CODEWITHSUNDEEP
javasocketsexceptionbinddatagram
dvaccam
dvaccam

Reputation: 53

Why different DatagramSocket constructors behave so different? (BindException)

I'm building a program that will connect every computer in my LAN to my computer, through DatagramSocket's and DatagramPacket's (in other words, I'll be the server and the others will be clients). I made some research and read the documentation, and found out how it worked, and I've actually been able to send and recieve data across the network. I had two options for the data sending method in the client class, which I thought were equivalent, but it seems not. The first one (Client1):

DatagramSocket socket = null;
DatagramPacket packet = null;

try
{
     byte[] data= "test".getBytes();
     packet = new DatagramPacket(data, data.length, InetAddress.getByName("192.168.0.26"), 325);
     socket = new DatagramSocket();
     socket.send(packet);
 }
 catch (Exception ex)
 {
     ...
 }

And the second one (Client2):

DatagramSocket socket = null;
DatagramPacket packet = null;

try
{
     byte[] data= "test".getBytes();
     packet = new DatagramPacket(data, data.length);
     socket = new DatagramSocket(325, InetAddress.getByName("192.168.0.26"));
     socket.send(packet);
 }
 catch (Exception ex)
 {
     ...
 }

These two options came when I was making tests sending data to and from my own PC. My IP address is 192.168.0.26. With the first one I had no problems, but the second one throws two kinds of exceptions. I can verify that Client1 works running the next code in the computer at the same time

(Server)

byte data[] = null;
try
{
    data= new byte[1000];
    socket = new DatagramSocket(325);
    packet = new DatagramPacket(datos, datos.length);
    socket.receive(packet);
    System.out.println(datos);
}
catch (Exception ex)
{
    ...
}

If I run Server first and then the Client1 option, it actually recieves the data. When I run the Client1 first and then the Server, obviously don't get any data but neither get any exceptions. The real problem (and actual question, sorry about so many, perhaps useless, info) is with Client2. If this is ran before Server, I get NullPointerException:

java.lang.NullPointerException: null address || null buffer
at java.net.DualStackPlainDatagramSocketImpl.send(DualStackPlainDatagramSocketImpl.java:115)
at java.net.DatagramSocket.send(DatagramSocket.java:676)

generated when method send() is invoked. I know maybe I should ignore this; even thought I'm a little bit newbie with Socket's, it seems wrong to send data if you know nobody will recieve it. In the other hand, when Server runs first and then Client2, I get BindException:

java.net.BindException: Address already in use: Cannot bind
at java.net.DualStackPlainDatagramSocketImpl.socketBind(Native Method)
at java.net.DualStackPlainDatagramSocketImpl.bind0(DualStackPlainDatagramSocketImpl.java:65)
at java.net.AbstractPlainDatagramSocketImpl.bind(AbstractPlainDatagramSocketImpl.java:95)
at java.net.DatagramSocket.bind(DatagramSocket.java:376)
at java.net.DatagramSocket.<init>(DatagramSocket.java:231)
at java.net.DatagramSocket.<init>(DatagramSocket.java:284)

from the line when the socket is initialized. Due to that Client1 works, I'll use it; but I'd really like to know why does it work but Client2 doesn't. I read in another forum here about that BindException, and I understood it was caused when you attempt to use one port which is already in use, but in this sense Client1 should fail too. Could someone explain me the difference between Client1 and Client2?

Upvotes: 2

Views: 2452

Answers (1)

Diego C Nascimento
Diego C Nascimento

Reputation: 2801

You are binding to the same port, if you are running it on the same machine/address, you need to bind to different ports.

Like on the server:

socket = new DatagramSocket(325);    //server binds its UDP/IP socket to port 325

On the client 2:

socket = new DatagramSocket(325, InetAddress.getByName("192.168.0.26"));

you are binding to the same port on the same machine/address that the "server" is using

On client 1:

packet = new DatagramPacket(data, data.length, InetAddress.getByName("192.168.0.26"), 325);
//you are specifying the destination address and port the packet should be sent.
socket = new DatagramSocket();
//but in the DatagramSocket contructor you don't pass any address or port to bind to, so it will use an available port, not 325 as it is in use by the server.

As you said you are newly to network communication, be in mind that this is UDP/IP sockets, there's TCP/IP and others. UDP/IP is not connection oriented by specification, and does not guarantee the packet to be delivered.

Upvotes: 2

Related Questions