Why should I override hashCode() when I override equals() method?

Question

Ok, I have heard from many places and sources that whenever I override the equals() method, I need to override the hashCode() method as well. But consider the following piec

Answer 1

Most of the other comments already gave you the answer: you need to do it because there are collections (ie: HashSet, HashMap) that uses hashCode as an optimization to "index" object instances, an those optimizations expects that if: a.equals(b) ==> a.hashCode() == b.hashCode() (NOTE that the inverse doesn't hold).

But as an additional information you can do this exercise:

class Box {
     private String value;
     /* some boring setters and getters for value */
     public int hashCode() { return value.hashCode(); }
     public boolean equals(Object obj) { 
           if (obj != null && getClass().equals(obj.getClass()) { 
               return ((Box) obj).value.equals(value); 
            } else { return false; }
     }
}

The do this:

Set s = new HashSet();
Box b = new Box();
b.setValue("hello");
s.add(b);
s.contains(b); // TRUE
b.setValue("other");
s.contains(b); // FALSE
s.iterator().next() == b // TRUE!!! b is in s but contains(b) returns false

What you learn from this example is that implementing equals or hashCode with properties that can be changed (mutable) is a really bad idea.