Always override hashCode when you override equals

Here is the contract, copied from the Objectspecification [JavaSE6]:
• Whenever it is invoked on the same object more than once during an execution of an application, the hashCodemethod must consistently return the
same integer, provided no information used in equalscomparisons on the
object is modified. This integer need not remain consistent from one execution of an application to another execution of the same application.
• If two objects are equal according to the equals(Object)method, then calling the hashCodemethod on each of the two objects must produce the same
integer result. 
• It is not required that if two objects are unequal according to the equals(Object)method, then calling the hashCodemethod on each of the two objects
must produce distinct integer results. However, the programmer should be
aware that producing distinct integer results for unequal objects may improve
the performance of hash tables.

Here is a recipe.

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

2. For each significant field fin your object (each field taken into account by the equalsmethod, that is), do the following:

  a. Compute an inthash code cfor 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 longas 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 hashCodeon the field. If a more complex comparison is required, compute a “canonical representation” for this field and invoke hashCodeon the canonical representation. If the value of the field is null, return 0(or some other constant, but 0is 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.hashCodemethods added in release 1.5.

  b. Combine the hash code ccomputed in step 2.a into resultas follows:
result = 31 * result + c;

3. Return result.

4. When you are finished writing the hashCodemethod, 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.

For example:

// Lazily initialized, cached hashCode
private volatile int hashCode; // (See Item 71)
@Override public int hashCode() {
int result = hashCode;
if (result == 0) {
result = 17;
result = 31 * result + areaCode;
result = 31 * result + prefix;
result = 31 * result + lineNumber;
hashCode = result;
}
return result;
}