I have the following classes.
public class Super{
public static void useSubClass(){
//I want to access the sub class object here, how.
}
}
public class Sub1 extends Super{
}
public class Sub2 extends Super{
}
I want to access the sub-class object from a static method in super-class. i.e. When I call Sub1.useSubClass() the method has access to Sub1.class and when I use Sub2.useSubClass(), I can access the Sub2.class.
Is there any way to access the sub-class object from super-class.
In general, you cannot do that from a superclass (and shouldn't!) because you won't know (and shouldn't assume anything about!) what classes will inherit from your superclass.
Depending on exactly what you want to do, there are alternatives, such as:
Use the template pattern to define "filler methods" that your subclasses must implement; these filler methods will be called by the template method in your superclass.
Define methods to be overridden by your subclass.
Define interfaces to be implemented by your subclass.
Update: As #JB Nizet has pointed out, I might have misread the question.
Here's something (very similar to the Observer Pattern) you can do if you wish to access subclasses from the static method in your superclass:
Define a static listener list in your superclass, call it List observerList
In the constructor of your superclass, add the class instance itself to that static observerList
For all subclasses, it is their responsibility to call super() from their constructors in order to register themselves to observerList (and unregister in deconstructor)
Then in your superclass's static useSubClass() method, you can iterate through that list of subclass instances, find the particular one you care about (maybe specified by some argument), and then do something with it.
Static methods are not inherited, and calling Sub2.useSubClass() is strictly equivalent to calling Super.useSubclass().
There is no way to get this information, because it doesn't exist. The compiler allows calling Sub2.useSubclass(), but translates it into Super.useSubclass().
public static void useSubClass(Super sub) {
if (sub instanceof Sub1) {
// Do something
} else if (sub instanceof Sub2) {
// Do something else
} else {
// Something else is extending Super
}
}
However, a better question is why? Can't you simply override the method in your subclass?
No you cannot because the super-class cannot know the methods of the sub-classes.
You should consider to create a new class which sees both super-class and sub-classes and implement the static method inside this new class
For the record, you could do this in Python, using class methods:
class super(object):
#classmethod
def usesubclass(cls):
print cls
class sub1(super):
pass
class sub2(super):
pass
Using this code, you could call sub1.usesubclass() or sub2.usesubclass(), and that would print the representations of the sub1 and sub2 classes, respectively:
>>> sub1.usesubclass()
<class '__main__.sub1'>
>>> sub2.usesubclass()
<class '__main__.sub2'>
Java, however, does not support such mechanisms, unfortunately. When you compile Sub1.useSubClass() in your example, the compiler will simply use Sub1 as the basic namespace to look up the the useSubClass() method in Super, but no information on that is actually compiled into code. In the resulting bytecode, the call is simply one directly to Super.useSubClass() and nothing more.
I sympathize with your plight, but Java is what it is. The closest thing you could come, I think, would be the following code:
public class Super {
public static <T extends Super> void useSubClass(Class<T> sub) {
}
}
And then call that method explicitly as either Super.useSubClass(Sub1.class) or Super.useSubClass(Sub2.class).
I figured something out. It works if implemented with care.
/** SuperClass.java **/
public abstract class SuperClass {
public static void printClass(){
System.out.println(new ImplementingClassRetriever().getCallingClass());
}
static class ImplementingClassRetriever extends SecurityManager{
public Class getCallingClass() {
Class[] classes = getClassContext();
for (Class clazz : classes) {
if (SuperClass.class.isAssignableFrom(clazz) && clazz != null
&& !clazz.equals(SuperClass.class)) {
return clazz;
}
}
return null;
}
}
}
/** Main.java **/
public class Main{
public static void main(String[] args) {
Sub.printClass(); //this does not work
Sub.testStaticCall(); //this works!! :)
}
}
class Sub extends SuperClass{
public static void testStaticCall(){
Sub.printClass(); //calling the method in the super class
}
}
This is just a toy example. The super class contains a static class that contains a method to retrieve the calling class.
In the subclass I have another static method which calls the superclass's method for printing the class name.
The Main class/function contains two calls to Sub's inherited and locally implemented method. The first call prints null, because the calling context (i.e. Main) is not a subclass of Super However the delegate method in Sub works because the calling context is now a subclass of SuperClass and hence the calling class can be determined.
Although You can create a reference to the super class and point it to any sub-class. This can also be done dynamically during run-time. This is a way of run-time polymorphism.
Related
I have these interface and classes:
public interface AlternateLight {
public String change(long time);
}
public abstract class AbstractLight {
public String change(int time) {
return "AbstractLight Method was used";
}
}
public class DistinctAlternateLight extends AbstractLight implements AlternateLight {
public String change(long time) {
return "DistinctAlternateLight Method was used";
}
}
Now I call the method change() with following main-method:
public class WhatUsedTest {
public static void main(String[] args) {
AlternateLight al = new DistinctAlternateLight();
System.out.println(al.change(100));
}
}
It prints "DistinctAlternateLight Method was used", but why? I thought since I didn't put a "L" behind the 100 as argument, it would call the method of the abstract class, because its method takes integers. With missing out on the "L", I guessed the compiler wouldn't handle the 100 as a long value and call the method taking longs, but it does. Why is that so?
It's due to the polymorphism, if you declare your variable with AlternateLight class and this type has only acces to change(long time).
be careful. If you use interface as reference type and assign an object of implementing class to it then you can call only those methods that are declared inside the interface. This is quite obvious because the implementing class can define methods of its own that are not a part of the contract between the interface and class. So, to call those methods you have to use the class as reference type as following:
DistinctAlternateLight al =new DistinctAlternateLight();
The method with the closest match to your type arguments will be used.
But for compatible numerical, the deepest declared one.
You can also see that the type of "al" object is an interface type, so you can`t call the method from super class without casting, you also can`t call any method from AbstractLight without casting. You can only call methods there were declared in interface class. In that case, compiler will favour the method from the interface class.
You can force the compiler to call a method from abstract class if you write something like that:
System.out.println(((AbstractLight)al).change(100));
In dynamic method binding a superclass reference can only call a subclass method which is inherited and overrode by it. However, the otherwise can be implemented.
abstract class in
{
abstract void print();
}
class a extends in
{
String name=this.getClass().getSimpleName();
void show()
{
System.out.println("class "+name);
}
void print()
{
show();
}
}
class Main
{
public static void main(String args[])
{
in x;
x = new a();
x.print();
}
}
Here, it prints successfully
class a
Also getClass() returns the subclass name instead of superclass name as this refers to the superclass object in main method.
A parent object reference is just constrained by the methods that it has in its class definition. If those methods are overridden by subclass, and at run time, if the actual object referred by the parent reference is of subclass type, then that overridden method is invoked. It doesn't matter if the overridden method invokes methods that are not originally present in the parent class or accesses the variables that are not present in the parent class.
This is what polymorphism is all about. It is by design meant to be this way as it makes program extension easier in case if we have different specific inheritance hierarchies where the parent class need not know the exact implementation of certain methods and can make things implemented by the subclasses as some sort of contract.
Future is unknown A developer writing a class A.java today can never predict in future the names or signatures of the methods which any other developer may include in his class extending A.java. Also such classes may be numerous with each having separate methods.
Base class should never be coupled with its sub classes. It must not care about how the sub classes are implemented.
Although it is not recommended but still if you wish to invoke the method defined in one of the sub class you may do it by typecasting like below.
public class Parent {
public class someMethod(){
if( this instanceof Child1){
((Child1)this).someAdditionalMethod();
}
}
}
public class Child1 extends Parent{
public class someAdditionalMethod(){
}
}
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Use of ‘super’ keyword when accessing non-overridden superclass methods
I'm new to Java and have been reading a lot about it lately to get more knowledge and experience about the language. I have a question about inherited methods and extending classes when the compiler inserts automatic code.
I've been reading that if I create class A with some methods including lets say a method called checkDuePeriod(), and then create a class B which extends class A and its methods.
If I then call the method checkDuePeriod() within class B without using the super.checkDuePeriod() syntax, during compilation will the compiler include the super. before checkDuePeriod() or will the fact that the compiler includes the super() constructor automatically when compiling the class imply the super. call of the methods that class B calls from class A?
I'm a little confused about this. Thanks in advance.
The super class's implementation of regular methods is not automatically invoked in sub classes, but a form of the super class's constructor must be called in a sub class's constructor.
In some cases, the call to super() is implied, such as when the super class has a default (no-parameter) constructor. However, if no default constructor exists in the super class, the sub class's constructors must invoke a super class constructor directly or indirectly.
Default constructor example:
public class A {
public A() {
// default constructor for A
}
}
public class B extends A {
public B() {
super(); // this call is unnecessary; the compiler will add it implicitly
}
}
Super class without default constructor:
public class A {
public A(int i) {
// only constructor present has a single int parameter
}
}
public class B extends A {
public B() {
// will not compile without direct call to super(int)!
super(100);
}
}
If you call checkDuePeriod() in B without super., means you want to invoke the method that belongs to the this instance (represented by this within B) of B. So, it equivalent to saying this.checkDuePeriod(), so it just doesn't make sense for the compiler to add super. in the front.
super. is something that you must explicitly add to tell the compiler that you want to call the A's version of the method (it is required specially in case B has overridden the implementation provided by A for the method).
Call of super() as a default constructor (constructor with no args) can be direct or non direct but it garants that fields of extendable class are properly initialized.
for example:
public class A {
StringBuilder sb;
public A() {
sb = new StringBuilder();
}
}
public class B extends A {
public B() {
//the default constructor is called automatically
}
public void someMethod(){
//sb was not initialized in B class,
//but we can use it, because java garants that it was initialized
//and has non null value
sb.toString();
}
}
But in case of methods:
Methods implement some logic. And if we need to rewrite logic of super class we use
public class B extends A {
public B() {
}
public boolean checkDuePeriod(){
//new check goes here
}
}
and if we want just implement some extra check, using the value returned from checkDuePeriod() of superclass we should do something like this
public class B extends A {
public B() {
}
public boolean checkDuePeriod(){
if(super.checkDuePeriod()){
//extra check goes here
}else{
//do something else if need
}
return checkResult;
}
}
First about the Constructors:
- When ever an object of a class is created, its constructor is initialized and at that time immediately the constructor of its super-class is called till the Object class,
- In this process all the instance variables are declared and initialized.
- Consider this scenario.
Dog is a sub-class of Canine and Canine is a sub-class of Animal
Now when Dog object is initialized, before the object actually forms, the Canine class object must be form, and before Canine object can form the Animal class object is to be formed, and before that Object class object must be form,
So the sequence of object formed is:
Object ---> Animal ---> Canine ---> Dog
So the Constructor of the Super-Class is Called before the Sub-Class.
Now with the Method:
The most specific version of the method that class is called.
Eg:
public class A{
public void go(){
}
}
class B extends A{
public static void main(String[] args){
new B().go(); // As B has not overridden the go() method of its super class,
// so the super-class implementation of the go() will be working here
}
}
I have the following problem in Java:
I have a base class and a derived class and I have a method in the base class. When I call the Base's foo method through Derived I want to get the Derived's class. The foo method can be generic if it can be done that way.
class Base
{
static void foo()
{
// I want to get Derived class here
// Derived.class
}
}
class Derived extends Base
{
}
Derived.foo();
Thanks for your help!
David
That's not the way that static methods work. You'll have to implement Derived.foo(), do whatever it is that's special to Derived, and that method then calls Base.foo(). If you really need the type information, you could create Base.foo0(Class klass).
But to be honest, any static method that needs to know that type of the class that it's invoked on should probably be an instance method.
Well, the caller of Derived.foo() knows what they are calling, so you could alter your methods thus:
class Base
{
static void foo(Class< T > calledBy)
{
// I want to get Derived class here
// Derived.class
}
}
class Derived extends Base
{
}
Derived.foo(Derived.class);
static methods are not inheritated. Static methods with the same signature can only hide similar methods in the superclass. This means that you never will see the result you probably want - you always exactly know the enclosing class. It is never possible that the static method is somehow "within" another class. So it is just impossible to produce the desired result. Calling a static method from a subclass or an instance is a bad idea for this reason as it just hides the real class. (IDEs and static code analysis tools can mark or correct this.)
Sources:
JLS http://java.sun.com/docs/books/jls/second_edition/html/classes.doc.html#227961
http://docs.oracle.com/javase/tutorial/java/IandI/override.html
So what works with inherited methods does not work with static methods that are not inherited.
class Base {
static void foo() {
// Only the static context is available here so you can't get class dynamic class information
}
void bar() {
System.out.println(getClass());
}
}
class Derived extends Base {
}
class Another extends Base {
static void foo() {
// No super call possible!
// This method hides the static method in the super class, it does not override it.
}
void bar() {
super.bar();
}
}
Derived derived = new Derived();
derived.bar(); // "class Derived"
Base base = new Base();
base.bar(); // "class Base"
// These are only "shortcuts" for Base.foo() that all work...
derived.foo(); // non-static context
Derived.foo(); // could be hidden by a method with same signature in Derived
base.foo(); // non-static context
Base.foo(); // Correct way to call the method
Another a = new Another();
a.foo(); // non-static context
Another.foo();
Is it good idea that the language allows this? - Hm. I think it is telling that IDEs and code analysis tools warn and can even correct this automatically.
Not possible, Derived.foo() will simply give code for Base.foo().
Derived.foo();
This will go to foo defined in Derived, if one is defined there:
class Base {
static void foo() {
System.out.println("Base");
}
}
class Der extends Base {
static void foo() {
System.out.println("Der");
}
}
class Check {
public static void main(String[] args) {
Base.foo();
Der.foo();
}
}
When I run it:
javac -g Check.java && java Check
Base
Der
So what is your question? If you to require that each derived class implement foo that is not possible to enforce in Java.
First things first, please be aware I am trying to express my question as best I can with my current knowledge and vocabulary, so please excuse this...
I have an abstract class in which I want to make a method where it instantiates itself.... Of course this is impossible in an abstract class, however, what I really want is for the concrete children (those classes that "extends") to inherit this instantiation so that they then can instantiate themselves....
Basically what I want to do is this:
MyAbstract a = new this();
However this isn't allowed... Is there any way I can do what I want?
Here is some non-compiling dream-code (i.e. code I wish worked). Basically I am wanting the ConcreteChild to call a method in which it create an object of itself. The method is inherited from it's parent.
public class Abstract {
public void instantiateMyConcreteChild()
{
Abstract a = new this();
}
}
public class ConcreteChild extends Abstract{
public static void main(String[] args) {
ConcreteChild c = new ConcreteChild();
c.instantiateMyConcreteChild();
}
}
* Additional info **
Thanks for the replies but I think I missed something vital....
Basically I wanted to pass an object's self ( "this" ) into some methods of some other classes. However, creating instantiating another object within an object is a bit backwards, I can just pass "this", right...
You can do this using reflection, something like :
Abstract a = getClass().newInstance();
This is because getClass() always returns the concrete class, so this.getClass() will return the real subclass and not the current class.
However, beware that if the subclass defines a custom constructor, having more or less parameters than your abstract class, it could fail. Unless you specify in the documentation that subclasses must have a constructor with such given parameters ... but it's fragile anyway.
You can inspect it, using getClass().getConstructors() and see which constructors are there, and if there is the one you are expecting, or even search for a viable one, otherwise you can catch the exception thrown by newInstance(..), and wrap it in a more descriptive exception for the users, so that they understand better what they missed ... but it would still be a kind of a hack, cause there is no explicit language support for such a situation.
Another approach could be to implement Cloneable in your abstract class, and then use the clone method, but it could be overkill or even wrong if what you want is a new, clean instance.
You can't do this using an instance method. Because as the name implies an instance methods requires that the instance has already instantiated.
What you actually need to do here is to separate the non-changing internal functionality from the abstract class itself. So what I could do is to ,for e.g., have an inner class that really encapsulates the non-changing functionality like so:
public class Abstract {
public void instantiateMyConcreteChild()
{
Abstract a = new NonChangingOperations();
}
class NonChangingOperations
{
public void operationA() {}
}
}
Infact you really dont need to keep the class NonChangingOperations as an inner class, you could make it as an external utility class with its own class hierarchy.
Are you trying to define a constructor that the subclasses of Abstract can use? If so you could simply do it the same way you define any other constructor.
public class Abstract {
Abstract() {
//set fields, etc. whatever you need to do
}
}
public class ConcreteChild extends Abstract{
ConcreteChild() {
//call superclass's constructor
super();
}
}
Could you just have this ?
public abstract class AbstractClassWithConstructor {
public AbstractClassWithConstructor() {
init();
}
protected abstract void init();
}
FYI
In the objective-c you need to set this by calling method init. The the method init() would look like this:
protected AbstractClassWithConstructor init() {
return this;
}