This week's book giveaway is in the OO, Patterns, UML and Refactoring forum. We're giving away four copies of Refactoring for Software Design Smells: Managing Technical Debt and have Girish Suryanarayana, Ganesh Samarthyam & Tushar Sharma on-line! See this thread for details.
I saw this question on a mock exam, but I don't always trust their answers. I hear that hashCode questions are big on the certification exam, so a thorough explanation of this question would be helpful. Given two instances of x and y, of a class that correctly implements equals() and hashCode() methods, which two are always equal? (Choose two) A. (x.equals(y) == false) implies (x.hashCode(y) != y.hashCode()) B. (x.hashCode(y) != y.hashCode()) implies (x.equals(y) == false) C. (x.hashCode(y) == y.hashCode()) implies (x.equals(y) == y.equals(x)) D. (x.equals(y) == true) implies (x.hashCode() == y.hashCode())
The equals/hashcode contract says: 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. 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. Following this, options B and D should be correct.