Question

Eclipse source menu has a “generate hashCode / equals method” which generates functions like the one below.

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

@Override
public boolean equals(Object obj)
{
    if (this == obj)
        return true;
    if (obj == null)
        return false;
    if (getClass() != obj.getClass())
        return false;
    CompanyRole other = (CompanyRole) obj;
    if (name == null)
    {
        if (other.name != null)
            return false;
    } else if (!name.equals(other.name))
        return false;
    return true;
}

If I select multiple fields when generating hashCode() and equals() Eclipse uses the same pattern shown above.

I am not an expert on hash functions and I would like to know how “good” the generated hash function is? What are situations where it will break down and cause too many collisions?

Answer 1

You can see the implementation of hashCode function in java.util.ArrayList as

public int hashCode() {
    int hashCode = 1;
    Iterator<E> i = iterator();
    while (i.hasNext()) {
        E obj = i.next();
        hashCode = 31*hashCode + (obj==null ? 0 : obj.hashCode());
    }
    return hashCode;
}

It is one such example and your Eclipse generated code follows a similar way of implementing it. But if you feel that you have to implement your hashCode by your own, there are some good guidelines given by Joshua Bloch in his famous book Effective Java. I will post those important points from Item 9 of that book. Those are,

1. Store some constant nonzero value, say, 17, in an int variable called result.

2. For each significant field f in your object (each field taken into account by the equals method, that is), do the following:

   a. Compute an int hash code c for the field:

   i. If the field is a boolean, compute (f ? 1 : 0).

   ii. If the field is a byte, char, short, or int, compute (int) f.

   iii. If the field is a long, compute (int) (f ^ (f >>> 32)).

   iv. If the field is a float, compute Float.floatToIntBits(f).

   v. If the field is a double, compute Double.doubleToLongBits(f), and then hash the resulting long as in step 2.a.iii.

   vi. If the field is an object reference and this class’s equals method compares the field by recursively invoking equals, recursively invoke hashCode on the field. If a more complex comparison is required, compute a “canonical representation” for this field and invoke hashCode on the canonical representation. If the value of the field is null, return 0 (or some other constant, but 0 is traditional)

   vii. If the field is an array, treat it as if each element were a separate field.
   That is, compute a hash code for each significant element by applying
   these rules recursively, and combine these values per step 2.b. If every
   element in an array field is significant, you can use one of the
   Arrays.hashCode methods added in release 1.5.

   b. Combine the hash code c computed in step 2.a into result as follows:

      result = 31 * result + c;
   

3. Return result.

4. When you are finished writing the hashCode method, ask yourself whether
   equal instances have equal hash codes. Write unit tests to verify your intuition!
   If equal instances have unequal hash codes, figure out why and fix the problem.

Java language designers and Eclipse seem to follow similar guidelines I suppose. Happy coding. Cheers.