How to ensure hashCode() is consistent with equals()?

Question

When overriding the equals() function of java.lang.Object, the javadocs suggest that,

it is generally necessary to override the hashCode method whenever this method is overridden, so as to maintain the general contract for the hashCode method, which states that equal objects must have equal hash codes.

The hashCode() method must return a unique integer for each object (this is easy to do when comparing objects based on memory location, simply return the unique integer address of the object)

How should a hashCode() method be overriden so that it returns a unique integer for each object based only on that object's properities?

public class People{
   public String name;
   public int age;

   public int hashCode(){
      // How to get a unique integer based on name and age?
   }
}
/*******************************/
public class App{
   public static void main( String args[] ){
       People mike = new People();
       People melissa = new People();
       mike.name = "mike";
       mike.age = 23;
       melissa.name = "melissa";
       melissa.age = 24;
       System.out.println( mike.hasCode() );  // output?
       System.out.println( melissa.hashCode(); // output?
   }
}

Answer 1

It doesn't say the hashcode for an object has to be completely unique, only that the hashcode for two equal objects returns the same hashcode. It's entirely legal to have two non-equal objects return the same hashcode. However, the more unique a hashcode distribution is over a set of objects, the better performance you'll get out of HashMaps and other operations that use the hashCode.

IDEs such as IntelliJ Idea have built-in generators for equals and hashCode that generally do a pretty good job at coming up with "good enough" code for most objects (and probably better than some hand-crafted overly-clever hash functions).

For example, here's a hashCode function that Idea generates for your People class:

public int hashCode() {
    int result = name != null ? name.hashCode() : 0;
    result = 31 * result + age;
    return result;
}