immutable objects and lazy initialization.

Question

http://www.javapractices.com/topic/TopicAction.do?Id=29

Above is the article which i am looking at. Immutable objects greatly simplify your program, since they:

Answer 1

What that line means is, since the object is immutable, then the hashCode has to only be computed once. Further, it doesn't have to be computed when the object is constructed - it only has to be computed when the function is first called. If the object's hashCode is never used then it is never computed. So the hashCode function can look something like this:

@Override public int hashCode(){
    synchronized (this) {
        if (!this.computedHashCode) {
            this.hashCode = expensiveComputation();
            this.computedHashCode = true;
        }
    }
    return this.hashCode;
}