Can I use default method in is override? [duplicate] - java

Java 8 introduces default methods to provide the ability to extend interfaces without the need to modify existing implementations.
I wonder if it's possible to explicitly invoke the default implementation of a method when that method has been overridden or is not available because of conflicting default implementations in different interfaces.
interface A {
default void foo() {
System.out.println("A.foo");
}
}
class B implements A {
#Override
public void foo() {
System.out.println("B.foo");
}
public void afoo() {
// how to invoke A.foo() here?
}
}
Considering the code above, how would you call A.foo() from a method of class B?

As per this article you access default method in interface A using
A.super.foo();
This could be used as follows (assuming interfaces A and C both have default methods foo())
public class ChildClass implements A, C {
#Override
public void foo() {
//you could completely override the default implementations
doSomethingElse();
//or manage conflicts between the same method foo() in both A and C
A.super.foo();
}
public void bah() {
A.super.foo(); //original foo() from A accessed
C.super.foo(); //original foo() from C accessed
}
}
A and C can both have .foo() methods and the specific default implementation can be chosen or you can use one (or both) as part of your new foo() method. You can also use the same syntax to access the default versions in other methods in your implementing class.
Formal description of the method invocation syntax can be found in the chapter 15 of the JLS.

This answer is written mainly for users who are coming from question 45047550 which is closed.
Java 8 interfaces introduce some aspects of multiple inheritance. Default methods have an implemented function body. To call a method from the super class you can use the keyword super, but if you want to make this with a super interface it's required to name it explicitly.
class ParentClass {
public void hello() {
System.out.println("Hello ParentClass!");
}
}
interface InterfaceFoo {
public default void hello() {
System.out.println("Hello InterfaceFoo!");
}
}
interface InterfaceBar {
public default void hello() {
System.out.println("Hello InterfaceBar!");
}
}
public class Example extends ParentClass implements InterfaceFoo, InterfaceBar {
public void hello() {
super.hello(); // (note: ParentClass.super could not be used)
InterfaceFoo.super.hello();
InterfaceBar.super.hello();
}
public static void main(String[] args) {
new Example().hello();
}
}
Output:
Hello ParentClass!
Hello InterfaceFoo!
Hello InterfaceBar!

The code below should work.
public class B implements A {
#Override
public void foo() {
System.out.println("B.foo");
}
void aFoo() {
A.super.foo();
}
public static void main(String[] args) {
B b = new B();
b.foo();
b.aFoo();
}
}
interface A {
default void foo() {
System.out.println("A.foo");
}
}
Output:
B.foo
A.foo

You don't need to override the default method of an interface. Just call it like the following:
public class B implements A {
#Override
public void foo() {
System.out.println("B.foo");
}
public void afoo() {
A.super.foo();
}
public static void main(String[] args) {
B b=new B();
b.afoo();
}
}
Output:
A.foo

It depends on your choice whether you want to override the default method of an interface or not. Because default are similar to instance method of a class which can be directly called upon the implementing class object. (In short default method of an interface is inherited by implementing class)

Consider the following example:
interface I{
default void print(){
System.out.println("Interface");
}
}
abstract class Abs{
public void print(){
System.out.println("Abstract");
}
}
public class Test extends Abs implements I{
public static void main(String[] args) throws ExecutionException, InterruptedException
{
Test t = new Test();
t.print();// calls Abstract's print method and How to call interface's defaut method?
}
}

Related

How to properly use a parameter of type class?

In java, I'd like to do something like this
public class Tata{
public static void f(){
//something
}
public static void g(){
//something
}
}
public class Titi{
public static void f(){
//something
}
public static void g(){
//something
}
}
public class Toto{
private Class c = Tata.class; //or Titi.class
public static void main(String[] args) {
c.f();
c.g();
}
}
To be precise, I'd like to be able to freely switch between classes Tata and Titi, to use their respective methods f or g.
This doesn't work as intended, as I get the cannot resolve method 'f()' error. Simply replacing c.f(); and c.g(); with Tata.f(); and Tata.g(); works fine, but defeats the purpose of using a parameter. How to solve this?
Will turn the comment into answer after all.. The correct (Java) way to deal with what you want is the use of interface. So in your demo code the implementation would be the following:
public interface TheFGFunctions {
void f();
void g();
}
public class Tata implements TheFGFunctions {
#Override
public void f() {
//something
}
#Override
public void g() {
//something
}
}
public class Titi implements TheFGFunctions {
#Override
public void f() {
//something
}
#Override
public void g() {
//something
}
}
public class Toto {
private TheFGFunctions c;
public Toto(TheFGFunctions c) {
this.c = c;
}
public void notStaticFunction() {
c.f();
c.g();
}
}
This way is totally typesafe with zero exceptions to deal with!
You cannot access a static method polymorphically. The Java language doesn't support it.
The reason your current approach fails is that c is an instance of the class Class, and the class Class doesn't define methods f() or g().
(The methods that it does define are listed in the javadoc for Class. Note that Class is final so you can't create a custom subclass with extra methods.)
The simple alternative is to use reflection; e.g.
Class c =
Method f = c.getMethod("f");
f.invoke(null); // because it is static
But note:
This is not statically type-safe. The compiler cannot tell when you make the mistake of trying to use a static f() on a class that doesn't have such a method.
There are a few exceptions that you need to deal with; e.g. missing methods, incorrect signatures, methods that are not static, methods that don't have the correct access.
Other answers have proposed creating an interface and wrapper classes to make certain static methods dispatchable. It will work and it will be compile-time type-safe (!) but there is a lot of boiler plate code to write.
#Michael Michailidis commented:
Thus interfaces!
Yea ... kind of. You can only dispatch polymorphically on instance methods declared on an interface. That implies that you must have an instance of Tata or Titi, and call the methods on it. My reading of the Question is that the author wants to avoid that.
(IMO, the avoidance is the real problem. You are better of not trying to avoid instance methods.)
FWIW, you can declare static methods in an interface (since Java 8), but they would behave the same as if you declared them in a class. You wouldn't be able to dispatch ...
You could use reflections:
private Class c = Tata.class;
public Toto() throws Exception {
c.getMethod("f").invoke(null);
c.getMethod("g").invoke(null);
}
Here my Tata class
public class Tata {
public static void f() {
System.out.println("ffff");
}
public static void g() {
System.out.println("gggg");
}
}
Output on new Toto() call:
ffff
gggg
Update (call with parameters):
public Toto() throws Exception {
c.getMethod("f", String.class).invoke(null, "paramValue1");
c.getMethod("g", String.class).invoke(null, "paramValue2");
}
public class Tata {
public static void f(String param1) {
System.out.println("ffff " + param1);
}
public static void g(String param2) {
System.out.println("gggg " + param2);
}
}
Output:
ffff paramValue1
gggg paramValue2
Write a wrapper interface
interface CWrapper {
void f();
void g();
}
and wrapper class factory method for each Class containing the methods
class CWrappers {
CWrapper forTiti(Class<Titi> titiClass) {
return new CWrapper() {
void f() { Titi.f(); }
void g() { Titi.g(); }
}
}
// another factory method for Tata
}
Then you can use that:
public class Toto {
private CWrapper c = CWrappers.forTata(Tata.class); //or forTiti(Titi.class)
public static void main(String[] args) {
c.f();
c.g();
}
}

C# equivalent to implement interface inside super keyword in Java [duplicate]

I am not sure how am I suppose to go about my question. It is about Android can Instantiate Interface. I am trying to do in C#. Now I am pretty sure that the rules for both Java and C# is you can't create an Instance of abstract and Interface as being said.
But I would really like to know how Android does this practice.
In Android you can do this.
public interface Checkme{
void Test();
void Test2();
}
public void myFunc(Checkme my){
//do something
}
// Now this is the actual usage.
public void Start(){
myFunc(new Checkme(){
#Override
public void Test()
{
}
#Override
public void Test2()
{
}
});
}
Actually once you press Enter on new Checkme() You will automatically get the Override methods of the Interface. Like auto Implement method of an Interface in C#.
I hope my question make sense.
C# doesn't support anonymously auto-implemented interfaces because it has delegates:
public void Foo(Func<string> func, Action action) {}
// call it somewhere:
instance.Foo(() => "hello world", () => Console.WriteLine("hello world"));
With delegates you can fill the gap and it can be even more powerful than implementing interfaces with anonymous classes.
Learn more about delegates.
This is an Anonymous Class:
public void Start(){
myFunc(new Checkme() {
#Override
public void Test() {
}
#Override
public void Test2() {
}
});
}
An anonymous class is an unnamed class implemented inline.
You could also have done it using a Local Class, but those are rarely seen in the wild.
public void Start(){
class LocalCheckme implements Checkme {
#Override
public void Test() {
}
#Override
public void Test2() {
}
}
myFunc(new LocalCheckme());
}
These both have the advantage that they can use method parameters and variables directly, as long as they are (effectively) final.
As a third option, you could do it with an Inner Class.
private class InnerCheckme implements Checkme {
#Override
public void Test() {
}
#Override
public void Test2() {
}
}
public void Start(){
myFunc(new InnerCheckme());
}
An inner class cannot access method variables (obviously because it's outside the method), but can be used by multiple methods.
Any local values from the method can however be passed into the constructor and stored as fields of the inner class, to get the same behavior. Just requires a bit more code.
If the inner class doesn't need access to fields of the outer class, it can be declared static, making it a Static Nested Class.
So, all 3 ways above a very similar. The first two are just Java shorthands for the third, i.e. syntactic sugar implemented by the compiler.
C# can do the third one, so just do it that way for C#.
Of course, if the interface only has one method, using a Java lambda or C# delegate is much easier than Anonymous / Local / Inner classes.
If I understand correcly, you're defining a class that implements an interface, and when you specify that the class implements an interface, you want it to automatically add the interface's methods and properties.
If you've declared this:
public interface ISomeInterface
{
void DoSomething();
}
And then you add a class:
public class MyClass : ISomeInterface // <-- right-click
{
}
Right-click on the interface and Visual Studio will give you an option to implement the interface, and it will add all the interface's members to the class.
you mean something like this?
pulic interface Foo{
void DoSomething();
}
public class Bar : Foo {
public void DoSomething () {
//logic here
}
}
myFunc(new Checkme(){
#Override
public void Test()
{
}
#Override
public void Test2()
{
}
});
You're passing into myFunc() something that is called an anonymous class. When it says "new Checkme() { .... }", it is defining an anonymous implementation of the Checkme interface. So, it's not an instance of the interface itself, just an instance of a type that implements it.
In C# anonymously implemented classes for Interface are not auto generated just like in java, you need to follow the below procedure to workout.
public class MyClass {
public void someMethod (string id, IMyInterface _iMyInterface) {
string someResponse = "RESPONSE FOR " + id;
_iMyInterface.InterfaceResponse (someResponse);
}
}
public interface IMyInterface {
void InterfaceResponse (object data);
void InterfaceResponse2 (object data, string x);
}
public class MyInterfaceImplementor : IMyInterface {
private readonly Action<object> actionname;
private readonly Action<object, string> actionInterfaceResponse2;
public MyInterfaceImplementor (Action<object> InterfaceResponse) {
this.actionname = InterfaceResponse;
}
public MyInterfaceImplementor(Action<object> interfaceResponseMethod, Action<object, string> interfaceResponseMethod1) {
this.actionname = interfaceResponseMethod ?? throw new ArgumentNullException(nameof(interfaceResponseMethod));
this.actionInterfaceResponse2 = interfaceResponseMethod1 ?? throw new ArgumentNullException(nameof(interfaceResponseMethod1));
}
public void InterfaceResponse (object data) {
this.actionname (data);
}
public void InterfaceResponse2(object data, string x) {
this.actionInterfaceResponse2(data, x);
}
}
Gist Source : https://gist.github.com/pishangujeniya/4398db8b9374b081b0670ce746f34cbc
Reference :

Java: invoke a default method in another default method of the same interface

I'm very new to the java 8 features and try to understand default methods. Is there an easier way to invoke a default method by another default method of the same interface than using an anonymous class?
For example:
public class Frame{
public static void main(String... args){
Frame.C c= new Frame.C();
c.doSomething();
}
public interface A{
public default void doSomething(){
System.out.println("A");
}
}
public interface B extends A {
#Override
public default void doSomething(){
System.out.println("B");
//is there an easier way to invoke that method??
new B(){}.other();
}
default public void other(){
//doSomething();
System.out.println("other");
}
}
public static class C implements B{
#Override
public void other(){
Lambda.B.super.other();
System.out.println("C");
}
}
}
Your intention is not entirely clear, but the construct new B(){}.other(); implies two things:
You don’t want to invoke an overriding method implementation
The instance on which you invoke other() is obviously irrelevant when invoking it on an entirely different instance (new B(){}) is a viable solution
These two things together imply that you should use a static method instead:
public interface B extends A {
#Override
public default void doSomething(){
System.out.println("B");
otherInB();
}
default public void other(){
otherInB();
}
static void otherInB() {
//doSomething();
System.out.println("other");
}
}
Since your original method names did not carry useful information, it’s not possible to suggest a useful name for that static method either.
Note that Java 9 is going to introduce support for private methods in interfaces which allows hiding otherInB() to other classes and even making it non-static in case it has to use other methods on the same instance.
If the visibility of the method in Java 8 is an issue, consider that the actual place of a non-polymorphic method is irrelevant, so you can always use a companion class:
public interface B extends A {
#Override
public default void doSomething(){
System.out.println("B");
BHelper.other();
}
default public void other(){
BHelper.other();
}
}
…
/* not public */ class BHelper {
/* not public */ static void other() {
//doSomething();
System.out.println("other");
}
}
This even works if the implementation needs the actual B instance as you may pass it as a parameter.
public interface B extends A {
#Override
public default void doSomething(){
System.out.println("B");
BHelper.other(this);
}
default public void other(){
BHelper.other(this);
}
}
…
/* not public */ class BHelper {
/* not public */ static void other(B instance) {
//doSomething();
System.out.println("other");
}
}

How to invoke an interface's default implementation of a method in an overriding implementation?

Suppose I have the following code:
interface HumanoidForm {
default HumanoidForm reproduce() {
<appropriate code for humanoid form reproduction>
}
}
class Android extends Machine implements HumanoidForm {
public HumanoidForm reproduce() {
<appropriate code for android reproduction> // how to use HumanoidForm's default implementation here?
}
}
Now suppose "appropriate code for android reproduction" is best described by using "appropriate code for humanoid form reproduction" as a sub-routine. How can I access "appropriate code for humanoid form" from within "appropriate code for android reproduction"? I can think of three ways, but none of them works:
Simply invoking reproduce() invokes the overriding implementation.
Writing ((HumanoidForm) this).reproduce() still invokes the overriding implementation.
Mimicking the re-use of implementations of methods in super classes by overriding methods, one may think of writing super.reproduce(). However, that refers to Machine's implementation of reproduce, which may not even exist.
So it seems there is no way to re-use the code in the default method for overriding. Is that really so?
HumanoidForm.super.reproduce();
Actually, you can choose freely the existing implementation. Let me give you a scenario slightly more complicated than yours. To make things worse, all A,B & C has the same method signature.
interface A {
default void doWork() {
System.out.println("Default implementation From A");
}
}
interface B{
default void doWork() {
System.out.println("Default implementation From B");
}
}
class C{
void doWork(){
System.out.println("Default implementation From C");
}
}
Now, I create a subclass to C which implements A & B:
class Tester extends C implements A, B
{
#Override public void doWork(){
A.super.doWork(); //Invoke A's implementation
B.super.doWork(); //Invoke B's implementation
super.doWork(); //Invoke C's implementation
}
}
The output will be:
Default implementation From A
Default implementation From B
Default implementation From C
when you run:
new Tester().doWork();
Take a look at this: https://blog.idrsolutions.com/2015/01/java-8-default-methods-explained-5-minutes/
In particular the section where it says "If we want to specifically invoke one of the sayHi() methods in either InterfaceA or InterfaceB, we can also do as follows:"
public class MyClass implements InterfaceA, InterfaceB {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
}
#Override
public void saySomething() {
System.out.println("Hello World");
}
#Override
public void sayHi() {
InterfaceA.super.sayHi();
}
}
interface InterfaceA {
public void saySomething();
default public void sayHi() {
System.out.println("Hi from InterfaceA");
}
}
interface InterfaceB {
default public void sayHi() {
System.out.println("Hi from InterfaceB");
}
}
So it seems there is no way to re-use the code in the default method for overriding. Is that really so?
No, you can reuse the code. You can easily test it and you will see that the following code works:
public class Test implements HumanoidForm
{
public static void main(String[] args)
{
new Test().reproduce();
}
#Override
public void reproduce(){
HumanoidForm.super.reproduce(); //Invoking default method
System.out.println("From implementing class");
}
}
interface HumanoidForm {
default void reproduce() {
System.out.println("From default interface");
}
}
OUTPUT:
From default interface
From implementing class

interface - Overridden methods are not inherited

I was reading Java SCJP book by Khalid A. Mughal (for JE6), and in topic 7.6 Interfaces and Page number 313, it is given that
A subinterface can override abstract method declarations from its superinterfaces. Overridden methods are not inherited.
I could not quite understand what "Overridden methods are not inherited." means. I tried to do this:
interface A
{
void abc();
}
interface B extends A
{
#Override
void abc();
}
interface C extends B
{
void abc();
}
And I did not get any error. What am I not understanding?
This simply means that overridden methods can have a slightly different signature than the superinterface's methods. For example:
public interface Foo {
Object doSomething(String input) throws IOException;
}
public interface Bar extends Foo {
#Override
String doSomething(String input);
}
As you can see, in the subinterface, I no longer throw a checked exception, and I guarantee that the returned object is a more specific type. The method that did throw a checked exception is not inherited, because it is overridden.
I don't have the context of the entire paragraph, but there's something else related that applies only to implementations and not interfaces, which is that if you override a method without explicitly calling super, the superclass's implementation won't occur.
For example, if I have:
public class Example {
public static class Foo {
public void printSomething() {
System.out.println("Foo");
}
}
public static class Bar extends Foo {
#Override
public void printSomething() {
System.out.println("Bar");
}
}
public static void main(String[] args) {
Bar bar = new Bar();
bar.printSomething();
}
}
This program will simply output:
Bar
but NOT Foo. If I add a call to super.printSomething(), then it will print both.

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