Can we override only one method while implementing Java interface?
An interface in Java is a specification of method prototypes. Whenever you need to guide the programmer or, make a contract specifying how the methods and fields of a type should be you can define an interface.
To create an object of this type you need to implement this interface, provide body for all the abstract methods of the interface and obtain the object of the implementing class.
Example
interface Sample < void demoMethod1(); void demoMethod2(); void demoMethod3(); >public class InterfaceExample implements Sample < public void demoMethod1() < System.out.println("This is demo method-1"); >public void demoMethod2() < System.out.println("This is demo method-2"); >public void demoMethod3() < System.out.println("This is demo method-3"); >public static void main(String args[]) < InterfaceExample obj = new InterfaceExample(); obj.demoMethod1(); obj.demoMethod1(); obj.demoMethod3(); >> Output
This is demo method-1 This is demo method-2 This is demo method-3
While implementing an interface it is mandatory to override all the abstract methods of it, if you skip overriding any of the abstract methods a compile time error will be generated.
Example
interface Sample < void demoMethod1(); void demoMethod2(); void demoMethod3(); >public class InterfaceExample implements Sample < public void demoMethod1() < System.out.println("This is demo method-1"); >public void demoMethod2() < System.out.println("This is demo method-2"); >public static void main(String args[]) < InterfaceExample obj = new InterfaceExample(); obj.demoMethod1(); obj.demoMethod2(); >> Output
InterfaceExample.java:6: error: InterfaceExample is not abstract and does not override abstract method demoMethod3() in Sample public class InterfaceExample implements Sample< ^ 1 error
But, still if you want to override only one abstract method −
You can leave remaining methods unimplemented as −
Example
interface Sample < void demoMethod1(); void demoMethod2(); void demoMethod3(); >public class InterfaceExample implements Sample < public void demoMethod1() < System.out.println("This is demo method-1"); >public void demoMethod2() < >public void demoMethod3() < >public static void main(String args[]) < InterfaceExample obj = new InterfaceExample(); obj.demoMethod1(); obj.demoMethod2(); obj.demoMethod3(); >> Output
Since int is not mandatory to implement default methods of an interface, you can declare remaining methods default as −
Example
interface Sample < void demoMethod1(); default void demoMethod2() < System.out.println("Default demo method 2"); >default void demoMethod3() < System.out.println("Default demo method 3"); >> public class InterfaceExample implements Sample < public void demoMethod1() < System.out.println("This is demo method-1"); >public static void main(String args[]) < InterfaceExample obj = new InterfaceExample(); obj.demoMethod1(); obj.demoMethod2(); obj.demoMethod3(); >> Output
This is demo method-1 Default demo method 2 Default demo method 3
Overriding and Hiding Methods
An instance method in a subclass with the same signature (name, plus the number and the type of its parameters) and return type as an instance method in the superclass overrides the superclass's method.
The ability of a subclass to override a method allows a class to inherit from a superclass whose behavior is "close enough" and then to modify behavior as needed. The overriding method has the same name, number and type of parameters, and return type as the method that it overrides. An overriding method can also return a subtype of the type returned by the overridden method. This subtype is called a covariant return type.
When overriding a method, you might want to use the @Override annotation that instructs the compiler that you intend to override a method in the superclass. If, for some reason, the compiler detects that the method does not exist in one of the superclasses, then it will generate an error. For more information on @Override , see Annotations .
Static Methods
If a subclass defines a static method with the same signature as a static method in the superclass, then the method in the subclass hides the one in the superclass.
The distinction between hiding a static method and overriding an instance method has important implications:
- The version of the overridden instance method that gets invoked is the one in the subclass.
- The version of the hidden static method that gets invoked depends on whether it is invoked from the superclass or the subclass.
Consider an example that contains two classes. The first is Animal , which contains one instance method and one static method:
public class Animal < public static void testClassMethod() < System.out.println("The static method in Animal"); >public void testInstanceMethod() < System.out.println("The instance method in Animal"); >> The second class, a subclass of Animal , is called Cat :
public class Cat extends Animal < public static void testClassMethod() < System.out.println("The static method in Cat"); >public void testInstanceMethod() < System.out.println("The instance method in Cat"); >public static void main(String[] args) < Cat myCat = new Cat(); Animal myAnimal = myCat; Animal.testClassMethod(); myAnimal.testInstanceMethod(); >> The Cat class overrides the instance method in Animal and hides the static method in Animal . The main method in this class creates an instance of Cat and invokes testClassMethod() on the class and testInstanceMethod() on the instance.
The output from this program is as follows:
The static method in Animal The instance method in Cat
As promised, the version of the hidden static method that gets invoked is the one in the superclass, and the version of the overridden instance method that gets invoked is the one in the subclass.
Interface Methods
Default methods and abstract methods in interfaces are inherited like instance methods. However, when the supertypes of a class or interface provide multiple default methods with the same signature, the Java compiler follows inheritance rules to resolve the name conflict. These rules are driven by the following two principles:
- Instance methods are preferred over interface default methods. Consider the following classes and interfaces:
public interface Mythical < default public String identifyMyself() < return "I am a mythical creature."; >>
public class Pegasus extends Horse implements Flyer, Mythical < public static void main(String. args) < Pegasus myApp = new Pegasus(); System.out.println(myApp.identifyMyself()); >>
public interface EggLayer extends Animal < default public String identifyMyself() < return "I am able to lay eggs."; >>
public interface FireBreather extends Animal
public class Dragon implements EggLayer, FireBreather < public static void main (String. args) < Dragon myApp = new Dragon(); System.out.println(myApp.identifyMyself()); >>
If two or more independently defined default methods conflict, or a default method conflicts with an abstract method, then the Java compiler produces a compiler error. You must explicitly override the supertype methods.
Consider the example about computer-controlled cars that can now fly. You have two interfaces ( OperateCar and FlyCar ) that provide default implementations for the same method, ( startEngine ):
public interface OperateCar < // . default public int startEngine(EncryptedKey key) < // Implementation >>
A class that implements both OperateCar and FlyCar must override the method startEngine . You could invoke any of the of the default implementations with the super keyword.
public class FlyingCar implements OperateCar, FlyCar < // . public int startEngine(EncryptedKey key) < FlyCar.super.startEngine(key); OperateCar.super.startEngine(key); >>
The name preceding super (in this example, FlyCar or OperateCar ) must refer to a direct superinterface that defines or inherits a default for the invoked method. This form of method invocation is not restricted to differentiating between multiple implemented interfaces that contain default methods with the same signature. You can use the super keyword to invoke a default method in both classes and interfaces.
Inherited instance methods from classes can override abstract interface methods. Consider the following interfaces and classes:
public class Mustang extends Horse implements Mammal < public static void main(String. args) < Mustang myApp = new Mustang(); System.out.println(myApp.identifyMyself()); >>
The method Mustang.identifyMyself returns the string I am a horse. The class Mustang inherits the method identifyMyself from the class Horse , which overrides the abstract method of the same name in the interface Mammal .
Note: Static methods in interfaces are never inherited.
Modifiers
The access specifier for an overriding method can allow more, but not less, access than the overridden method. For example, a protected instance method in the superclass can be made public, but not private, in the subclass.
You will get a compile-time error if you attempt to change an instance method in the superclass to a static method in the subclass, and vice versa.
Summary
The following table summarizes what happens when you define a method with the same signature as a method in a superclass.
| Superclass Instance Method | Superclass Static Method | |
|---|---|---|
| Subclass Instance Method | Overrides | Generates a compile-time error |
| Subclass Static Method | Generates a compile-time error | Hides |
Note: In a subclass, you can overload the methods inherited from the superclass. Such overloaded methods neither hide nor override the superclass instance methodsthey are new methods, unique to the subclass.