Java disable stack trace on a specific call?

Question

I want to disable one error message produces by a socket timeout, because it's filling my console (Linux, terminal).

The code:

public ThreadListenResponseQuery(DatagramSocket socket){
   this.socket = socket;
   try {
      socket.setSoTimeout(1500);
   } catch (SocketException e) {
      // TODO Auto-generated catch block
      e.printStackTrace();
   }
}

@Override
public void run(){
   System.out.println("Waiting for response...");

   // Waiting for 60 seconds...
   while(System.currentTimeMillis() < startTime + 60000){

      socket.receive(receive_packet);   
      // .. Additional work

   }
}

The program is waiting for 60 seconds to get any response. I am setting a timeout on the socket because the while cycle is freezing out if no response message is coming.

Error:

java.net.SocketTimeoutException: Receive timed out at java.net.DualStackPlainDatagramSocketImpl.socketReceiveOrPeekData(Native Method) at java.net.DualStackPlainDatagramSocketImpl.receive0(Unknown Source) at java.net.AbstractPlainDatagramSocketImpl.receive(Unknown Source) at java.net.DatagramSocket.receive(Unknown Source) at thread.ThreadListenResponseQuery.run(ThreadListenResponseQuery.java:46) at java.lang.Thread.run(Unknown Source)

Answer 1

Surround socket.receive with try-catch:

try {
    socket.receive()
} catch (SocketTimeoutException e) {
    // Log an error, abort an communication, or just ignore
}

Answer 2

Take a look at this.

As written by Alex W, it's possible to catch a Throwable object in Java without a stack trace:

 Throwable(String message, Throwable cause, boolean enableSuppression,boolean 
writableStackTrace) 

Answer 3

If you just suppress the entire exception will let really hard to figure out what is really going on. Instead I would just print a log output that could be something like this:

try {
   // invocation that throws the exception
} catch (SocketException e) {
   System.out.println(e.getMessage());
}