Again hashcode() question ?

Question

thanks for your several answers,
but i want a clear and definitive answer for this assertion ?
i want to use instance of ObjectX in a HashTable.
if i overide the equals() methode for objectX, have i to do the same for hashCode(), and if yes how can i generate an integer hascode for each distinct object X.
class ObjectX {
private int x;
private int y;
public boolean equals(ObjectX ox) {
return (this.x == ox.x);
public int hasCode() {
.....how to generate an integer hashCode
}

Answer 1

There is no clear and definitive answer for your question. The only answer is that the hashcode calculation should involve the same variables that are involved in the equals calculation. In some way.
So in your example, two ObjectXes are "equal" if and only if their x variables are equal. So their hashCode() method should return the same value for all ObjectXes that have the same value of x. The easiest way to do this ispublic int hashCode() {
  return x;
}You can see (if you look at it for a bit) that if two ObjectXes are equal, then their x variables are equal, and hence their hashCode() methods return the same value (namely x). Clearly other methods would work just as well -- for example you could return 2x, or x squared.
Now if the equals() method compared both the x and y variables, then your hashCode() method should return some function involving both x and y. As setmuss said, x^y would do. So would x+y and x-y.
As I said, there's no definitive answer for this.