Java 7
First of all, I'm going to simplify the example to avoid posting unnecesary code. My specific concrete example a little bit complicated, but I' try to preserve the point.
public class Test {
public static void main(String[] args){
Test t = new Test(){ //<---------------------------------------------------------
public void m(){ // |
Test t = new Test(){// |
public void m(){// |
//Here I need to invoke the most inclosing class's m() method
}
//other actions
};
}
public void someMethod(){
//action
}
};
}
public void m(){
}
}
Is it possible to do in Java? I mean, to invoke the method of anonymous class that way?
No it's impossible because there is no reference to the anonymous classes.
This is the only possible way to call the instance m() method :
new Test(){
public void m(){
}
}.m();
By definition according to the oracle documentation here :
Anonymous classes enable you to make your code more concise. They
enable you to declare and instantiate a class at the same time. They
are like local classes except that they do not have a name. Use them
if you need to use a local class only once
So if you have to use one of the methods of your class you have to create a local one.
You cannot access the methods of the anonymous class using normal java, but you are able using reflection:
Test t = new Test{
public void m() {
System.out.println("Welcome to my class");
}
};
Class<?> c = t.getClass();
Method m = c.getDeclaredMethod("m");
// m.setaccessible(true); // if private
m.invoke(t);
Here is a way to do it:
public class Test
{
public static void main(String[] args)
{
Test t = new Test()
{
public void m() // this one will be called
{
Runnable r = new Runnable()
{
#Override
public void run()
{
m();
}
};
Test t = new Test()
{
public void m()
{
r.run();
}
};
}
};
}
public void m()
{
}
}
If the method returns a value, use Callable<V> instead.
Related
I'm new to Java and is trying to learn the concept of anonymous class. Could someone please tell me how I can invoke the 'awesomeMethod' from the main method of the LocallClassExample?
public class LocalClassExample {
interface Awesome {
public void awesomeMethod();
}
class AwesomeClass {
public int finalInt= 10;
Awesome a1 = new Awesome() {
#Override
public void awesomeMethod() {
System.out.println(finalInt);
}
};
}
public static void main(String[] args) {
}
}
Consider this:
new AwesomeClass().a1.awesomeMethod();
will invoke the method awesomeMethod() on the member variable a1 (which is something Awesome) of the newly created instance of AwesomeClass.
It will get more tricky once your main is outside of your AwesomeClass - and more so once it's outside of the package. In these cases you'd have to provide a getter like
public Awesome getAwesome() {
return a1;
}
Which would when invoked still execute the method as defined in your anonymous class.
Try to use this to create inner class object as:
public static void main(String[] args) {
LocalClassExample.AwesomeClass oi = new LocalClassExample().new AwesomeClass();
oi.awesomeMethod();
}
Imagine we have a class:
class A {
public void m() {
System.out.println("A - > m()");
}
}
...and I want to override the method m on class creation without making a second subclass B to extend A.
public static void main(String[] args) {
A a = new A() {
#Override
public void m() {
System.out.println("Override - > m()");
new Thread(new Runnable() {
#Override
public void run() {
// I want to be able to call the super method.
// This is illegal!
A.super.m();
}
}).start();
}
};
a.m();
}
Currently my solution is to create a private method that calls the super.m()
A a = new A() {
private void superMethod() {
super.m();
}
#Override
public void m() {
System.out.println("Overrided - > m()");
new Thread(new Runnable() {
#Override
public void run() {
superMethod();
}
}).start();
}
};
a.m();
I want to know why I am not able to write A.super.m() and if there another way to perform this task.
I want to know why I am not able to write A.super.m()...
This is because A is in fact not a directly enclosing class. The directly enclosing class of the Runnable is new A() { ... } which is an anonymous subclass of A.
In other words, if you would have had
class A extends Base {
new Runnable() { ... }
}
then A.super would have worked, but now you have
class <Anonymous subclass of A> extends A {
new Runnable() { ... }
}
which means that something like A.super isn't possible, since there's no syntax for <Anonymous subclass of A>.super.m.
...and, is there another way to perform this task.
The way you've solved it is reasonable in my opinion. Another way would be to create a local subclass of A (just to introduce an identifier to use in ____.super.m) as follows:
public static void main(String[] args) {
class SubA extends A {
#Override
public void m() {
System.out.println("Override - > m()");
new Thread(new Runnable() {
#Override
public void run() {
SubA.super.m();
// ^^^^ we now have a name of the directly enclosing class
}
}).start();
}
}
A a = new SubA();
a.m();
}
Writing A.super.m(), suppose that A has a superclass with a m method.
But in your code, you don't specify a superclass, and by default, the only superclass you have is Object.
But Object class doesn't have a 'm' method, so you could not call it.
A good way to do something like that is to use design pattern, like decorator.
I don't think there would be a simpler way to do it other than the way you already have it.
The problem is that the anonymous class A itself (not the base class A) cannot be referenced inside the Runnable class. The anonymous class is represented as something like package.A$1 when compiled to its own class file. For example, when you call superMethod inside the run of the thread, the following bytecode is executed:
getfield mypackage/Test$1$1/this$1 Lmypackage/Test$1;
invokestatic mypackage/Test$1/access$0(Lmypackage/Test$1;)V
In order to reference its base class A, there is no reference to this inner class instance so that you call the super.m() expression.
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?
}
}
Suppose I have a class A:
public class A {
public A(){....}
public void method1() {...}
};
And an instance of that class:
A anA = new A();
Is there any way to override the method1() only for anA?
This question arises when I write a small painting program in which I have to extend the JPanel class several times just to make minor changes to the different panels that have slightly different characteristics.
You can do the following:
A anA = new A() {
public void method1() {
...
}
};
This is the same as:
private static class myA extends A {
public void method1() {
...
}
}
A anA = new myA();
Only with the exception that in this case myA can be reused. That's not possible with anonymous classes.
You can create a an new anonymous class on the fly, as long as you are using the no-arg constructor of your class A:
A anA = new A() {
#Override
public void method1() {
...
}
};
Note that what you want to do is very close to what is known as a lambda, which should come along the next release 8 of Java SE.
I like to do this kind of thing with a delegate, or "strategy pattern".
public interface ADelegate {
public void method1();
}
public class A {
public A(){....}
public ADelegate delegate;
public final void method1() { delegate.method1(); }
};
A anA = new A();
anA.delegate = new ADelegate() {
public void method1() { ... }
};
Is there any Java syntax to access new methods defined within anonymous inner classes from outer class? I know there can be various workarounds, but I wonder if a special syntax exist?
For example
class Outer {
ActionListener listener = new ActionListener() {
#Override
void actionPerformed(ActionEvent e) {
// do something
}
// method is public so can be accessible
public void MyGloriousMethod() {
// viva!
}
};
public void Caller() {
listener.MyGloriousMethod(); // does not work!
}
}
MY OWN SOLUTION
I just moved all methods and members up to outer class.
Once the anonymous class instance has been implicitly cast into the named type it can't be cast back because there is no name for the anonymous type. You can access the additional members of the anonymous inner class through this within the class, in the expression immediate after the expression and the type can be inferred and returned through a method call.
Object obj = new Object() {
void fn() {
System.err.println("fn");
}
#Override public String toString() {
fn();
return "";
}
};
obj.toString();
new Object() {
void fn() {
System.err.println("fn");
}
}.fn();
identity(new Object() {
void fn() {
System.err.println("fn");
}
}).fn();
...
private static <T> T identity(T value) {
return value;
}
A student in my class asked our professor if this could be done the other day. Here is what I wrote as a cool proof of concept that it CAN be done, although not worth it, it is actually possible and here is how:
public static void main(String[] args){
//anonymous inner class with method defined inside which
//does not override anything
Object o = new Object()
{
public int test = 5;
public void sayHello()
{
System.out.println("Hello World");
}
};
//o.sayHello();//Does not work
try
{
Method m = o.getClass().getMethod("sayHello");
Field f = o.getClass().getField("test");
System.out.println(f.getInt(o));
m.invoke(o);
} catch (Exception e)
{
e.printStackTrace();
}
}
By making use of Java's Method class we can invoke a method by passing in the string value and parameters of the method. Same thing can be done with fields.
Just thought it would be cool to share this!
Your caller knows listener as an ActionListener and therefore it doesn't know anything about that new method. I think the only way to do this (other than doing reflection gymnastics, which really would defeat the purpose of using an anonymous class, i.e. shortcut/simplicity) is to simply subclass ActionListener and not use an anonymous class.
Funny enough, this is now allowed with var construct (Java 10 or newer). Example:
var calculator = new Object() {
BigDecimal intermediateSum = BigDecimal.ZERO;
void calculate(Item item) {
intermediateSum = Numbers.add(intermediateSum, item.value);
item.sum= intermediateSum;
}
};
items.forEach(calculator::calculate);
Here with method reference, but works with dot method call as well, of course. It works with fields as well. Enjoy new Java. :-)
I found more tricks with var and anonymous classes here: https://blog.codefx.org/java/tricks-var-anonymous-classes/
No, it's imposible. You would need to cast the ActionListener to its real subclass name, but since it's anonymous, it doesn't have a name.
The right way to do it is using reflection:
import java.lang.reflect.InvocationTargetException;
public class MethodByReflectionTest {
public static void main(String[] args) throws IllegalAccessException, IllegalArgumentException, InvocationTargetException, NoSuchMethodException, SecurityException {
Object obj = new Object(){
public void print(){
System.out.println("Print executed.");
}
};
obj.getClass().getMethod("print", null).invoke(obj, null);
}
}
You can check here: How do I invoke a Java method when given the method name as a string?
Yes you can access the method see the example below if any doubt please comment
package com;
interface A
{
public void display();
}
public class Outer {
public static void main(String []args)
{
A a=new A() {
#Override
public void display() {
System.out.println("Hello");
}
};
a.display();
}
}