UDP Broadcast in Java

Question

Morning.

I'm pretty new in Java and socket connections but I'm trying to send out a UDP packet/broadcast on 255.255.255.255 on port 8001 to a device. I can get the data to send just fine, however when it comes time to receive the data the connection times out. I have a packet sniffer and I can see the packet send and then the device respond.

I'm pretty sure it is a rookie mistake that I'm missing in my code but I've been stuck on it for awhile and any help would be appreciated.

 m_Socket = new DatagramSocket(m_SERVERPORT);
 InetAddress address = InetAddress.getByName(m_SERVERIP);


 m_DataPack = new DatagramPacket(m_SERVERCMD.getBytes(), m_SERVERCMD.getBytes().length,
 address, m_SERVERPORT);
 m_Socket.setBroadcast(true);
 m_Socket.connect(address, m_SERVERPORT);

 m_Socket.send(m_DataPack);
 m_DataPack = new DatagramPacket(data, data.length,
 address, m_SERVERPORT);


 m_Socket.receive(m_DataPack); // This is where it times out


 data = m_DataPack.getData();
 String received = data.toString();
 System.out.println("Received: " + received);
 m_Socket.close();

Thanks and Gig'Em.

EDIT:

I'm not sure if this helps but when I watch the m_Socket object I can see the following right before it sends:

bound = true;
close = false;
connectedAddress = Inet4Address (id = 32) (-1,-1,-1,-1);
connectedPort = 8001;
connectState = 1;
created = true;
impl = PlainDatagramSocketImpl;
oldImpl = false;

and the m_DataPack object is the following:

address = Inet4Address (id = 32) (-1,-1,-1,-1);
bufLength = 6 (size of packet I'm sending is 6 char long);
offset = 0;
port = 8001;

Answer 1

If you want to receive a datagram you need to bind() to the local endpoint (address + port).

Answer 2

This doesn't make sense. You are broadcasting, which is 1-to-many, and you are also connecting, which is 1-to-1. Which is it?

Lose the connect. And lose the 255.255.255.255. This has been heavily deprecated for about 20 years. Use a subnet-local broadcast address, e.g. 192.168.1.255.