Question

This program is correct and does compile and run. But why does method ‘a’ not have a throws declaration?

class Exception1 {
      public void a() 
        {
            int array[] = new int[5];
            try
            {
                System.out.println("Try");
                array[10]=1;
            }
            catch (Exception e)
            {
               System.out.println("Exception");

                throw e;
            }
            finally
            {
                System.out.println("Finally");
                return;
            }
        }
    public static void main(String[] args) 
        {
            Exception1 e1 = new Exception1();

            try {
                e1.a();
            } 
            catch(ArrayIndexOutOfBoundsException e)
            {
                System.out.println("Catch");
            }
            System.out.println("End of main");
        }
}

Answer 1

The problem is the return in the finally block:

Since the finally will always be executed and it will always complete abruptly (either with an unchecked exception or with a return), there is no way that the throw e in the catch-block (or any unchecked exception in the try block) could ever be propagated downwards on the call stack.

If you remove the return, then you’ll notice that the compiler will not accept the code, stating that Exception is not declared to be thrown on the method a().