Option #2 fulfills the conditions of the hashcode contract:
(1) Whenever it is invoked on the same object more than once during an execution of a Java application, the hashCode method must consistently return the same integer, provided no information used in equals comparisons on the object is modified. This integer need not remain consistent from one execution of an application to another execution of the same application.
(2) If two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce the same integer result.
(3) It is not required that if two objects are unequal according to the equals(java.lang.Object) method, then calling the hashCode method 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 hashtables.
So '0' response will work here since its consistent (condition 1), it always produces same hashcode for equal objects (condition 2), and though it produces same hashcode for unequal objects it is tolerated by the hashcode contract (condition 3).
For 2nd question there are no arguments to be passed in hashcode() method.
Joined: Nov 28, 2009
Thanks for your quick and elaborate reply.Can you please elaborate condition 1 further.It is still unclear to me.
Joined: Jan 26, 2011
Condition 1 really means when you create an object instance, you will always get the same value in the hashcode provided the object's hasn't been changed in a way its equal method comparisons are done. So if object's members (instance variables) have been changed and they are used in the equals method to calculate object equality, the hashcode should reflect that, otherwise it should stay same. Note that hashcode need not be same from one execution of the program to another brand new execution of the same program.