Java Exception Handling understanding issue
<p>I'm not able to understand this program. I expect it to output "Hello World", but instead it prints only the "World".  I thought that first the <code>try</code> block would execute, printing "Hello" and " ", then afterward when it encounters a <code>1/0</code>, it would throw an <code>ArithmeticException</code>.  The exception would be caught by <code>catch</code> block, then "World" would be printed.</p>

<p>The program is as follows.     </p>

<pre><code> import java.util.*;
 class exception{
     public static void main(String args[]) 
     {
         try
         {
             System.out.println("Hello"+" "+1/0);
         } 
         catch(ArithmeticException e) 
         {
             System.out.println("World");
         }
     }
 }
</code></pre>