Reputation: 309

Java static function

I have some classes.

class A{    
    static void v(){
        System.out.println("A");
    }
}

class A1 extends A{
    static void v(){
        System.out.println("A1");
    }
}

class B <T extends A>{
     void v(){
         T.v();
     }
}

Why following code outputs "A"?

B b = new B<A1>();
b.v();

I thought that the code should output "A1" because B<A1>().

Upvotes: 3

Answers (6)

roll1987

Reputation: 183

You didn't override the function v() for class T. Since T is a subclass of A, the effect of calling T.v() equals A.v() which outputs "A".

Upvotes: 0

Bohemian

Reputation: 424983

The issue here is that calls to static methods are statically bound at compile time to the declared type of the variable. In this case, the compiler knows that T is an A.

It doesn't matter that the generic type of the local object is A1 - the binding is made inside B, where A is the only information the compiler has about the type.

Upvotes: 0