Question

I have something similar to this.

void func() {
  try {
    //socket disconnects in middle of ..parsing packet..
  } catch(Exception ex) {
   if(!ex.getMessage().toString().equals("timeout") || !ex.getMessage().toString().equals("Connection reset")) {
     debug("Exception (run): " + ex.getMessage());
     ex.printStackTrace();
  }
}

Why is it that when I get a connection reset exception or a timeout exception, it still goes inside the condition. I tried without toString and with no luck.

Answer 1

You shouldn’t catch all exceptions and then test the error message of the exception. Instead only catch those exceptions that you intend to handle – for example SocketTimeoutException.

catch (SocketTimeoutException ex)
{
    // Do something...
}

With your current code you may be catching some other type of exception that you weren’t expecting. Currently you will just ignore this exception, not even logging it. This can make it very difficult to debug what is going on. If you have an exception that you can’t handle you should either rethrow it or log it.

I want to catch all exceptions

If you really want to do that then you can write your code as follows:

catch (SocketTimeoutException ex)
{
    // Do something specific for SocketTimeoutException.
}
catch (Exception ex)
{
    // Do something for all other types of exception.
}

---

Regarding your specific error, you have written:

!a.equals(b) || !a.equals(c)

This expression always evaluates to true. What you meant was:

!a.equals(b) && !a.equals(c)

Or equivalently:

!(a.equals(b) || a.equals(c))

Note that by rewriting your code as I suggested above you completely avoid having to write this complicated boolean expression.