Why do I receive twice the data sent over UDP broadcast?

Question

Hello,

I have a problem with the code below. What it does is send the current date/time using UDP broadcast and listens to this message. My current use of this code is local, I use the Send and Receive in the same application, on the same computer. I didn't try it between 2 computers yet.

public class Agent
{
    public static int Port = 33333;
    public delegate void OnMessageHandler(string message);

    Socket socketSend;
    Socket socketReceive;

    bool receiving = false;
    Thread receiveThread;

    public event OnMessageHandler OnMessage;

    public Agent()
    {
        socketSend = new Socket(AddressFamily.InterNetwork, SocketType.Dgram, ProtocolType.Udp);
        socketSend.EnableBroadcast = true;

        socketSend.Connect(new IPEndPoint(IPAddress.Broadcast, Port));

        socketReceive = new Socket(AddressFamily.InterNetwork, SocketType.Dgram, ProtocolType.Udp);
        socketReceive.EnableBroadcast = true;

        socketReceive.Bind(new IPEndPoint(IPAddress.Any, Port));
    }

    public void Start()
    {
        Console.WriteLine("Agent started!");

        receiving = true;
        receiveThread = new Thread(new ThreadStart(Receive));
        receiveThread.IsBackground = true;
        receiveThread.Start();
    }

    public void Stop()
    {
        receiving = false;

        socketSend.Close();
        socketReceive.Close();

        receiveThread.Join();

        Console.WriteLine("Agent ended.");
    }

    public void Send(string message)
    {
        Console.WriteLine("Sending : {0}", message);
        byte[] bytes = Encoding.Convert(Encoding.Unicode, Encoding.UTF8, Encoding.Unicode.GetBytes(message));
        socketSend.Send(bytes);
        Console.WriteLine("Message sent.");
    }

    protected void Receive()
    {
        Console.WriteLine("Thread started!");
        while (receiving)
        {
            try
            {
                if (socketReceive.Available > 0)
                {
                    byte[] bytes = new byte[socketReceive.Available];

                    Console.WriteLine("Waiting for receive...");

                    int bytesReceived = socketReceive.Receive(bytes, SocketFlags.None);
                    Console.WriteLine("Bytes received : {0}", bytesReceived);

                    string message = Encoding.UTF8.GetString(bytes);
                    Console.WriteLine("Received message : {0}", message);
                    OnMessage(message);
                }
                else
                {
                    Console.Write(".");
                    Thread.Sleep(200);
                }
            }
            catch (SocketException ex)
            {
                if (ex.SocketErrorCode == SocketError.Interrupted)
                {
                    break;
                }
                else
                {
                    throw;
                }
            }
        }
        Console.WriteLine("Thread ended.");
    }
}

The problem I have is that it sends the data once (when I click on a button) but it receives twice the amount expected. For example, a normal date/time is 19 bytes long but the first time Available is > 0, its value is 38. The Receive call following only grabs 19 bytes and the loop continues with the next 19 bytes. That means I receive my message and twice and I want it only once of course.

Example of an output :

Agent started!
Thread started!
..........Sending : 29/07/2009 12:05:04
Message sent.
Waiting for receive...
Bytes received : 19
Received message : 29/07/2009 12:05:04
Waiting for receive...
Bytes received : 19
Received message : 29/07/2009 12:05:04
......Thread ended.
Agent ended.

Thanks for any help or insight.

Answer 1

I'd advise you to try this on a specified target IP first. This way the network topography will have less of an effect, and you can just move it to a broadcast address when everything works the way you expect.

Answer 2

socketReceive.Bind(new IPEndPoint(IPAddress.Any, Port));

By binding to all your IP addresses, you'll be getting local loopback and via the network as well.