Question

我有一个Java类，旨在促进计算机之间的连接，而无需显式定义对等地址。

import java.io.IOException;
import java.net.DatagramPacket;
import java.net.DatagramSocket;
import java.net.MulticastSocket;
import java.net.SocketException;

public class Broadcasts {

    private final Runnable receiver;
    private final Runnable sender;
    private boolean run = true;

    public Broadcasts(TheThing parent) {
        receiver = new Runnable() {
            public void run() {
                byte data[] = new byte[0];
                DatagramPacket packet = new DatagramPacket(data, data.length);
                try {
                    DatagramSocket socket = new DatagramSocket();
                    while (run) {
                        socket.receive(packet);
                        parent.newAddress(packet.getAddress());
                    }
                } catch (SocketException ex) {
                    ex.printStackTrace();
                    parent.quit();
                } catch (IOException ex) {
                    ex.printStackTrace();
                    parent.quit();
                }
            }
        };
        sender = new Runnable() {
            public void run() {
                byte data[] = new byte[0];
                DatagramPacket packet = new DatagramPacket(data, data.length);
                try {
                    MulticastSocket socket = new MulticastSocket();
                    socket.setBroadcast(true);
                    socket.joinGroup(Globals.publicAddress);
                    while (run) {
                        socket.send(packet);
                        Thread.sleep(Globals.UDPINTERVAL);
                    }
                } catch (IOException ex) {
                    ex.printStackTrace();
                    parent.quit();
                } catch (InterruptedException ex) {
                    ex.printStackTrace();
                    parent.quit();
                }
            }
        };
        new Thread(receiver).start();
        new Thread(sender).start();
    }

    public void quit() {
        run = false;
    }
}

我收到以下错误：

java.net.SocketException: Not a multicast address
    at java.net.MulticastSocket.joinGroup(MulticastSocket.java:310)
    at the.thing.Broadcasts$2.run(Broadcasts.java:42)
    at java.lang.Thread.run(Thread.java:745)

我面临的问题是定义为Globals.publicAddress 的地址是多播地址。我在这里缺少什么？

使用以下方法获得

Globals.publicAddress。应该是DHCP分配的地址：

private static InetAddress getPublicIface() {
    InetAddress result = null;
    Enumeration e;
    try {
        e = NetworkInterface.getNetworkInterfaces();
    } catch (SocketException ex) {
        return null;
    }
    while (e.hasMoreElements() && result == null) {
        NetworkInterface n = (NetworkInterface) e.nextElement();
        Enumeration ee = n.getInetAddresses();
        while (ee.hasMoreElements()) {
            InetAddress i = (InetAddress) ee.nextElement();;
            if (i.getHostAddress().startsWith("192.168.")) {
                result = i;
                break;
            }
        }
    }
return result;

我还尝试了以下方法来获取有效的广播地址：

private static InetAddress getPublicIface() {
    InetAddress result = null;
    Enumeration e;
    try {
        e = NetworkInterface.getNetworkInterfaces();
    } catch (SocketException ex) {
        return null;
    }
    while(e.hasMoreElements() && result == null) {
        NetworkInterface iface = (NetworkInterface) e.nextElement();
        try {
            if(iface == null || iface.isLoopback() || !iface.isUp() ||
                    iface.isVirtual() || !iface.supportsMulticast())
                continue;
        } catch (SocketException ex) {
            continue;
        }
        Iterator it = iface.getInterfaceAddresses().iterator();
        while (it.hasNext()) {
            InterfaceAddress address = (InterfaceAddress) it.next();
            if(address == null || address.getBroadcast() == null)
                continue;
            result = address.getBroadcast();
            break;
        }
    }
    return result;
}

Answer 1

我不知道NetworkInterface，所以对我来说这是一个很酷的新事物。所以我匆匆忙忙地把一些似乎有用的代码放在一起。它需要清理，但它在我的系统上运行，似乎实际上是广播。

突出点：

您无需收听广播地址。由于它是广播，无论你听什么IP地址，你都会得到它。
您必须发送到广播地址，由于某种原因，这似乎仅作为DatagramPacket的参数，而不是DatagramSocket。我不知道为什么，但这就是API似乎想要它的方式。

public class BroadcastTest
{
   private static final int RANDOM_PORT = 4444;

   public static void main( String[] args )
           throws Exception
   {
      InetAddress addr = getBroadcastAddrs().get(0);
      System.err.println( addr );
      new Thread( new BroadcastServer( RANDOM_PORT ) ).start();
      DatagramSocket dsock = new DatagramSocket();
      byte[] send = "Hello World".getBytes( "UTF-8" );
      DatagramPacket data = new DatagramPacket( send, send.length, addr, RANDOM_PORT );
      dsock.send( data );
   }

   public static List<InetAddress> getBroadcastAddrs() throws SocketException {
      Set<InetAddress> set = new LinkedHashSet<>();
      Enumeration<NetworkInterface> nicList = NetworkInterface.
              getNetworkInterfaces();
      for( ; nicList.hasMoreElements(); ) {
         NetworkInterface nic = nicList.nextElement();
         if( nic.isUp() && !nic.isLoopback() )  {
            for( InterfaceAddress ia : nic.getInterfaceAddresses() )
               set.add( ia.getBroadcast() );
         }
      }
      return Arrays.asList( set.toArray( new InetAddress[0] ) );

   }
}

class BroadcastServer implements Runnable {
   private final int port;

   public BroadcastServer( int port )
   {
      this.port = port;
   }

   @Override
   public void run()
   {
      try {
         DatagramSocket dsock = new DatagramSocket( port );
         DatagramPacket data = new DatagramPacket( new byte[2048], 2048 );
         dsock.receive( data );
         System.out.println( new String( data.getData(), "UTF-8" ) );
      } catch( SocketException ex ) {
         Logger.getLogger( BroadcastServer.class.getName() ).
                 log( Level.SEVERE, null, ex );
      } catch( IOException ex ) {
         Logger.getLogger( BroadcastServer.class.getName() ).
                 log( Level.SEVERE, null, ex );
      }
   }
}

UDP广播和接收

1 个答案: