CODEWITHSUNDEEP
c#network-programmingmulticast
skataben
skataben

Reputation: 2152

Multicast listener port in use

I am new to multicast programming. So far I can successfully send and receive multicast messages from two separate processes (a sender and a receiver). My problem is with the receiver...

ReceiverCode:

private static void ReceiveMulticastMessages()
{
    var groupEndPoint = new IPEndPoint(IPAddress.Parse("238.8.8.8"), 23888);
    var localEndPoint = new IPEndPoint(IPAddress.Any, 23888);

    using (var udpClient = new UdpClient())
    {
        udpClient.Client.SetSocketOption(SocketOptionLevel.Socket, SocketOptionName.ReuseAddress, true);
        udpClient.Client.Bind(localEndPoint);
        udpClient.JoinMulticastGroup(groupEndPoint.Address, localEndPoint.Address);

        while (true)
        {
            var remoteEndPoint = new IPEndPoint(IPAddress.Any, 0);
            var bytes = udpClient.Receive(ref remoteEndPoint);
            var message = Encoding.ASCII.GetString(bytes);
            Console.WriteLine(message);
        }
    }
}

The above code works as long as I specify port 23888 for the localEndPoint. If I change the local port number, no messages are received. I would prefer to set it to 0 so the OS could choose the port. Why can I not specify a different local port than that of multicast group?

Assuming the local endpoint port must match the multicast group port, how does a client deal with a local port conflict?

On the flip side, how can an application (a multicast sender) choose a multicast group port such that any subscribers will not have a port conflict?

Upvotes: 0

Views: 837

Answers (1)

dbush
dbush

Reputation: 223739

When sending any UDP message (not just muticast messages), the port that the sender sends to must match the port that the receiver is listening on. That's how messages get to the right place. If a message is sent to a port other that the one the receiver is bound to, the receiver won't get it.

So a port number needs to be defined that the receiver(s) will listen on and the server will send to.

Upvotes: 2

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