Question

How is the hashcode computed for a card object which consists of enum suit and enum rank ?

Answer 1

If you’re using Eclipse, it can generate a “good enough” hashCode() implementation for you:

public class Card
{
    private Suit suit;
    private Rank rank;

    @Override
    public int hashCode()
    {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((rank == null) ? 0 : rank.hashCode());
        result = prime * result + ((suit == null) ? 0 : suit.hashCode());
        return result;
    }

    @Override
    public boolean equals(Object obj)
    {
        if (this == obj) return true;
        if (!(obj instanceof Card)) return false;
        Card other = (Card) obj;
        if (rank != other.rank) return false;
        if (suit != other.suit) return false;
        return true;
    }
}

I can only imagine that NetBeans, IntelliJ IDEA, etc., can do this as well.

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That said, since the domain is small, this implementation will work equally well (I think…):

public int hashCode()
{
    int rankHash = ((rank == null) ? 0 : (1+rank.ordinal()));
    int suitHash = ((suit == null) ? 0 : (1+suit.ordinal()));
    return rankHash + 31*suitHash;
}

This assumes that Rank ordinals are 0-12 inclusive and Suit ordinals are 0-3 inclusive. Note that most of the ugliness there comes from the null checks. If the values can never be null, then:

public int hashCode()
{
    return rank.ordinal() + 31*suit.ordinal();
}