Enthuware questions

[k reeta]

From what I have understood if a risky method in a try block throws an exception, the rest of the try block is not executed. Instead, the catch/finally blocks are executed. But what about the rest of the code in the same method as the try block.

public class TestClass29
{
   public static void main(String args[])
   {
      int k = 0;
      try{
         int i = 5/k;

      }
      catch (ArithmeticException e){
         System.out.println("1");
      }
      catch (RuntimeException e){
         System.out.println("2");
         return ;
      }
      catch (Exception e){
         System.out.println("3");
      }
      finally{
         System.out.println("4");
      }
      System.out.println("5");
   }
}

The output of the above code is 1 4 5, which implies that the code after "finally" is executed.

BUT here,

public class TestClass28
{
 static String j = "";
 public static void method( int i)
 {
  try
  {
   if(i == 2)
   {
     throw new Exception();
   }
   j += "1";
  }
  catch (Exception e)
  {
   j += "2";
   return;
  }
  finally
  {
   j += "3";
  }
  j += "4";
 }
 public static void main(String args[])
 {
  method(1);
  method(2);
  System.out.println(j);
 }
}

The output is 13423, which implies that the code after finally is not executed.

What am I missing?

Thanks!

[Sanjeev Ba]

From Kathy Sierra book.

A finally block encloses code that is always executed at some point after the try block, whether an exception was thrown or not.
Even if there is a return statement in the try block, the finally block executes right after the return statement is encountered, and before the return executes!

Based on the code sample provided, I think the behaviour is right.

Hit if(i==2) -> throw exception, handle exception print "2" and return, but just before returning, execute finally -> print 3.

Hope this clarifies.

[k reeta]

I completely missed the return statement! Makes sense now, thank you!