method overriding using Interface

Question

 interface I { void show(); } class A implements I { void show() { System.out.println("class A"); } public static void main(String s[]) { I i=new A(); i.show(); i.toString(); } }

Q> As interface I does not contain the abstract method toString() but still The following code gets compiled. How?

when super class variable is used to refer sub class obj then compiler first searches the similar method in the super class if not found gives error. here Interface does not contain the method toString().

ex=>

class A { void show() { System.out.println("show"); } } class B { void show() { System.out.println("show B"); } void display() { System.out.println("display B"); } public static void main(String s[]) { A a=new B(); a.show(); //will execute a.display(); //give error }

Michael Lloyd Lee mlk · Accepted Answer · 2009-09-21 11:31:10Z

All classes inherit from Object. Object has a toString.

To use any interface it must be backed by a actual class. So the Java compiler knows that it can use any method defined in java.lang.Object when dealing with an Interface.

To put it a slightly different way:

interface I { ... }

has an "magic"

interface I extends Object { ... }

So you can use Objects methods when detail with I. However you can not use any methods in the concrete class that do not appear in the interface. So to combine you two examples:

interface Car { void drive(); } class Convertible implements Car { void drive() {} void openRoof() {} public static void main() { Car porscheBoxster = new Convertible(); porscheBoxster.drive(); // OK - exists in interface porscheBoxster.toString(); // OK - exists in java.lang.Object. porscheBoxster.openRoof(); // Error. All we know is the porscheBoxster is of type Car. // We don't know if it is a Convertible or not. } }

when super class variable is used to refer sub class obj then compiler first searches the similar method in the super class if not found gives error. here Interface does not contain the method toString(). ex=> class A { void show() { System.out.println("show A"); } } class B { void show() { System.out.println("show B"); } void display() { System.out.println("display B"); } public static void main(String s[]) { A a=new B(); a.show(); //will execute a.display(); //give error }
All classes inherit from Object in Java. So Java knows when it encounters an interface it will be backed by a concrete class and so will have the methods defined in java.lang.Object

Romain Linsolas · Accepted Answer · 2009-09-21 10:31:01Z

Every class in Java is an Object, thus, they are always able to run the following methods:

clone()
equals(Object)
finalize()
getClass()
hashCode()
notify()
notifyAll()
toString()
wait()
wait(long)
wait(long, int)

NawaMan · Accepted Answer · 2009-09-21 10:28:43Z

Because 'toString()' is in the class Object which every non-primitive data is derived from. So every object has this method.

Chii · Accepted Answer · 2009-09-21 10:35:14Z

In Java, every class you construct, inherits from the base class Object. This means that your class by default will have a lot of useful methods, amongst others toString().

Collectives™ on Stack Overflow

method overriding using Interface

4 Answers 4

2 Comments

Comments

Comments

Comments

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

2 Comments

Comments

Comments

Comments

Related