Simplest possible advertise/listen arrangement through sockets

主宰稳场 提交于 2019-12-02 04:08:49

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.

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!