Friday, June 19, 2009

A rose by any other name...

Martin Odersky, Lex Spoon, and Bill Venners have collaborated to venture where everyone has gone before. Yes, I mean the age old story about right way to override the equals method in Java. I see you rolling your eyes with the "I know, I know I must override hashcode method as well" bored refrain.

Spare me a few more minutes and I will convince you that you must give the article a serious once over. I for one got a better appreciation of the complexity of the problem -- if you want it addressed fully.

The article warms up around the 3rd of the four pitfalls they identify
  1. Defining equals with the wrong signature
  2. Changing equals without also changing hashCode
  3. Defining equals in terms of mutable fields
  4. Failing to define equals as an equivalence relation
Pitfall #3 discusses what happens when the state of an object that has been placed in a collection is modified. Typical equalsmethod implementation will result in it disappearing from the collection.

public class Point {
private int x, y;
public Point(int i, int j) { x = i; y = j; }
public int getX() { return x; }
public void setX(int i) { x = i; }
public int getY() { return y; }
public void setY(int i) { y = i; }
@Override public int hashCode() { return 41 * (41 + getX()) + getY(); }
@Override public boolean equals(Object other) {
boolean result = false;
if (other instanceof Point) {
Point that = (Point) other;
result = (this.getX() == that.getX() && this.getY() == that.getY());
return result;

HashSet hashSet = new HashSet();
Point p = new Point(1, 2);
System.out.println(hashSet.contains(p)); // <-- Prints false!!

Yep! A forehead smacking moment alright. The article suggests a way around this problem. Which leads us to the 4th pitfall -- that of equivalence. This means that for non-null values, the following is true about the equals method:
  • Reflexivity: x.equals(x)
  • Symmetry: x.equals(y) => y.equals(x)
  • Transitivity:x.equals(y) and y.equals(z) => x.equals(z)
  • Consistence: x.equals(y) always returns the same result
  • x.equals(null) is always false
With clear and concise examples, the article explains the complications of these contracts and how to implement them.


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