Java static function

Question

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>().

`T extends A` <- Right there – Engineer2021 Mar 27 '14 at 12:42 — Engineer2021, Mar 27 '14 at 12:42
Why should it output something else? – Maroun Mar 27 '14 at 12:43 — Maroun, Mar 27 '14 at 12:43
I thought that the code should output "A1" because B() – dr_yand Mar 27 '14 at 12:47 — dr_yand, Mar 27 '14 at 12:47
@dr_yand I added it to your question. – Maroun Mar 27 '14 at 12:48 — Maroun, Mar 27 '14 at 12:48

score 3 · Answer 1 · answered Mar 27 '14 at 12:47

3

Your T.v() is a static method call and compiles into A.v() because T erases to A according to its upper type bound.

answered Mar 27 '14 at 12:47

Marko Topolnik

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score 2 · Answer 2 · edited Mar 27 '14 at 12:48

2

You can't override static methods in Java, you can only hide them.

Read this: Overriding and Hiding Methods

edited Mar 27 '14 at 12:48

Maroun

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answered Mar 27 '14 at 12:47

Zsolt Süli

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score 0 · Answer 3 · edited May 23 '17 at 11:57

0

Shortly java does not support overwritting static methods:

here is a possible duplicate explaining better the case: Why doesn't Java allow overriding of static methods?

edited May 23 '17 at 11:57

Community

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answered Mar 27 '14 at 12:49

pezetem

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score 0 · Answer 4 · answered Mar 27 '14 at 12:49

0

Static methods are not dispatched dynamically. The exact method to be called is resolved on compile time and it's A due to erasure.

answered Mar 27 '14 at 12:49

aljipa

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score 0 · Answer 5 · answered Mar 27 '14 at 12:52

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.

score 0 · Answer 6 · answered Mar 27 '14 at 13:02

0

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".

answered Mar 27 '14 at 13:02

roll1987

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Java static function

6 Answers6