Simplest possible advertise/listen arrangement through sockets

微笑、不失礼 提交于 2019-12-20 04:17:15

问题


What's wrong with the following simple arrangement. All I'm doing is to create a UDP advertiser that multicasts a message, and a listener that joins the multicast group to receive this message, both running on the same machine.

string Port = "54153";
HostName Host = new HostName("224.3.0.5"); //a multicast range address

//listener
var L = new DatagramSocket();
L.MessageReceived += (sender2, args) => { /*something*/ };
await L.BindServiceNameAsync(Port);
L.JoinMulticastGroup(Host);

//advertiser
var AdvertiserSocket = new DatagramSocket();
AdvertiserSocket.Control.MulticastOnly = true;

Stream outStream = (await AdvertiserSocket.GetOutputStreamAsync(Host, Port)).AsStreamForWrite();
using (var writer = new StreamWriter(outStream))
{
  await writer.WriteLineAsync("MESSAGE");
  await writer.FlushAsync();
}

The listener doesn't receive anything at all (MessageReceived never invoked). I have tried the following variations without success:

  1. Calling and not calling BindServiceNameAsync() on advertiser.
  2. Using MulticastOnly on advertiser, listener or both
  3. Waiting for a few seconds after creating one object before the other.
  4. Using 255.255.255.255 as host.

回答1:


It seems like the advertiser is multicasting data correctly (TCPView shows sent packets), but the receiving port is not getting anything.

Thank you for sharing this problem. I made a demo and did some tests. I found out the listener socket won't receive any message until it has sent one message in the group.

So, currently the workaround is to send an empty message immediately after register for listening:

private async void btnListen_Click(object sender, RoutedEventArgs e)
{
        socket = new DatagramSocket();
        socket.MessageReceived += Socket_MessageReceived;
        socket.Control.MulticastOnly = true;
        await socket.BindServiceNameAsync(serverPort);
        socket.JoinMulticastGroup(serverHost);
        SendWithExistingSocket(socket, "");//send an empty message immediately
}

private async void SendWithExistingSocket(DatagramSocket socket, String text)
{
    if (socket != null)
    {
        Stream stream = (await socket.GetOutputStreamAsync(serverHost, serverPort)).AsStreamForWrite();
        using (var writer = new StreamWriter(stream))
        {
            writer.WriteLine(text);
            await writer.FlushAsync();
        }
    }
}

As for the root cause of this problem, I'll consult with related team and let you know once I got response.



来源:https://stackoverflow.com/questions/39722217/simplest-possible-advertise-listen-arrangement-through-sockets

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