Here is the code:
interface hi
{
public void meth1();
}
abstract class Hullo
{
public abstract void meth1();
}
public class Hello extends Hullo implements hi
{
public void meth1(){}
}
Question:The code compiles and everything. I wanted to know the meth1() in class Hello is overriding which meth1()?
The ont in the interface or the one in the abstract class and why?
The answer is short: Both.....
In fact, to be correct: You are overriding none of them, you are implementing them both, with one method.
Generally we override a existing method which already has some definition. I mean we are adding a extra bit of feature in child class compared to super class. Since both methods are abstract, so we can say that we implementing the unimplemented method.
You can take the reference of creating threads in Java where we prefer to implement Runnable interface compared to extending Thread class.
Absolutely correct question.
Here both the interface and abstract class have same method.
You have one class name is hello and extends abstract class and implement interface its true and you override meth1 method on hello class its fine and its compile correctly and not given any error but her you can't identify which class method is override like abstract class or interface.
This is runtime polymorphism you can't create object of abstract class and interface but you can create reference variable of that. Here solution is you can't identify that on compile time its actual override at runtime.
interface hi
{
public void meth1();
}
abstract class Hullo
{
public abstract void meth1();
}
public class Hello extends Hullo implements hi
{
public void meth1(){
System.out.println("hello");
}
hi h= new Hello();
h.meth1();//its means interface method is override. and its decide when we call method.
hullo hu= new Hello();
hu.meth1();//its means abstract class method is override.
}
Related
this is my inteface.
public interface ConnectionListener{
public void onConnectionReady();
public void onConnectionDown();
}
I implement this interfaces in HomeActivity class.I would like to learn what is the differences between using #Override in implemented methods and not using #Override annotation...
public class HomeActivity implements ConnectionListener
{
#Override
public void onConnectionReady() {
}
#Override
public void onConnectionDown() {
}
}
#Override only shows the compiler that you would like to override a method. If the method signature is not known in a super class or an implemented interface you get a compile time error.
At runtime there is no difference.
For more Information see the javadoc.
If you have class B which extends class A (B extends A) can you use #Override to "replace" some method. You can modify how methods work in class B without replace it in class A. In overriden method you can invoked method from class A too by using "super" method and modify how it will work after super usage. In life you can use class Animal to extend class Dog and Cat. If Animal class had method getSound() you can for Dog override it for dog sound and for Cat for a cat sound. But Dog and Cat will have all artifacts which have animal like legs or other parts of body.
Interface you can use when you try to standardize some group class there will be same methods name but in interface you didn't implement a code of a method.
You can read more about it here:
Implements vs extends: When to use? What's the difference?
You will find the best usage of #Override tag when using it with static methods during inheritance.
Suppose class A has a static method EAT, class B extends class A and creates a static method EAT in class B. Now class A and B both have static method EAT with different implementation but it is not overridden, the moment you try to tag it with #Override, it shows you an error which means the sub-class has the same signature and different implementations but not overridden.
I am just confused about abstract class concept. Please clear my doubt. Definition of Abstract class says we can not create object of such class, then what we called like A a = new A() { }. Example is below:
public abstract class AbstractTest {
public abstract void onClick();
public void testClick() {
}
}
public class A {
AbstractTest test = new AbstractTest() {
#Override
public void onClick() {
}
};
}
Then test is a object or what?
test is an object of an anonymous concrete sub-class of AbstractTest (note that it implements all the abstract methods of AbstractTest), which is why this sub-class can be instantiated.
On the other hand,
AbstractTest test = new AbstractTest();
wouldn't pass compilation, since that would be an attempt to instantiate an abstract class.
You are mixing up object and reference.
AbstractTest test = new AbstractTest() {
#Override
public void onClick() {
}
};
test here is a reference to a anonymous class that extends AbstractTest, the above code is like saying:
class MyClass extends AbstractTest {
#Override
public void onClick() {
}
}
AbstractTest test = new MyClass(); // test is a reference to a MyClass object
Abstract Class in my opinion needs to be explained together with Interface.
Interface allows you to specify operations that are supported/allowed on objects with that interface. More specifically Objects are instances of a Class which implements that Interface. Defining an interface allows you to describe a group of different classes of objects so that other objects can interact with them in the same manner.
Abstract Class is one step between interface and a Class (loosely speaking). Abstract Class allows you to specify operations that are supported by classes that extend it, but it also allows you to implement (some of) those operations. This way you can implement common methods for a group of classes in that abstract class. Other methods in the abstract class that are not implemented (aka abstract methods) need to be implemented by the class that extends it. The fact that you didn't implement all methods on an Abstract class naturally means you can't instantiate it (create an object of such class). There are other useful implementations for Abstract classes (i.e. callbacks).
In your example what you see there is that you are not really trying to just create an object that Abstract class you are also providing implementation of abstract method onClick();
That is the only way you can "create an instance of the abstract class" - technically speaking you are creating an instance of an Anonymous class (that is extending your abstract class) for which you provide implementation of inherited abstract methods.
Below is the code snippet:
public abstract class MyAbstractClass {
public abstract void a();
public abstract void b();
}
public class Foo extends MyAbstractClass {
public void a() {
System.out.println("hello");
}
public void b(){
System.out.println("bye");
}
}
public class Bar extends MyAbstractClass {
public void a() {
System.out.println("hello");
}
public void delta() {
System.out.println("gamma");
}
}
There are couple of questions that I have:
Q-1 :- Should I implement ALL the methods in abstract class?
Q-2 :- Can the implementing class have its own methods?
When you extend an Interface or an Abstract class you are creating a contract of sorts with that superclass. In the contract you are saying:
"I will implement all unimplemented methods in my superclass"
If you do not, implement all the unimplemented methods, then you are breaking your contract. A way to not break your contract is make your subclass Abstract as well as a way of saying
"I have not implemented all the classes in my contract, I am going to
have my subclasses implement them".
For your class bar right now, you must implement b() or make bar an Abstract class or you are not fulfilling your contract with MyAbstractClass
The basic idea is:
Interface: None of my methods are implemented. A subclass must implement all my methods in order to implement me. (Note: I believe default interfaces have been added to Java 8 which may change this a bit)
Example:
public interface myInterface
{
//My subclasses must implement this to fulfill their contract with me
public void methodA();
//My subclasses must implement this to fulfill their contract with me
public void methodB();
}
Abstract: I may implement some of my methods, but I will also leave methods as abstract so that my subclasses must implement because they can implement those classes to suit their needs better than I can.
Example:
public abstract class myAbstractClass
{
//My subclasses must implement this to fulfill their contract with me
public abstract void methodC();
public void helloWorld()
{
System.out.println("Hello World");
}
}
Abstract classes can also extend interfaces so they can implement some of their methods. But they can also leave some of the methods unimplemented so the subclass can implement them. If you leave an interface method unimplemented, there is not need to declare it abstract, it is already in the contract.
Example:
public abstract class myAbstractClass2 implement myInterface
{
#Override
public void methodA()
{
// this fulfills part of the contract with myInterface.
// my subclasses will not need to implement this unless they want to override
// my implementation.
}
//My subclasses must implement this to fulfill their contract with me
public abstract void methodD();
}
So in essence, an abstract class doesn't have as strict a contract with it's superclass because it can delegate its methods to its subclasses.
Regular Class: (I use regular to mean non-interface, and non-abstract). I must implement all unimplemented methods from all of my superclasses. These classes have a binding contract.
Example:
public class mySubClass extends myAbstractClass2
{
#Override
public void methodB()
{
//must be implemented to fulfill contract with myInterface
}
#Override
public void methodD()
{
//must be implemented to fulfill contract with myAbstractClass2
}
public void myMethod()
{
//This is a method specifically for mySubClass.
}
}
Q-1:- Should I implement all methods in abstract class?
Yes, you must implement all abstract methods.
Q-2 :- Can the implementing class have its own methods?
Yes, you can declare own (more specfic) methods.
You not only should, but have to implement all abstract methods (if the subclass is non-abstract). Otherwise an object of that subclass wouldn't know what to do if that method was called!
The only way to prevent this is if the subclass is also declared abstract, so that it cannot be instantiated in the first place.
You don't have to implement all methods of an abstract class. But you must implement all abstract methods of it.
In fact extending an abstract class has no difference then extending a normal class. It's not like implementing interfaces. Since you're extending you are creating a subclass thus you can add as many methods and attributes as you need.
Ya definately implementing class can define its own method as well and if are not implementing all the methods of your abstract class in the derived class then mark this derived class also as Abstract
but at the end of chain you have to make one concrete class which implements all the method that was not implement in abstract sub-parent
public interface I{
public void m();
}
public abstract class A1 implements I{
//No need to implement m() here - since this is abstract
}
public class B1 extends A1{
public void m(){
//This is required, since A1 did not implement m().
}
}
public abstract class A11 extends A1{
//Again No need to implement m() here - since this is abstract
public abstract void newA11Method()
}
public class B11 extends A11{
//This class needs to implement m() and newA11Method()
}
Yes, the implementing class need only implement the methods labeled as abstract in the abstract class.
Yes you must implement all the methods present in an abstract class. As the purpose of abstract class is purely to create a template for the functions whose implementation is decided by the class implementing them. So if you don't implement them, then you are breaking the concept of abstract class.
To answer your second question, yes you can create any number of your own methods irrespective of the abstract class you are extending.
Yes, When you are implementing an Interface you will have to implement all its methods. That is the purpose of defining an Interface. And yes, you can have your own methods in the class where you implement an Interface.
Yes, when you extends abstract you should give implementation for all abstract methods which are presents in abstract class. Othrewise you should make it implementation class as abtract class.
You must implement all the abstract methods you have in the abstract class. But you can directly use the concrete methods that are implemented.
For more information please refer to the :
https://www.javatpoint.com/abstract-class-in-java
Is there any way to NOT implement all of the methods of an interface in an inheriting class?
The only way around this is to declare your class as abstract and leave it to a subclass to implement the missing methods. But ultimately, someone in the chain has to implement it to meet the interface contract. If you truly do not need a particular method, you can implement it and then either return or throw some variety of NotImplementedException, whichever is more appropriate in your case.
The Interface could also specify some methods as 'default' and provide the corresponding method implementation within the Interface definition (https://docs.oracle.com/javase/tutorial/java/IandI/defaultmethods.html). These 'default' methods need not be mentioned while implementing the Interface.
The point of an interface is to guarantee that an object will outwardly behave as the interface specifies that it will
If you don't implement all methods of your interface, than you destroy the entire purpose of an interface.
We can override all the interface methods in abstract parent class and in child class override those methods only which is required by that particular child class.
Interface
public interface MyInterface{
void method1();
void method2();
void method3();
}
Abstract Parent class
public abstract class Parent implements MyInterface{
#Override
public void method1(){
}
#Override
public void method2(){
}
#Override
public void method3(){
}
}
In your Child classes
public class Child1 extends Parent{
#Override
public void method1(){
}
}
public class Child2 extends Parent{
#Override
public void method2(){
}
}
I asked myself the same question, and then learned about Adapters. It solved my problem, maybe it can solve yours. This explains it very well : https://blogs.oracle.com/CoreJavaTechTips/entry/listeners_vs_adapters
You can do that in Java8. Java 8 introduces “Default Method” or (Defender methods) new feature, which allows a developer to add new methods to the Interfaces without breaking the existing implementation of these interfaces.
It provides flexibility to allow Interface define implementation which will use as default in the situation where a concrete Class fails to provide an implementation for that method.
interface OldInterface {
public void existingMethod();
default public void DefaultMethod() {
System.out.println("New default method" + " is added in interface");
}
}
//following class compiles successfully in JDK 8
public class ClassImpl implements OldInterface {
#Override
public void existingMethod() {
System.out.println("normal method");
}
public static void main(String[] args) {
ClassImpl obj = new ClassImpl ();
// print “New default method add in interface”
obj.DefaultMethod();
}
}
Define that class as an abstract class. However, you must implement those unimplemented methods when you want to create an instance of it (either by using a subclass or an anonymous class).
It is possible and it is easy. I coded an example.
All you have to do is inherit from a class that does implement the method. If you don't mind a class that is not instantiable, then you can also define an abstract class.
If you want an instantiable class, it is not possible. You may try to define an abstract class, though.
If you try to implement an interface and you find yourself in a situation where there is no need to implement all of them then, this is a code smell. It indicates a bad design and it violates Liskov substitution principle. Often this happens because of using fat interface.
Also sometimes this happens because you are trying to implement an interface from an external dependency. In this case, I always look inside the source code to see if there is any implementation of that interface which I can either use it directly or subclass it and override methods to my needs.
We can use Adapter classes ,which reduces complexcity by not making mandatory to implement all the methods present in the interface
Adapter class is a simple java class that implements an interface with only EMPTY implementation .
Instead of implementing interface if we extends Adapter class ,we provide implementation only for require method
ex--- instead of implementing Servlet(I) if we extends GenericServlet(AC) then we provide implementation for Service()method we are not require to provide implementation for remaining meyhod..
Generic class Acts as ADAPTER class for Servlet(I).
yes possible below shown is the way
interface Test {
void m() throws NullPointerException;
}
class Parent {
// Parent class doesn't implements Test interface
public void m() {
System.out.println("Inside Parent m()");
}
}
class Child extends Parent implements Test {
}
public class Program {
public static void main(String args[]) {
Child s = new Child();
s.m();
}
}
I was wondering if the next thing is possible for implementation:
Lets say I've got 2 interfaces while each one of them has 1 function header.
For example, iterface1 has function g(...) and interface2 has function f(...)
Now, I make a class and declaring that this class is implementing these 2 interfaces.
In the class I try doing the next thing:
I start implementing function g(...) and in it's implementation I make a local class that implements interface2 and I add to this class the implementation of f(...).
I'm not quite sure what you mean. I am picturing something like this:
interface Interface1
{
public void g();
}
interface Interface2
{
public void f();
}
class MyClass implements Interface1, Interface2
{
#Override
public void g()
{
class InnerClass implements Interface2
{
#Override
public void f()
{
}
}
}
}
Is that what you meant?
In this case, the answer is no. The inner class (InnerClass) works fine, but it doesn't count as an implementation of f for the outer class. You would still need to implement f in MyClass:
MyClass.java:11: MyClass is not abstract and does not override abstract method
f() in Interface2
Yes it is legal. However a class does not implement an interface because an inner class implements it. The class must implement the interface explicitly or declare itself as abstract.
Yes, it's legal. In the example you've given, your class should implement all methods of both interfaces, and your local class should implement all methods of interface2.