Handing Arguments To Custom Exception

Question

Hi,
I've written the simple custom exception class below
public class JAMException extends Exception
     public void JAMException(String area, String type)
          System.out.println("There is a " + type + " problem in " + area);
}As you can see I want to be able to take in two string arguments which will be used in the output message to the user.
I utilise the exception like so..
try
      //something
catch (JAMException("Virtual Band","InterruptedException"))     //Area, Type
}However the code above results in the following error
<identifier> expected
catch (JAMException("Virtual Band","InterruptedException"))I've never really seem a exception take in arguments so Im wondering if this is possible? Also I want the JAMException class to be a generic exception class to handle IOExceptions, EOFExceptions etc.. Is the above approach correct?
Help would be really appreciated!

Answer 1

Thanks for the advice, I've got my JAMException working as you suggested above but when it comes to implementing it, there are certain places where I can't throw it, for example the run() method in a thread. when I attempt to do this i get "overridden method does not throw JAMException". I understand the logic behind this I was just wondering how I can get around it...
Below is an example of a method that works
private boolean connect() throws JAMException
     exl.addMessage("Virtual Band");
     try
          data_group = InetAddress.getByName(data_address);
          data_socket = new MulticastSocket(data_port);
          data_socket.setTimeToLive(10);
          System.out.println("TTL: " + data_socket.getTimeToLive());
          data_socket.joinGroup(data_group);
          data_socket.setSoTimeout(500); // 500 ms timeout on functions like receive and send..
          Random r = new Random();
          int x = (int)(r.nextFloat() * 255.0);
          voice_address = voice_address + x;
          System.out.println("Sending my audio to multicast ip=" + voice_address);
          System.out.println("Connect ok!");
          return true;
     catch (Exception e)
          throw new JAMException("Virtual Band", "General Exception");
//          e.printStackTrace();
          //return false;
This is the way I was doing the run() method, doesn't work
public void run() throws JAMException
     System.out.println(">>>>>>> THREAD RUNNING");
          long last = System.currentTimeMillis() - send_intervall;           while (thread != null) {
          if (System.currentTimeMillis() - last > send_intervall)
               try
                    String msg = "alive:" + my_name + ":" + voice_address + ":";
                    DatagramPacket hi = new DatagramPacket(msg.getBytes(), msg.length(),
                    data_group, data_port);
                    data_socket.send(hi);
                    last = System.currentTimeMillis();
               catch(Exception e)
                    throw new JAMException("Virtual Band", "General Exception");
}Anybody got any idea how to get around this problem??