Two interfaces with same method names and signatures. But implemented by a single class then how the compiler will identify the which method is for which interface?
Ex:
interface A{
int f();
}
interface B{
int f();
}
class Test implements A, B{
public static void main(String... args) throws Exception{
}
#Override
public int f() { // from which interface A or B
return 0;
}
}
If a type implements two interfaces, and each interface define a method that has identical signature, then in effect there is only one method, and they are not distinguishable. If, say, the two methods have conflicting return types, then it will be a compilation error. This is the general rule of inheritance, method overriding, hiding, and declarations, and applies also to possible conflicts not only between 2 inherited interface methods, but also an interface and a super class method, or even just conflicts due to type erasure of generics.
Compatibility example
Here's an example where you have an interface Gift, which has a present() method (as in, presenting gifts), and also an interface Guest, which also has a present() method (as in, the guest is present and not absent).
Presentable johnny is both a Gift and a Guest.
public class InterfaceTest {
interface Gift { void present(); }
interface Guest { void present(); }
interface Presentable extends Gift, Guest { }
public static void main(String[] args) {
Presentable johnny = new Presentable() {
#Override public void present() {
System.out.println("Heeeereee's Johnny!!!");
}
};
johnny.present(); // "Heeeereee's Johnny!!!"
((Gift) johnny).present(); // "Heeeereee's Johnny!!!"
((Guest) johnny).present(); // "Heeeereee's Johnny!!!"
Gift johnnyAsGift = (Gift) johnny;
johnnyAsGift.present(); // "Heeeereee's Johnny!!!"
Guest johnnyAsGuest = (Guest) johnny;
johnnyAsGuest.present(); // "Heeeereee's Johnny!!!"
}
}
The above snippet compiles and runs.
Note that there is only one #Override necessary!!!. This is because Gift.present() and Guest.present() are "#Override-equivalent" (JLS 8.4.2).
Thus, johnny only has one implementation of present(), and it doesn't matter how you treat johnny, whether as a Gift or as a Guest, there is only one method to invoke.
Incompatibility example
Here's an example where the two inherited methods are NOT #Override-equivalent:
public class InterfaceTest {
interface Gift { void present(); }
interface Guest { boolean present(); }
interface Presentable extends Gift, Guest { } // DOES NOT COMPILE!!!
// "types InterfaceTest.Guest and InterfaceTest.Gift are incompatible;
// both define present(), but with unrelated return types"
}
This further reiterates that inheriting members from an interface must obey the general rule of member declarations. Here we have Gift and Guest define present() with incompatible return types: one void the other boolean. For the same reason that you can't an void present() and a boolean present() in one type, this example results in a compilation error.
Summary
You can inherit methods that are #Override-equivalent, subject to the usual requirements of method overriding and hiding. Since they ARE #Override-equivalent, effectively there is only one method to implement, and thus there's nothing to distinguish/select from.
The compiler does not have to identify which method is for which interface, because once they are determined to be #Override-equivalent, they're the same method.
Resolving potential incompatibilities may be a tricky task, but that's another issue altogether.
References
JLS 8.4.2 Method Signature
JLS 8.4.8 Inheritance, Overriding, and Hiding
JLS 8.4.8.3 Requirements in Overriding and Hiding
JLS 8.4.8.4 Inheriting Methods with Override-Equivalent Signatures
"It is possible for a class to inherit multiple methods with override-equivalent signatures."
This was marked as a duplicate to this question https://stackoverflow.com/questions/24401064/understanding-and-solving-the-diamond-problems-in-java
You need Java 8 to get a multiple inheritance problem, but it is still not a diamon problem as such.
interface A {
default void hi() { System.out.println("A"); }
}
interface B {
default void hi() { System.out.println("B"); }
}
class AB implements A, B { // won't compile
}
new AB().hi(); // won't compile.
As JB Nizet comments you can fix this my overriding.
class AB implements A, B {
public void hi() { A.super.hi(); }
}
However, you don't have a problem with
interface D extends A { }
interface E extends A { }
interface F extends A {
default void hi() { System.out.println("F"); }
}
class DE implement D, E { }
new DE().hi(); // prints A
class DEF implement D, E, F { }
new DEF().hi(); // prints F as it is closer in the heirarchy than A.
As far as the compiler is concerned, those two methods are identical. There will be one implementation of both.
This isn't a problem if the two methods are effectively identical, in that they should have the same implementation. If they are contractually different (as per the documentation for each interface), you'll be in trouble.
There is nothing to identify. Interfaces only proscribe a method name and signature. If both interfaces have a method of exactly the same name and signature, the implementing class can implement both interface methods with a single concrete method.
However, if the semantic contracts of the two interface method are contradicting, you've pretty much lost; you cannot implement both interfaces in a single class then.
Well if they are both the same it doesn't matter. It implements both of them with a single concrete method per interface method.
As in interface,we are just declaring methods,concrete class which implements these both interfaces understands is that there is only one method(as you described both have same name in return type). so there should not be an issue with it.You will be able to define that method in concrete class.
But when two interface have a method with the same name but different return type and you implement two methods in concrete class:
Please look at below code:
public interface InterfaceA {
public void print();
}
public interface InterfaceB {
public int print();
}
public class ClassAB implements InterfaceA, InterfaceB {
public void print()
{
System.out.println("Inside InterfaceA");
}
public int print()
{
System.out.println("Inside InterfaceB");
return 5;
}
}
when compiler gets method "public void print()" it first looks in InterfaceA and it gets it.But still it gives compile time error that return type is not compatible with method of InterfaceB.
So it goes haywire for compiler.
In this way, you will not be able to implement two interface having a method of same name but different return type.
Try implementing the interface as anonymous.
public class MyClass extends MySuperClass implements MyInterface{
MyInterface myInterface = new MyInterface(){
/* Overrided method from interface */
#override
public void method1(){
}
};
/* Overrided method from superclass*/
#override
public void method1(){
}
}
The following two approaches can also be taken to implement both the duplicate methods and avoid ambiguity -
APPROACH 1:
App.java -
public class App {
public static void main(String[] args) {
TestInterface1 testInterface1 = new TestInterface1();
TestInterface2 testInterface2 = new TestInterface2();
testInterface1.draw();
testInterface2.draw();
}
}
TestInterface1.java -
public class TestInterface1 implements Circle {
}
TestInterface2.java -
public class TestInterface2 implements Rectangle {
}
Circle.java -
public interface Circle extends Drawable {
#Override
default void draw() {
System.out.println("Drawing circle");
}
}
Rectangle.java -
public interface Rectangle extends Drawable {
#Override
default void draw() {
System.out.println("Drawing rectangle");
}
}
Drawable.java -
public interface Drawable {
default void draw() {
System.out.println("Drawing");
}
}
Output -
Drawing circle
Drawing rectangle
APPROACH 2:
App.java -
public class App {
public static void main(String[] args) {
Circle circle = new Circle() {
};
Rectangle rectangle = new Rectangle() {
};
circle.draw();
rectangle.draw();
}
}
Circle.java -
public interface Circle extends Drawable {
#Override
default void draw() {
System.out.println("Drawing circle");
}
}
Rectangle.java -
public interface Rectangle extends Drawable {
#Override
default void draw() {
System.out.println("Drawing rectangle");
}
}
Drawable.java -
public interface Drawable {
default void draw() {
System.out.println("Drawing");
}
}
Output -
Drawing circle
Drawing rectangle
To access sub-class method down-casting is needed, is there is a way to achieve this using generic without type-casting in same manner.
public class Main {
public static void main(String[] args){
Animal parrot = new Bird();
((Bird)parrot).fly();
}
}
interface Animal{
void eat();
}
class Bird implements Animal{
#Override
public void eat() {}
public void fly(){}
}
public class Main {
public static void main(String[] args){
Animal parrot = new Bird();
parrot.move();
}
}
interface Animal{
void eat();
void move();
}
class Bird implements Animal{
#Override
public void eat() {}
public void move(){fly();}
public void fly(){}
}
It could work with something like this I guess
Interfaces are meant to add methods to implemented classes without defining actual code for this method, meaning that any implemented class will definitely have the same methods but 2 implemented methods with the same name won't necessarily perform the same action.
To explain it with the current thread it would be:
interface Animal {
void move();
}
class Bird implements Animal{
public void move(){
fly();
}
}
class Dog implements Animal{
public void move(){
walk();
}
}
This way, each class will have its own definition of the move method while in main each method will be called by object.move().
This way of doing things allows to go from a code like this
for (object tmp:objList){
if(tmp.class=="Bird")
tmp.fly();
}
else if (tmp.class=="Dog"){
tmp.walk();
}
...
}
to
for (object tmp:objList){
tmp.move();
}
The parrot class should extend the Bird class, then you can call the fly method from a parrot object directly and without need to cast.
public class Parrot extends Bird{
//have some filed or method
}
Anyone provide suggestion for below mentioned:
in java8 consider an interface having two methods (eg interface1 ,interface 2)
implementing those to many subclass later i want to remove one method interface1 from one of my subclass without affecting other is any possible solution is there?
If your subclass declares that it implements this interface, then you have no choice but to provide implementations for all methods, or declare the class abstract. If you want a concrete class which however does not functionally implement all methods in the interface, then here is one option:
public interface YourInterface {
void method1();
void method2();
}
public class YourSubClass implements YourInterface {
#Override
public void method1() {
// actually do something
}
#Override
public void method2() {
throw new UnsupportedOperationException("method2() is not supported here.");
}
}
Here while we do implement all methods, we throw a runtime exception should a caller try to access method2().
You can do this by providing a default method implementation for the interface1 method in the interface itself.
interface Interface {
default void interface1() {
System.out.println("interface1");
}
void interface2();
}
class Clazz implements Interface {
#Override
public void interface2() {
System.out.println("interface2");
}
}
Depends how you define 'remove one method'.
If you have interface
interface Interface{
void interface1();
void interface2();
}
And for example two subclasses that extend it:
class Class1 implementes Interface {
#Override
public void interface1(){ ... }
#Override
public void interface2(){ ... }
}
class Class2 implementes Interface {
#Override
public void interface1(){ ... }
#Override
public void interface2(){ ... }
}
Then there are two scenarios:
You don't want to implement for example interface1() method in Class2:
You don't want to have interface1() method in Class2
In case of 1. as Robby Cornelissen mentioned, you can simply provide default implementation in Interface:
default void interface1() { /*do default thing*/ }
In case of 2. you need to remove the interface1() method from the Interface.
You can do that by simple moving definition of method interface1() to Class1 (and any other sublass that needs to have it). but that is not really generic approach.
Best is to extract for example Interface1 with method interface1() and use that interface in classes that need to have that method. You will end up with this situation:
interface Interface{
void interface2();
}
interface Interface1{
void interface2();
}
And for example two subclasses that extend it:
class Class1 implementes Interface, Interface1 {
#Override
public void interface1(){ ... }
#Override
public void interface2(){ ... }
}
class Class2 implementes Interface {
#Override
public void interface2(){ ... }
}
I have a 1 Question asked in 1 Interview
class Birds {
public void fly();
public void eat();
public void sleep();
}
Have several other classes extending it like -
Sparrow extends Birds{
overriding fly();// Sparrow can have its own style to fly
overriding eat();
overriding sleep();
}
Eagle extends Birds{
overriding fly();// Eagle can have its own style to fly
}
Ostrich extends Birds {
overriding fly();// Ostrich could not fly !!!
}
How can I restrict such situation ?? Since parent class doesn't aware of its Child class here.
Birds class is extended by several classes some could override Some could not be able to override.
You need to use composition:
Define an interface called, say, Flyable which has a fly() method. Sparrow and Eagle implement that interface. Ostrich does not.
Define an interface called Living, say which has eat() and sleep() methods. All your types implement that interface.
You could point out to your interviewer that you could then have an Aeroplane() object that implements Flyable but not Living.
You can simply add FlyingBird and GroundBird classes and make them abstract.
public class Birds {
public void eat() {
// default implementation
}
public void sleep() {
// default implementation
}
}
abstract class FlyingBird extends Birds {
public abstract void fly();
}
abstract class GroundBird extends Birds {
public abstract void walk();
}
class Sparrow extends FlyingBird {
#Override
public void eat() {
// different implementation
}
#Override
public void sleep() {
// different implementation
}
#Override
public void fly() {
// some impl
}
}
class Eagle extends FlyingBird {
#Override
public void fly() {
// some impl
}
}
class Ostrich extends GroundBird {
#Override
public void walk() {
// walk implementation
}
}
I have an abstract class Automobile which has an unimplemented method called move
so
public abstract class Automobile {
public void move() {
throw new UnsupportedOperationException();
}
}
I have a concrete class which extends my abstract class and implements the move method.My problem is the method keeps throwing an UnsupportedOperationException
public class Car extends Automobile{
int x;
public void move(){
x++;
}
}
It could be for many reasons in your concrete class: maybe your concrete doesn't actually extends Foo? Or maybe it calls super.move() somewhere in its body.
Instead of throwing an exception, the correct way is to define the class and method as abstract to force subclasses to override it.
public abstract class Foo {
public abstract void move();
}
Please note if Foo only has abstract methods, like in the example above, that's an interface that you want, not an abstract class. Also, you should name it to define a behaviour
public interface Moving {
void move();
}
And then:
public class MovingObject implements Moving {
....
#Override
public void move() {
// your implementation
}
....
}
Are you calling super.move() in your implementation class? Eclipse generates that call by default if you used Source->Override/Implement Methods...
Otherwise I think, that you did not override the method correctly.