Guess the Output ??

Question

Hi guys ,
Can anyone please explain me the out put of this code snippet ..
class A { }
class B extends A { }
class C extends B {
static void doSomething ( A x , A y ) {System.out.print("AA");}
static void doSomething ( A x , B y ) {System.out.print("AB");}
static void doSomething ( B x , A y ) {System.out.print("BA");}
static void doSomething ( B x , B y ) {System.out.print("BB");}
static void doSomething ( C x , C y ) { System.out.println("CC");}
public static void main(String[] args) {
doSomething(null,null);
Answer is : CC
I didnt understand how its working , please show some light .
Additional Notes : if u comment out ..static void doSomething ( C x , C y ) { System.out.println("CC");}
then it wud print "BB" and if u comment out the BB line , then it would throw error ???
Thnx
Suren

Answer 1

Suren_singh_khatana wrote:
Thanks for the quick turn around !! Could you please explain what do you mean by the term "more specific" here ?
How (c,c) is more specific then (B,B) because method is invoked from class C ??Like I said: "furthest" from object. The most derived class.
C is a subclass of B, so C is "more specific" or "more derived" or "further from object" than B. Likewise, B is a subclass of A, so it's more specific than A.
void foo(Object o) {}
void foo (String s) {}
Object o = new Object();
String s = "abc";
foo(o);
foo(s);In foo(o), it has to be the foo(Object) that's called, because o is not a string.
In foo(s), either one would work, since s is a String and s is also an Object. foo(String) is what gets called, because it's a more precise match than foo(Object).