通过套接字的最简单 advertise/listen 排列

Question

下面简单的排列有什么问题。我所做的就是创建一个多播消息的 UDP 广告商，以及一个加入多播组以接收此消息的侦听器，两者运行在同一台机器上。

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();
}

侦听器根本没有收到任何东西（MessageReceived 从未调用过）。我尝试了以下变体但没有成功：

调用和不调用广告商的 BindServiceNameAsync()。
对广告商、听众或两者使用MulticastOnly
先创建一个对象再等待几秒钟。
使用 255.255.255.255 作为主机。

Answer 1

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

感谢您分享这个问题。我做了一个演示并做了一些测试。我发现侦听器套接字在组中发送一条消息之前不会收到任何消息。

因此，目前的解决方法是在注册收听后立即发送一条空消息：

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();
        }
    }
}

关于这个问题的根本原因，我会咨询相关团队，得到回复后会告诉你。

通过套接字的最简单 advertise/listen 排列

Simplest possible advertise/listen arrangement through sockets

c#

sockets

udp

multicast

uwp