Is it possible to create an object for an interface? If yes, how is it done?
According to my view the following code says that we can:
Runnable r = new Runnable() {
// some implementation
}
This is not creating the instance of Interface, it is creating a class that implements interface. So when you write:
Runnable runnable = new Runnable() {
#Override
public void run() {
// TODO Auto-generated method stub
}
};
You are actually a creating a class that is implementing the Runnable interface.
You need to follow all rules here, here, we are overriding the run method for Runnable. There is similar thing for abstract class also. We can test using an example:
public abstract class AbstractClass {
public void someMethod() {
System.out.println("abstract class");
}
}
and another class i.e. TestClass:
public class TestClass {
public static void main(String[] args) {
AbstractClass abstractClass = new AbstractClass() {
public void someMethod() {
System.out.println("concrete class method");
}
};
abstractClass.someMethod();
}
}
This will create the instance of a subclass in which we are overriding someMethod();
This program prints:
concrete class method
This proves we are creating the instance of subclass.
You can't instantiate an interface directly, but you can instantiate a class that implements that interface:
public class RunClass implements Runnable {
// Class implementation
}
Runnable r = new RunClass();
This is basically the same as what you're doing inline. The brackets after new Runnable() will contain your implementation inline.
You can create an anonymous inner class:
Runnable r = new Runnable() {
#Override
public void run() {
}
};
Therefore you create a new class which implements the given interface.
Is it possible to creating object for an interface?
No. The code you've shown creates an object from an anonymous class, which implements the interface. Under the covers, the JVM actually creates a class implementing the interface, and then creates an instance of that class.
The "anonymous" class generated will actually have a name, based on the name of the class in which this code appears, for instance YourClass$1 or similar. E.g.:
public class AnonymousName {
public static final void main(String[] args) {
Runnable r = new Runnable() {
public void run() {
}
};
System.out.println(r.getClass().getName());
}
}
...outputs
AnonymousName$1
(At least on Oracle's JVM; I don't know if the naming convention is in the JLS or if it's JVM-specific behavior.)
Here's my understanding.
An interface
public Interface SomeInterface{
}
You can declare an Object for an interface.
SomeInterface anObject;
You cannot instantiate this object directly using this interface.
However, let's say you have a class that implements this interface.
public class SomeClass implements SomeInterface {}
Then you can do this,
anObject = new someClass();
(This is conceptually of course (like pseudocode) actual code may vary depending on your classes and access modifiers etc.)
I'll update as to why exactly we are doing this/what the point is as soon as I find out.
Note: Some of this has been mentioned in above answers, just want the OP to know this whole thing too.
we can not instatiate the interface (since do not have constructor).
What you are seeing is an anonymous inner class.
it’s creating an instance of a new, anonymous,implementer of Runnable class.
Because an anonymous class definition is an expression, it must be part of a statement.
Related
I'm learning abstract classes vs interfaces at the moment and trying to figure out situations where to use one over the other. I'm having trouble figuring out this example at the moment:
public interface Face {
public void test();
}
public abstract class Tract {
public void test() {
System.out.println("over here");
}
}
public class Thing extends Tract implements Face {
public void test() {
// what should print out?
}
}
Here, the test() function is implemented in the abstract class. If you don't implement it in the subclass, would it call the abstract class' method and print out "over here"? Does the interface accept implementations from an ancestor class or do you have to implement it in the subclass, therefore overriding the abstract class implementation?
All the interface cares about is that the class has implemented a method called test() that returns void. It does not matter whether the method is implemented in the class directly or in any ancestor (parent) class.
In your case, the Thing class has inherited its definition of test() from Tract, and therefore implements the Face interface without you having to provide a definition explicitly.
In the class "Tract" you have given an implementation for the method coming from the interface. Also you override it in "Thing" class so when calling this method on a Thing instance then this version(Thing version) is going to be called.
All java methods are virtual.
lets consider little bit modified code,
I hope, you will get the idea:
public interface Face {
public void test();
}
public abstract class Tract {
public void test() {
System.out.println("Tract here");
}
}
public class Thing extends Tract implements Face {
public void test() {
System.out.println("Thing here");
}
}
public class Thing2 extends Tract implements Face {
}
lets go to output:
Tract tr = new Tract();
tr.test();
will not compile because you can't instantiate abstract class.
Thing th = new Thing();
th.test();
will print "Thing here"
Thing2 th2 = new Thing2();
th2.test();
will print "Tract here",
because you not overwritten the test() method in abstract class.
Main idea of this approach - you can abstract implementation in the future use
class C {
void print(Face face) {
face.test();
}
}
new C(new Thing()).print();
will print "Thing here";
new C(new Thing2()).print();
will print "Tract here";
You can hide different implementations
But this is not main idea of abstract classes.
main idea abstract classes are:
public interface Face {
public void test();
}
public abstract class Abstract {
abstract public void test();
}
public class Thing1 extends Abstract implements Face {
public void test() {
System.out.println("Thing1 here");
}
}
public class Thing2 extends Abstract implements Face {
public void test() {
System.out.println("Thing2 here");
}
}
main idea - you can declare method without implementation
new C(new Thing1()).print();
will print "Thing1 here";
new C(new Thing2()).print();
will print "Thing2 here";
main idea - you declare the method in abstract class, that you MUST override to compile code.
I hope, this is enough explained answer.
I went to an interview. Interviewer asked me if one can instantiate an interface and abstract class? As per my knowledge I said "No". But he said "Yes, we can with the help of an anonymous class".
Can you please explain to me how?
This was a trick questions.
No you can not instantiate an interface or abstract class.
But you can instantiate an anonymous class that implements/extends the interface or abstract class without defining a class object. But it is just a shortcut to defining a fully named class.
So I would say technically your answer was correct.
I don't know what is "instantiation of interface and abstract class".
I think it's an inaccurate, improper expression of something,
we can only guess at the intended meaning.
You cannot create an instance of an interface or an abstract class in Java.
But you can create anonymous classes that implement an interface or an abstract class.
These won't be instances of the interface or the abstract class.
They will be instance of the anonymous class.
Here's an example iterator from the Iterator interface that gives you an infinity of "not really":
new Iterator<String>() {
#Override
public boolean hasNext() {
return true;
}
#Override
public String next() {
return "not really";
}
};
Or a funky AbstractList that contains 5 "not really":
List<String> list = new AbstractList<String>() {
#Override
public int size() {
return 5;
}
#Override
public String get(int index) {
return "yes";
}
};
Assume you have an abstract class: MyAbstractClass with abstract void method myAbstractMethod. Then you can make an "instance" of this class via this code:
MyAbstractClass myAbstractClassInstance = new MyAbstractClass() {
public void myAbstractMethod() {
// add abstract method implementation here
}
};
myAbstractClassInstance extends your MyAbstractClass in this case. When you instantiate this class you have to implement all abstract methods as you can see from the code above.
The same way works for interfaces, assume you have an interface MyInterface with a void method myInterfaceMethod inside, then you can create an "instance" (implementation of this instance) via this code:
MyInterface myInterfaceImpl = new MyInterface() {
public void myInterfaceMethod() {
// add method implementation here
}
}
myInterfaceImpl is an implemetation of MyInterface in this case. When you create an object using interface, you have to implement interface methods as it is shown above.
Interface :
interface Interface1 {
public void m1();
}
When you right
new Interface1() {
public void m1() {
}
}
Its not actually creating the instance of Interface. Its creating an instance of its subtype which doesnt have any name/reference. Hence we cannot create an instance of interface or abstract class
You cannot create instances of abstract classes or interfaces using the new operator. For example,
new AbstractSet(); // That's wrong.
You can, however, use them to declare reference variables. For example, You can do this:
AbstractSet set;
You can instantiate anonymous as well as declared implementing classes or subclass.
For example, Set extends AbstractSet, so you can instantiate Set.
Yes, we can create by having defining the abstract methods or the interface methods on the fly during instantiation. That's like a Named anonymous class.
//interface
Runnable r = new Runnable(){
public void run() {
System.out.println("Here we go");
}
};
//Abstract class
abstract class MyAbstract {
abstract void run();
}
MyAbstract ab = new MyAbstract(){
#Override
void run() {
System.out.println("Here we go");
}};
I have these 2 classes
class A {
public void foo1() {
...;
foo2();
...;
}
protected abstract foo2();
}
class B extends A {
public foo2() {
......
}
I need foo2 to be static so I can do B.foo2() but I also want the functionality in class A to remain.n
Any suggestions?
}
You can't override static methods or implement abstract methods as static.
Static methods are defined on a class definition, not on a class instance. Abstract methods are defined on a class instance.
What you said doesn't make sense in fact.
Although I don't quite get why you need to do it, there is a workaround:
class B {
#Override
public void foo() {
fooUtil();
}
public static void fooUtil() {
// your impl here
}
}
Then you can do B.fooUtil() instead, and using its behavior to override A.foo().
I am trying to wrap my head around interfaces, and I was hoping they were the answer to my question.
I have made plugins and mods for different games, and sometimes classes have onUpdate or onTick or other methods that are overridable.
If I make an interface with a method, and I make other classes which implement the method, and I make instances of the classes, then how can I call that method from all the objects at once?
You'll be looking at the Observer pattern or something similar. The gist of it is this: somewhere you have to keep a list (ArrayList suffices) of type "your interface". Each time a new object is created, add it to this list. Afterwards you can perform a loop on the list and call the method on every object in it.
I'll edit in a moment with a code example.
public interface IMyInterface {
void DoSomething();
}
public class MyClass : IMyInterface {
public void DoSomething() {
Console.WriteLine("I'm inside MyClass");
}
}
public class AnotherClass : IMyInterface {
public void DoSomething() {
Console.WriteLine("I'm inside AnotherClass");
}
}
public class StartUp {
private ICollection<IMyInterface> _interfaces = new Collection<IMyInterface>();
private static void Main(string[] args) {
new StartUp();
}
public StartUp() {
AddToWatchlist(new AnotherClass());
AddToWatchlist(new MyClass());
AddToWatchlist(new MyClass());
AddToWatchlist(new AnotherClass());
Notify();
Console.ReadKey();
}
private void AddToWatchlist(IMyInterface obj) {
_interfaces.Add(obj);
}
private void Notify() {
foreach (var myInterface in _interfaces) {
myInterface.DoSomething();
}
}
}
Output:
I'm inside AnotherClass
I'm inside MyClass
I'm inside MyClass
I'm inside AnotherClass
Edit: I just realized you tagged it as Java. This is written in C#, but there is no real difference other than the use of ArrayList instead of Collection.
An interface defines a service contract. In simple terms, it defines what can you do with a class.
For example, let's use a simple interface called ICount. It defines a count method, so every class implementing it will have to provide an implementation.
public interface ICount {
public int count();
}
Any class implementing ICount, should override the method and give it a behaviour:
public class Counter1 implements ICount {
//Fields, Getters, Setters
#Overide
public int count() {
//I don't wanna count, so I return 4.
return 4;
}
}
On the other hand, Counter2 has a different oppinion of what should count do:
public class Counter2 implements ICount {
int counter; //Default initialization to 0
//Fields, Getters, Setters
#Overide
public int count() {
return ++count;
}
}
Now, you have two classes implementing the same interface, so, how do you treat them equally? Simple, by using the first common class/interface they share: ICount.
ICount count1 = new Counter1();
ICount count2 = new Counter2();
List<ICount> counterList = new ArrayList<ICount>();
counterList.add(count1);
counterList.add(count2);
Or, if you want to save some lines of code:
List<ICount> counterList = new ArrayList<ICount>();
counterList.add(new Counter1());
counterList.add(new Counter2());
Now, counterList contains two objects of different type but with the same interface in common(ICounter) in a list containing objects that implement that interface. You can iterave over them and invoke the method count. Counter1 will return 0 while Counter2 will return a result based on how many times did you invoke count:
for(ICount current : counterList)
System.out.println(current.count());
You can't call a method from all the objects that happen to implement a certain interface at once. You wouldn't want that anyways. You can, however, use polymorphism to refer to all these objects by the interface name. For example, with
interface A { }
class B implements A { }
class C implements A { }
You can write
A b = new B();
A c = new C();
Interfaces don't work that way. They act like some kind of mask that several classes can use. For instance:
public interface Data {
public void doSomething();
}
public class SomeDataStructure implements Data {
public void doSomething()
{
// do something
}
}
public static void main(String[] args) {
Data mydataobject = new SomeDataStructure();
}
This uses the Data 'mask' that several classes can use and have certain functionality, but you can use different classes to actually implement that very functionality.
The crux would be to have a list that stores every time a class that implements the interface is instantiated. This list would have to be available at a level different that the interface and the class that implements it. In other words, the class that orchestrates or controls would have the list.
An interface is a contract that leaves the implementation to the classes that implements the interface. Classes implement the interface abide by that contract and implement the methods and not override them.
Taking the interface to be
public interface Model {
public void onUpdate();
public void onClick();
}
public class plugin implements Model {
#Override
public void onUpdate() {
System.out.println("Pluging updating");
}
#Override
public void onClick() {
System.out.println("Pluging doing click action");
}
}
Your controller class would be the one to instantiate and control the action
public class Controller {
public static void orchestrate(){
List<Model> modelList = new ArrayList<Model>();
Model pluginOne = new plugin();
Model plugTwo = new plugin();
modelList.add(pluginOne);
modelList.add(plugTwo);
for(Model model:modelList){
model.onUpdate();
model.onClick();
}
}
}
You can have another implementation called pluginTwo, instantiate it, add it to the list and call the methods specified by the interface on it.
Suppose that I have interface MyInterface and 2 classes A, B which implement MyInterface.
I declared 2 objects: MyInterface a = new A() , and MyInterface b = new B().
When I try to pass to a function - function doSomething(A a){} I am getting an error.
This is my code:
public interface MyInterface {}
public class A implements MyInterface{}
public class B implements MyInterface{}
public class Tester {
public static void main(String[] args){
MyInterface a = new A();
MyInterface b = new B();
test(b);
}
public static void test(A a){
System.out.println("A");
}
public static void test(B b){
System.out.println("B");
}
}
My problem is that I am getting from some component interface which can be all sorts of classes and I need to write function for each class.
So one way is to get interface and to check which type is it. (instance of A)
I would like to know how others deal with this problem??
Thx
Can you not just have a method on the interface which each class implements? Or do you not have control of the interface?
This would provide both polymorphism and avoid the need to define any external methods. I believe this is the intention of an interface, it allows a client to treat all classes implementing it in a non type specific manner.
If you cannot add to the interface then you would be best introducing a second interface with the appropriate method. If you cannot edit either the interface or the classes then you need a method which has the interface as a parameter and then check for the concrete class. However this should be a last resort and rather subverts the use of the interface and ties the method to all the implementations.
It sounds like you are after something like this:
public static void test(MyInterface obj){
if(obj instanceof A) {
A tmp = (A)obj;
} else if(obj instanceof B) {
B tmp = (B)obj;
} else {
//handle error condition
}
}
But please note this is very bad form and indicates something has gone seriously wrong in your design. If you don't have control of the interface then, as suggested by marcj, adding a second interface might be the way to go. Note you can do this whilst preserving binary compatibility.
I'm unclear on what you're actually asking, but the problem is that you don't have a method that takes a parameter of type MyInterface. I don't know what the exact syntax is in Java, but you could do something like if (b is B) { test(b as B) } but I wouldn't. If you need it to be generic, then use the MyInterface type as the variable type, otherwise use B as the variable type. You're defeating the purpose of using the interface.
I'm not sure if I fully understand the issue, but it seems like one way might be to move the test() methods into the child classes:
public interface MyInterface {
public void test();
}
public class A implements MyInterface{
public void test() {
System.out.println("A");
}
}
public class B implements MyInterface{
public void test() {
System.out.println("B");
}
}
public class Tester {
public static void main(String[] args){
MyInterface a = new A();
MyInterface b = new B();
b.test();
}
}
You could similarly use a toString() method and print the result of that. I can't quite tell from the question, though, if your requirements make this impossible.
I think visitor design pattern will help you out here. The basic idea is to have your classes (A and B) call the appropriate method themselves instead of you trying to decide which method to call. Being a C# guy I hope my Java works:
public interface Visitable {
void accept(Tester tester)
}
public interface MyInterface implements Visitable {
}
public class A implements MyInterface{
public void accept(Tester tester){
tester.test(this);
}
}
public class B implements MyInterface{
public void accept(Tester tester){
tester.test(this);
}
}
public class Tester {
public static void main(String[] args){
MyInterface a = new A();
MyInterface b = new B();
a.accept(this);
b.accept(this);
}
public void test(A a){
System.out.println("A");
}
public void test(B b){
System.out.println("B");
}
}
Use only one public class/interface in one .java file, otherwise it'll throw error. And call the object with the object name.. You declared two methos in Teater class only, then what the purpose of declaring class A,B.
I usually use an abstract class to get around this problem, like so:
public abstract class Parent {}
public class A extends Parent {...}
public class B extends Parent {...}
That allows you to pass Parent objects to functions that take A or B.
You have 3 options:
Visitor pattern; you'll need to be able to change the MyInterface type to include a method visit(Visitor) where the Visitor class contains lots of methods for visiting each subclass.
Use if-else inside your method test(MyInterface) to check between them
Use chaining. That is, declare handlers ATester, BTester etc, all of which implement the interface ITester which has the method test(MyInterface). Then in the ATester, check that the type is equal to A before doing stuff. Then your main Tester class can have a chain of these testers and pass each MyInterface instance down the chain, until it reaches an ITester which can handle it. This is basically turning the if-else block from 2 into separate classes.
Personally I would go for 2 in most situations. Java lacks true object-orientation. Deal with it! Coming up with various ways around it usually just makes for difficult-to-follow code.
Sounds like you need either a) to leverage polymorphism by putting method on MyInterface and implementing in A and B or b) some combination of Composite and Visitor design pattern. I'd start with a) and head towards b) when things get unwieldy.
My extensive thoughts on Visitor:
http://tech.puredanger.com/2007/07/16/visitor/
public interface MyInterface {}
public class A implements MyInterface{}
public class B implements MyInterface{}
public class Tester {
public static void main(String[] args){
MyInterface a = new A();
MyInterface b = new B();
test(b); // this is wrong
}
public static void test(A a){
System.out.println("A");
}
public static void test(B b){
System.out.println("B");
}
}
You are trying to pass an object referenced by MyInterface reference variable to a method defined with an argument with its sub type like test(B b). Compiler complains here because the MyInterface reference variable can reference any object which is a sub type of MyInterface, but not necessarily an object of B.There can be runtime errors if this is allowed in Java. Take an example which will make the concept clearer for you. I have modified your code for class B and added a method.
public class B implements MyInterface {
public void onlyBCanInvokeThis() {}
}
Now just alter the test(B b) method like below :
public static void test(B b){
b.onlyBCanInvokeThis();
System.out.println("B");
}
This code will blow up at runtime if allowed by compiler:
MyInterface a = new A();
// since a is of type A. invoking onlyBCanInvokeThis()
// inside test() method on a will throw exception.
test(a);
To prevent this, compiler disallows such method invocation techniques with super class reference.
I'm not sure what are you trying to achieve but it seems like you want to achieve runtime polymorphism. To achieve that you need to declare a method in your MyInterface and implement it in each of the subclass. This way the call to the method will be resolved at run time based on the object type and not on the reference type.
public interface MyInterface {
public void test();
}
public class A implements MyInterface{
public void test() {
System.out.println("A");
}
}
public class B implements MyInterface{
public void test() {
System.out.println("B");
}
}
public class Tester {
public static void main(String[] args){
MyInterface a = new A();
MyInterface b = new B();
b.test(); // calls B's implementation of test()
}
}