Static method hiding

Question

Look at the code below :

class Marsupial {
    public static boolean isBiped() {
        return false;
    }

    public void getMarsupialDescription() {
        System.out.println("Value of this : " + this.getClass().getName() + " , Marsupial walks on two legs: " + isBiped());
    }
}

public class Kangaroo extends Marsupial {
    public static boolean isBiped() {
        return true;
    }

    public void getKangarooDescription() {
        System.out.println("Value of this : " + this.getClass().getName() + ", Kangaroo hops on two legs: " + isBiped());
    }

    public static void main(String[] args) {
        Kangaroo joey = new Kangaroo();
        joey.getMarsupialDescription(); // Question here
        joey.getKangarooDescription();
    }

}

The output is :

Value of this : Kangaroo , Marsupial walks on two legs: false
Value of this : Kangaroo, Kangaroo hops on two legs: true

Now why in the call of getMarsupialDescription(), it will choose static method of Marsupial and not of Kangaroo especially when this points to Kangaroo ?

Static methods are never called virtually. They're always bound at compile-time. The point of static methods is that you don't *need* a `this` pointer to call it. — Silvio Mayolo, Apr 08 '18 at 17:47
Possible duplicate of [Overriding vs Hiding Java - Confused](https://stackoverflow.com/questions/10594052/overriding-vs-hiding-java-confused) — Bogdan Lukiyanchuk, Apr 08 '18 at 17:58
@SilvioMayolo does this mean static method hiding happens at compile time only as static binding will bind it at compile time ? — Number945, Apr 08 '18 at 20:56
I'm not sure I fully understand your question, but I believe you're correct. Static methods are always resolved at compile time, so if you used the same name in a subclass then that, too, would be resolved at compile time. — Silvio Mayolo, Apr 08 '18 at 20:57

Number945 · Accepted Answer · 2018-04-11T10:47:44.910

To put simply association of method call to the method body is type of binding. [What is binding ?] There are two types of binding: Static Binding that happens at compile time and Dynamic Binding that happens at runtime.

Static Binding or Early Binding

The binding which can be resolved at compile time by compiler is known as static or early binding. The binding of static, private and final methods is compile-time. Why? The reason is that the these method cannot be overridden and the type of the class is determined at the compile time. Lets see an example to understand this:

class Human{
   public static void walk()
   {
       System.out.println("Human walks");
   }
}
class Boy extends Human{
   public static void walk(){
       System.out.println("Boy walks");
   }
   public static void main( String args[]) {
       /* Reference is of Human type and object is
        * Boy type
        */
       Human obj = new Boy();
       /* Reference is of HUman type and object is
        * of Human type.
        */
       Human obj2 = new Human();
       // At compile time it gets decided which body of method it will call
       obj.walk(); 
       obj2.walk();
   }
}

Why at compile time ? As it is static. Static binding is done by Type of reference not by the type of object.

Output:

Human walks
Human walks

Similarly in the code in question above - isBiped() static method is linked to the body of the method during compile time only. Hence it calls Marsupial isBiped().

What is static and dynamic binding ?

Source of above answer

Static method hiding

1 Answers1