In Java a class can extend only one parent class but can implement multiple interfaces.
With
the introduction of default methods in Java 8 interface, there’s the possibility of a class inheriting more than one
method with the same signature by implementing 2 interfaces having the same default method
This can create diamond problem as in C++
Example in below code the output of
new C().hello(); is
This is Second
public interface First {
default void hello(){
System.out.println("This is First");
}
}
public interface Second extends First {
default void hello(){
System.out.println("This is Second");
}
}
public class MyClass implements First,Second {
public static void main(String[] args) {
new MyClass().hello();
}
}
What are the resolution rules that Java uses to resolve Diamond Problem?
A simple answer like who takes precedence and when will be great.
Following are the rules to follow when a class inherits a method with the same signature from multiple places (another class or interface):
Classes always win. A method declaration in the class or a superclass
takes priority over any default method declaration.
Otherwise, sub-interfaces win: the method with the same signature in the most specific defaultproviding
interface is selected. (for example in your case method from Second interface should run as Second extends First).
Finally, if the choice is still ambiguous, the class inheriting from multiple interfaces has to
explicitly select which default method implementation to use by overriding it and calling the
desired method explicitly.
There is the Oracle Tutorial that explains this pretty much in detail.
You have basically overridden your method from First interface and the most specific interface method is chosen.
Your snippet is no diamond problem as interface second extends interface first and overrides the method hello.
Without extending interface second the compiler throws an error
Duplicate default methods named hello with the parameters () and () are inherited from the types Second and First
Implementing two Interfaces that declare the same default Method leads to an compilation error:
MyClass inherits unrelated defaults for sayHi() from types InterfaceA and InterfaceB
So there's no Diamond-Problem ;)
To solve this you can override the Method in the implementing Class and either implement it there or defer to the correct Interface.
In your case it would look like that (in case you want to use the method of Interface First):
public class MyClass implements First,Second {
public void hello() {
First.super.hello();
}
}
Related
Here is an example of interface which is showing multiple inheritance.And i want to know how can we achieve multiple inheritance by interface and why can't we use it by class?
interface Printable // interface1
{
void print();
}
interface Showable //interface2
{
void print();
}
class TestTnterface1 implements Printable,Showable
{
public void print()
{
System.out.println("Hello");
}
public static void main(String args[])
{
TestTnterface1 obj = new TestTnterface1();
obj.print(); //which print method will be called now?
}
}
First question:
The implementation satisfies both contracts, so whether you cast your concrete class to Printable or Showable, it will be used just the same. You will notice that there is an "impossible" situation, like this:
public interface Printable{
String print();
}
public interface Showable{
void print();
}
public class Impl implements Printable,Showable{
/*impossible to implement because you cannot have the same method signature with two different return types*/
}
The multiple inheritance would usually imply there is something useful added by each parent. For example, if Printable would have the method #print and Showable would have the method #show, inheriting them both would add functionality to your program.
If you want to combine functionality from several concrete classes, you might want to look into composition instead of inheritance.
The answer to the second question is more tricky. You can find a longer discussion here. While it would have been possible for the creators of Java to put in such option, it would have opened the door to some pretty messed up code. Think of the example you gave: what if Printable and Showable had concrete implementations for the #print method? Which implementation should the inheriting class have chosen?
In any case the runtime will call the TestTnterface1#print() method, because it is the implementation (code that can be executed).
An interface is a contract that the class that implements must follow, and is used only at compile time. The compiler checks, if the implementor has (non abstract) methods with same name and signature.
There is only one concrete method that can be called
i.e.
class TestTnterface1 implements Printable,Showable
{
public void print()
{
System.out.println("Hello");
}
Question 1: Which method is called?
There is only one method-implementation that can actually be called. So, there is no ambiguity.
Question 2: How can we achieve multiple inheritance by interface?
You already gave an example for this. The special point about this kind of inheritance is that there can't be multiple implementations (until Java 7). Java 8 in contrast allows default methods in interfaces.
Question 3: Why can't we use it by class?
It avoids inheritance of state.
https://docs.oracle.com/javase/tutorial/java/IandI/multipleinheritance.html gives some more information.
Since both interfaces have print() method and TestInterface1 class has its implementation, it will call its implementation.
public void print()
{
System.out.println("Hello");
}
I understand that in Java static methods are inherited just like instance methods, with the difference that when they are redeclared, the parent implementations are hidden rather than overridden. Fine, this makes sense. However, the Java tutorial notes that
Static methods in interfaces are never inherited.
Why? What's the difference between regular and interface static methods?
Let me clarify what I mean when I say static methods can be inherited:
class Animal {
public static void identify() {
System.out.println("This is an animal");
}
}
class Cat extends Animal {}
public static void main(String[] args) {
Animal.identify();
Cat.identify(); // This compiles, even though it is not redefined in Cat.
}
However,
interface Animal {
public static void identify() {
System.out.println("This is an animal");
}
}
class Cat implements Animal {}
public static void main(String[] args) {
Animal.identify();
Cat.identify(); // This does not compile, because interface static methods do not inherit. (Why?)
}
Here's my guess.
Since Cat can only extend one class if Cat extends Animal then Cat.identify has only one meaning. Cat can implement multiple interfaces each of which can have a static implementation. Therefore, the compiler would not know which one to choose?
However, as pointed out by the author,
Java already has this problem, with default methods. If two interfaces
declare default void identify(), which one is used? It's a compile
error, and you have to implement an overriding method (which could
just be Animal.super.identify()). So Java already resolves this
problem for default methods – why not for static methods?
If I was to guess again, I'd say that with default the implementation is part of Cat's vtable. With static it cannot be. The main function must bind to something. At compile time Cat.identify could be replaced with Animal.identify by the compiler but the code wouldn't match reality if Cat was recompiled but not the class that contains main.
Before Java 8, you couldn't define static methods in an interface. This is heavily discussed in this question. I'm going to refer to this answer (by user #JamesA.Rosen) as to why the Java designers probably didn't want static methods in an interface initially:
There are a few issues at play here. The first is the issue of
declaring a static method without defining it. This is the difference
between
public interface Foo {
public static int bar();
}
and
public interface Foo {
public static int bar() {
...
}
}
Java doesn't allow either, but it could allow the second. The first is
impossible for the reasons that Espo mentions: you don't know which
implementing class is the correct definition.
Java could allow the latter, as long as it treated Interfaces as
first-class Objects. Ruby's Modules, which are approximately
equivalent to Java's Interfaces, allow exactly that:
module Foo
def self.bar
...
end
end
However, since the release of Java 8, you can actually add default and static methods inside an interface.
I'm going to be quoting this source a lot here. This is the initial problem:
Java's interface language feature lets you declare interfaces with
abstract methods and provide implementations of those methods in the
classes that implement the interfaces. You are required to implement
each method, which is burdensome when there are many methods to
implement. Also, after publishing the interface you cannot add new
abstract methods to it without breaking source and binary
compatibility.
This was the solution Java 8 provided default:
Java 8 addresses these problems by evolving the interface to support
default and static methods. A default method is an instance method
defined in an interface whose method header begins with the default
keyword; it also provides a code body. Every class that implements the
interface inherits the interface's default methods and can override
them
And for static:
A static method is a method that's associated with the class in which
it's defined, rather than with any object created from that class.
Every instance of the class shares the static methods of the class.
Java 8 also lets static methods be defined in interfaces where they
can assist default methods.
When you implement an interface that contains a static method, the
static method is still part of the interface and not part of the
implementing class. For this reason, you cannot prefix the method with
the class name. Instead, you must prefix the method with the interface
name
Example:
interface X
{
static void foo()
{
System.out.println("foo");
}
}
class Y implements X
{
}
public class Z
{
public static void main(String[] args)
{
X.foo();
// Y.foo(); // won't compile
}
}
Expression Y.foo() will not compile because foo() is a static member
of interface X and not a static member of class Y.
Static methods in interfaces could create a diamond of death if they were being inherited. So, calling a static method from the appropriate interface is good enough compared to the risk of calling it from a concrete class that may implement multiple interfaces that contain static methods of the same name.
Why are static methods any different?
Static methods are just functions unrelated to the objects. Instead of placing them in utility abstract classes (like calling Collections.sort() ) we move those functions (static methods) to their appropriate interfaces. They could be bound to the inherited objects like the default methods do, but that is not their job.
Static methods provide functionality which is unrelated to the instances of the class.
Example:
interface Floatable {
default void float() {
// implementation
}
static boolean checkIfItCanFloat(Object fl) {
// some physics here
}
}
class Duck implements Floatable { }
So, the point is that a Duck may float but the function that checks if an Object really floats is not something that a Duck can do. It is an irrelevant functionallity that we could pass to our Floatable interface instead of having it sit inside some utility class.
Let's begin with some background ...
Java doesn't support multiple inheritance (the ability to extend more than one class). This is because multiple inheritance is prone to the deadly diamond of death (also known as the diamond problem) which the designers of Java chose to preempt.
If B and C override a method inherited from A, which method does D inherit?
A class can implement multiple interfaces because interface methods are contracted for overriding; if a class C implements two interfaces A and B that declare the same method, then the same method in C will be invoked by clients of either interface (A or B). The introduction of default methods for interfaces in Java 8 was made possible by forcing the implementer to override the default in case of ambiguity. This was an acceptable compromise since default methods are intended to be defensive (to be used if no other implementation is explicitly provided by an implementer). However, since the compiler can’t force you to override a static method (static methods inherently can't be overridden), the introduction of static methods for interfaces in Java came with one restriction: the static methods of an interface are not inherited.
The answer is that static methods belong to a class/interface and there is only one instance for static blocks. that's the reason you can't override any static method.
Even in classes, you may override but only the super class' static method gets executed.
I have an interface, IfcBase which is implemented by another class Base. This class is further extended by a second class SubBase. Further SubBase class implements another interface IfcNew. Both these interfaces have a method declared that has the same signature. Now SubBase does not define the method from IfcNew. I now create an instance of SubBase and assign it to the reference type IfcNew. I then invoke the lone method and get an output. The method from IfcBase was executed in this case. I believe this should not be allowed at some stage, either during compilation or execution. I fail to understand the behavior and solicit help. The source is below. Thanks a lot!
public interface IfcBase
{
public void printString();
}
public class Base implements IfcBase
{
public void printString()
{
System.out.println("Base Class");
}
}
public interface IfcNew
{
public void printString();
}
public class SubBase extends Base implements IfcNew
{
//
}
public class Test
{
public static void main(String[] args)
{
IfcNew i = new SubBase();
i.printString(); //Output:Base Class
}
}
This is how inheritance works in Java.
You have a method called public void printString() implemented in Base, from which you extend SubBase. As a result this implementation will be implicitly available in this class.
For the implements IfcNew part in the SubBase declaration, compiler will only check to see if SubBase has a method implemented which has the same signature as public void printString(). Since it implicitly inherits this implementation from Base, it has nothing to complain about.
This behavior can be easily understood if you look at it from OO design point of view. Please take a look at this article which I wrote few years ago. Look under the section called Method signature, Object Interface, Types, Subtypes and Supertypes. By the definition of subtype that is provided there, SubBase is already a subtype of IfcNew, so compiler has no problem with it.
This is the correct behavior of inheritance in Java.
There is no warning when implementing multiple interfaces with same-signature methods, because resolution takes place at runtime.
In this case, your SubBase class does not need to implement printString as its parent does, even if it isn't the "same" printString: the identical signature allows resolution at runtime.
This is obvious and not strange. This is because the method with same signature is already implemented in the base class and when you extend the base class it is automatically inherited. Now when you try to assign the object to IfcNew reference the method implemented is called which is not overriden in subclass, meaning that the base class method is called which actually is the implementation of IfcBase.
Can Someone Explain how the methods of interface used in classes?
Note: My Doubt is "Methods are already defined in Class then why we should implement it ? "
For Example :
interface printable{
void print();
}
class A implements printable{
public void print(){System.out.println("Hello");}
public static void main(String args[]){
A obj = new A();
obj.print();
}
}
why print() is declared in interface??
You define a method by giving its implementation. They are the same thing, so you are right that once you define a method, you don't also need to implement it.
An interface declares that anything implementing this interface will defined those methods. This is part of the contract for interfaces. This allows you to call any method of an interface knowing than any concrete implementation will have such a method.
BTW In Java 8, it will support virtual extensions which means an interface can give a default implementation. This has to be defined in terms of other methods provided by the interface.
An Interface is a contract that all classes that implement it, should have a definition for the methods specified in the interface. An interface does not define the method body as such.
An interface defines a set of method which must be implemented. It says nothing on how they are implemented. This is where the class definition comes in, since it defines how these methods are implemented.
Thus, when you call a class which implements a particular interface, then you know, for sure, that you will find whatever set of methods the interface defines.
Interfaces are usually handy when you need to expose some endpoints to your application, without the need to expose the logic.
EDIT: As per your example, the printable interface defines what behaviour should a class which implements it expose, in this case print.
This will allow you to do something along the lines of printable p = new A(); p.print();.
Assuming you have something which yields an object which implements the printable interface, then, whoever is calling that method will not need to bother what is the actual implementation of the print method. The interface makes sure that whatever you are returning, will contain an implementation of that method.
#NarutoUzumaki
Welcome to Stack overflow!
I agree with Chris. You can replace the doSomething method with eat() method to get a better understanding. A dog may eat something different than a cat and to a giraffe.
Its up to you how you implement the eat method, and when using it create a reference of the interface Animal and point it to the instance of Dog, Cat or Giraffe which ever eat method you want to use. This makes your class design very extensible.
Hope you get a clear idea now.
Generally Interface based Programming is recommended, Because of the following reasons
1)Interface means rule , you should follow those rules while implementing those methods in Implemented class.
2) Dependency is less between classes while instancing your implemented class then call your methods from another class or some where.
3) You can publish your interface details only no need to disclose the implemented details of your methods to out side the world.
Defining an interface is the difference between:
public void doSomething(Dog d)
{
d.doSomething();
}
public void doSomething(Cat c)
{
c.doSomething();
}
public void doSomething(Giraffe g)
{
g.doSomething();
}
and
public void doSomething(Animal a)
{
a.doSomething();
}
Why?
Well, if all the classes just implement their own methods, there's no common reference between them. However, if they all implement the method from a common interface, they can be referred to by the same reference type; in this case Animal.
A new feature coming in JDK 8 allows you to add to an existing interface while preserving binary compatibility.
The syntax is like
public interface SomeInterface() {
void existingInterface();
void newInterface() default SomeClass.defaultImplementation;
}
This way for all existing implementations of SomeInterface when they upgrade to this new version they don't all suddenly have compiles errors around newInterface().
While this is neat, what happens when you are implementing two interfaces which both have added a new defaulted method which you did not implement? Let me explain with an example.
public interface Attendance {
boolean present() default DefaultAttendance.present;
}
public interface Timeline {
boolean present() default DefaultTimeline.present;
}
public class TimeTravelingStudent implements Attendance, Timeline {
}
// which code gets called?
new TimeTravelingStudent().present();
Has this been defined as part of JDK 8 yet?
I found the Java gods talking about something similar here http://cs.oswego.edu/pipermail/lambda-lib/2011-February/000068.html, but its part of private mailing list and I cannot ask them directly.
See this for more details on how defaults are going to be used in JDK 8 and extending the Collection interface to support lambdas:
https://oracleus.wingateweb.com/published/oracleus2011/sessions/25066/25066_Cho223662.pdf
The answer to the duplicate operation is:
To solve multiple inheritance issue a class implementing two interfaces providing a default implementation for the same method name and signature must provide an implementation of the method. [Full Article]
My answer to your question is: Yes, it is a form of multiple inheritance, because you can inherit behavior from different parents. What's missing is to inherit states, i. e., attributes.
I know this is a old post, but as i'm working with this stuff...
You will have an error from the compiler, telling you that:
 class TimeTravelingStudent inherits unrelated defaults for present() from types Attendance and Timeline
reference to present is ambiguous, both method present() in Timeline and method present() in Attendance match.
There are two scenarios:
1) First, that was mentioned, where there is no most specific interface
public interface A {
default void doStuff(){ /* implementation */ }
}
public interface B {
default void doStuff() { /* implementation */ }
}
public class C implements A, B {
// option 1: own implementation
// OR
// option 2: use new syntax to call specific interface or face compilation error
void doStuff(){
B.super.doStuff();
}
}
2) Second, when there IS a more specific interface:
public interface A {
default void doStuff() { /* implementation */ }
}
public interface B extends A {
default void doStuff() { /* implementation */ }
}
public class C implements A, B {
// will use method from B, as it is "closer" to C
}
In short: it's a compile time error, must override the method by hand in the implementation.
Purpose of default method
The major purpose to introduce default method in Java 8, is to make interface extendable, without breaking existing implementations (there are so many 3rd party Java libraries).
And multiple inheritance like in C++ is actually intended to be avoided, that's definitely not the purpose of default method in Java.
How to override
2 options:
Override the method, with its own logic.
Override the method, call one of the interface's method via super, format: <interface_name>.super.<method_name>();
Tips:
method from interface is default to public, so don't forget to add public keyword when override it.
My answer to your question is: Yes, it is a form of multiple inheritance, because you can inherit behavior from different parents. What's missing is to inherit states, i. e., attributes.
Yes, but you can add getters and setters to your interface that the implementing classes then must implement. Nevertheless, the implementing classes don't inherit attributes. So, AFAICS, it's more like a trait-style solution rather than a multiple inheritance style solution.
If anyone's still looking for an answer, in case a class implements two interfaces with the same default method then the class needs to resolve the disambiguity by providing an implementation of its own. Look at this tutorial for more details on how inheritance in default methods work.
"How will we distinguish the methods" was a question that was put on Stackoverflow and referred to this question concrete methods in interfaces Java1.8
The following is an example that should answer that question:
interface A{
default public void m(){
System.out.println("Interface A: m()");
}
}
interface B{
default public void m(){
System.out.println("Interface B: m()");
}
}
class C implements A,B {
public void m(){
System.out.println("Concrete C: m()");
}
public static void main(String[] args) {
C aC = new C();
aC.m();
new A(){}.m();
new B(){}.m();
}
}
Class C above must implement its own concrete method of the interfaces A and B. Namely:
public void m(){
System.out.println("Interface C: m()");
}
To call a concrete implementation of a method from a specific interface, you can instantiate the interface and explicitly call the concrete method of that interface
For example, the following code calls the concrete implementation of the method m() from interface A:
new A(){}.m();
The output of the above would be:
Interface A: m()
As far as I see it, it is no multiple inheritance because they are stateless.
So virtual extension methods don't support full object or class functionality.