How come this equals & hashCode override does NOT cause an exception or error?

Question

When I compile and run the code below, I get the following results:

o1==o2 ? true
Hash codes: 0 | 0
o1==o2 ? true
Hash codes: 1 | 8
o1==o2 ? true
Hash codes: 7 | 3
o1==o2 ? true
Hash codes: 68 | 10
o1==o2 ? true
Hash codes: 5 | 4

From what I've read, if two objects are equal, their hashCodes must also be equal. So, how does this code not cause an exception or error?

import java.io.*;
import java.lang.*;

public class EqualsAndHashCode {

    private int num1;
    private int num2;

    public EqualsAndHashCode(int num1, int num2) {
        this.num1 = num1;
        this.num2 = num2;
    }

    public static void main(String[] args) {
        for (int x=0; x < 5; x++) {
            EqualsAndHashCode o1 = new EqualsAndHashCode(x, x);
            EqualsAndHashCode o2 = new EqualsAndHashCode(x, x);
            System.out.println("o1==o2 ? " + o1.equals(o2));
            System.out.println("Hash codes: " + o1.hashCode() + " | " + o2.hashCode());
        }
    }

    public boolean equals(Object o) {
        return (this.getNum1() == ((EqualsAndHashCode)o).getNum1()) && (this.getNum2() == ((EqualsAndHashCode)o).getNum2());
    }

    public int hashCode() {
        return (int)(this.getNum1() / Math.random());
    }

    public int getNum1() { return num1; }
    public int getNum2() { return num2; }
}

EDIT I The premise behind my question was the wording surrounding the hashCode contract (http://docs.oracle.com/javase/6/docs/api/java/lang/Object.html#hashCode()):

If two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce the same integer result.

I assumed that this rule would have been enforced by the JVM at compile or run time and I would have seen errors or exceptions right away when the contract was violated...

I've read language in books that state if two objects are equal, their hashCodes MUST be equal...I guess I was expecting to see an error or exception thrown by the JVM with this code — Zack Macomber
As the answers state, this is a recommendation. How would the JVM enforce such a thing, though? Should it randomly call equals and hashCode on any object in the environment during the application run? What if the calculations for equality have nontrivial implementations and consume a lot of CPU time? — Tom G
@TomG - I like your points. I hadn't thought about those questions myself. — Zack Macomber

Darth Android Darth Android · Accepted Answer · 2013-07-01T15:59:48

Because the JVM does not check or validate that the method contract holds true. They're just methods, and they can return whatever they want.

However, any code which depends upon them supporting the method contract might or will fail. You will not be able to use your EqualsAndHashCode objects in a HashMap, for example. That will throw exceptions or will not return correct values in most cases.

This is the same thing with compareTo() and TreeMaps - compareTo() can return any int that it wants, but if it doesn't return a consistent ordering as defined by the method contract in the Comparable interface, then your TreeMap will throw exceptions as soon as it detects inconsistencies.

How come this equals & hashCode override does NOT cause an exception or error?

7 Answers