Question

Consider the following test case, is it a bad practice to use the hashCode method inside of equals as a convenient shortcut?

public class Test 
{    
    public static void main(String[] args){
        Test t1 = new Test(1, 2.0, 3, new Integer(4));
        Test t2 = new Test(1, 2.0, 3, new Integer(4));
        System.out.println(t1.hashCode() + "\r\n"+t2.hashCode());
        System.out.println("t1.equals(t2) ? "+ t1.equals(t2));
    }
    
    private int myInt;
    private double myDouble;
    private long myLong;
    private Integer myIntObj;
    
    public Test(int i, double d, long l, Integer intObj ){
        this.myInt = i;
        this.myDouble = d;
        this.myLong = l;
        this.myIntObj = intObj;
    }
    
    @Override
    public boolean equals(Object other)
    {        
        if(other == null) return false;
        if (getClass() != other.getClass()) return false;            
        
        return this.hashCode() == ((Test)other).hashCode();//Convenient shortcut?
    }

    @Override
    public int hashCode() {
        int hash = 3;
        hash = 53 * hash + this.myInt;
        hash = 53 * hash + (int) (Double.doubleToLongBits(this.myDouble) ^ (Double.doubleToLongBits(this.myDouble) >>> 32));
        hash = 53 * hash + (int) (this.myLong ^ (this.myLong >>> 32));
        hash = 53 * hash + (this.myIntObj != null ? this.myIntObj.hashCode() : 0);
        return hash;
    }   
}

Output from main method:

1097562307
1097562307
t1.equals(t2) ? true

Answer 1

In general, it’s not at all safe to compare the hashCode instead of using equals. When equals returns false, hashCode may return the same value, per the contract of hashCode.