new server connection.
• Is the server starting and stopping more frequently than the client? You could switch roles of who listens/connects

Answer 3

Personally I'd use a UDP broadcast from the server and have the "client" listening for it. The server could broadcast a UDP packet every X period whilst running and when the client gets one, if it's not already connected, it could connect.

This has the advantage that you can move the client onto a different machine without any issues (and since the main connection from client to server is sockets already it would be a pity to tie the client and server to the same machine simply because you selected a local IPC method for the initial bootstrap).