I have a class java ProductManager which extends another class with the same name,
located in another project with another package("com.services") .
I have to invoke a method deleteProduct(Long productId) located in the super-class.
try{
Object service = CONTEXT.getBean("ProductManager");
Method method = service.getClass().getDeclaredMethod("deleteProduct", Long.class);
method.invoke(service, productId);
} catch(Exception e){
log.info(e.getMessage());
}
I couldn't delete the product:
I get this info:
com.franceFactory.services.ProductManager.deleteProduct(java.lang.Long)
the product isn't deleted :(
The various getDeclaredMethod() and getDeclaredMethods() only return methods declared on the current class instance. From the javadoc:
This includes public, protected, default (package) access, and private methods, but excludes inherited methods.
The important part here is "but excludes inherited methods". This is why you are getting an exception with your code as it currently stands, it is not returning the deleteProduct() method from the parent class.
Instead if you wanted to continue using reflection you would need to use the getMethod method as this returns all public methods, "including those declared by the class or interface and those inherited from superclasses and superinterfaces."
If you have to use reflection then don't use getDeclaredMethod() because (as its name suggest) it can return only methods declared in current class, while you claim you want to invoke method declared in other class (to be precise declared in super class).
To get public method (including also inherited ones) use getMethod().
If your are overriding that method, just use the reserved word super (from the Oracle documents):
public class Superclass {
public void printMethod() {
System.out.println("Printed in Superclass.");
}
}
public class Subclass extends Superclass {
// overrides printMethod in Superclass
public void printMethod() {
super.printMethod(); // This calls to the method defined in the superclass
System.out.println("Printed in Subclass");
}
public static void main(String[] args) {
Subclass s = new Subclass();
s.printMethod();
}
}
This code will write:
Printed in Superclass.
Printed in Subclass
In other case (you are not overriding it, just using it), just write this.methodName(...). All methods inherited are directly available.
Disclaimer: I am not sure I totally understand your question. I will still try to answer what I think I understand.
The Product in package com.franceFactory.services (Lets call it A)
extends The Product class in package com.services (Lets call it B)
So A extends B.
B has method deleteProduct(java.lang.Long)
A overrides the method deleteProduct(java.lang.Long)
You have instance of Class A. So by OOPS concept method deleteProduct of object A is going to get called.
There is no way you can call the super method from outside unless you have the instance of class B.
EDIT
OPs Clarification yes, it's public, but it isn't overridden in my class
The method in super is getting called here. The product is not getting deleted for reason on what is written on the method.
Related
I'm dealing with a class which extends JFrame.
It's not my code and it makes a call to super before it begins constructing the GUI. I'm wondering why this is done since I've always just accessed the methods of the superclass without having to call super();
There is an implicit call to super() with no arguments for all classes that have a parent - which is every user defined class in Java - so calling it explicitly is usually not required. However, you may use the call to super() with arguments if the parent's constructor takes parameters, and you wish to specify them. Moreover, if the parent's constructor takes parameters, and it has no default parameter-less constructor, you will need to call super() with argument(s).
An example, where the explicit call to super() gives you some extra control over the title of the frame:
class MyFrame extends JFrame
{
public MyFrame() {
super("My Window Title");
...
}
}
A call to your parent class's empty constructor super() is done automatically when you don't do it yourself. That's the reason you've never had to do it in your code. It was done for you.
When your superclass doesn't have a no-arg constructor, the compiler will require you to call super with the appropriate arguments. The compiler will make sure that you instantiate the class correctly. So this is not something you have to worry about too much.
Whether you call super() in your constructor or not, it doesn't affect your ability to call the methods of your parent class.
As a side note, some say that it's generally best to make that call manually for reasons of clarity.
None of the above answers answer the 'why'.
Found a good explanation here:
A subclass can have its own private data members, so a subclass can
also have its own constructors.
The constructors of the subclass can initialize only the instance
variables of the subclass. Thus, when a subclass object is
instantiated the subclass object must also automatically execute one
of the constructors of the superclass.
You might also want to read everything about the super keyword here or watch everything about the super keyword here.
We can access super class elements by using super keyword
Consider we have two classes, Parent class and Child class, with different implementations of method foo. Now in child class if we want to call the method foo of parent class, we can do so by super.foo(); we can also access parent elements by super keyword.
class parent {
String str="I am parent";
//method of parent Class
public void foo() {
System.out.println("Hello World " + str);
}
}
class child extends parent {
String str="I am child";
// different foo implementation in child Class
public void foo() {
System.out.println("Hello World "+str);
}
// calling the foo method of parent class
public void parentClassFoo(){
super.foo();
}
// changing the value of str in parent class and calling the foo method of parent class
public void parentClassFooStr(){
super.str="parent string changed";
super.foo();
}
}
public class Main{
public static void main(String args[]) {
child obj = new child();
obj.foo();
obj.parentClassFoo();
obj.parentClassFooStr();
}
}
It simply calls the default constructor of the superclass.
We use super keyword to call the members of the Superclass.
As a subclass inherits all the members (fields, methods, nested classes) from its parent and since Constructors are NOT members (They don't belong to objects. They are responsible for creating objects), they are NOT inherited by subclasses.
So we have to explicitly give the call for parent constructor so that the chain of constructor remains connected if we need to create an object for the superclass. At the time of object creation, only one constructor can be called. Through super, we can call the other constructor from within the current constructor when needed.
If you are thinking why it's there for a class that is not extending any other class, then just remember every class follows object class by default. So it's a good practice to keep super in your constructor.
Note: Even if you don't have super() in your first statement, the compiler will add it for you!
We can Access SuperClass members using super keyword
If your method overrides one of its superclass's methods, you can invoke the overridden method through the use of the keyword super. You can also use super to refer to a hidden field (although hiding fields is discouraged). Consider this class, Superclass:
public class Superclass {
public void printMethod() {
System.out.println("Printed in Superclass.");
}
}
// Here is a subclass, called Subclass, that overrides printMethod():
public class Subclass extends Superclass {
// overrides printMethod in Superclass
public void printMethod() {
super.printMethod();
System.out.println("Printed in Subclass");
}
public static void main(String[] args) {
Subclass s = new Subclass();
s.printMethod();
}
}
Within Subclass, the simple name printMethod() refers to the one declared in Subclass, which overrides the one in Superclass. So, to refer to printMethod() inherited from Superclass, Subclass must use a qualified name, using super as shown. Compiling and executing Subclass prints the following:
Printed in Superclass.
Printed in Subclass
as constructor is not a part of class,
so while calling it cannot be implemented,
by using SUPER() we can call the members and memberfunctions in constructor.
I'm dealing with a class which extends JFrame.
It's not my code and it makes a call to super before it begins constructing the GUI. I'm wondering why this is done since I've always just accessed the methods of the superclass without having to call super();
There is an implicit call to super() with no arguments for all classes that have a parent - which is every user defined class in Java - so calling it explicitly is usually not required. However, you may use the call to super() with arguments if the parent's constructor takes parameters, and you wish to specify them. Moreover, if the parent's constructor takes parameters, and it has no default parameter-less constructor, you will need to call super() with argument(s).
An example, where the explicit call to super() gives you some extra control over the title of the frame:
class MyFrame extends JFrame
{
public MyFrame() {
super("My Window Title");
...
}
}
A call to your parent class's empty constructor super() is done automatically when you don't do it yourself. That's the reason you've never had to do it in your code. It was done for you.
When your superclass doesn't have a no-arg constructor, the compiler will require you to call super with the appropriate arguments. The compiler will make sure that you instantiate the class correctly. So this is not something you have to worry about too much.
Whether you call super() in your constructor or not, it doesn't affect your ability to call the methods of your parent class.
As a side note, some say that it's generally best to make that call manually for reasons of clarity.
None of the above answers answer the 'why'.
Found a good explanation here:
A subclass can have its own private data members, so a subclass can
also have its own constructors.
The constructors of the subclass can initialize only the instance
variables of the subclass. Thus, when a subclass object is
instantiated the subclass object must also automatically execute one
of the constructors of the superclass.
You might also want to read everything about the super keyword here or watch everything about the super keyword here.
We can access super class elements by using super keyword
Consider we have two classes, Parent class and Child class, with different implementations of method foo. Now in child class if we want to call the method foo of parent class, we can do so by super.foo(); we can also access parent elements by super keyword.
class parent {
String str="I am parent";
//method of parent Class
public void foo() {
System.out.println("Hello World " + str);
}
}
class child extends parent {
String str="I am child";
// different foo implementation in child Class
public void foo() {
System.out.println("Hello World "+str);
}
// calling the foo method of parent class
public void parentClassFoo(){
super.foo();
}
// changing the value of str in parent class and calling the foo method of parent class
public void parentClassFooStr(){
super.str="parent string changed";
super.foo();
}
}
public class Main{
public static void main(String args[]) {
child obj = new child();
obj.foo();
obj.parentClassFoo();
obj.parentClassFooStr();
}
}
It simply calls the default constructor of the superclass.
We use super keyword to call the members of the Superclass.
As a subclass inherits all the members (fields, methods, nested classes) from its parent and since Constructors are NOT members (They don't belong to objects. They are responsible for creating objects), they are NOT inherited by subclasses.
So we have to explicitly give the call for parent constructor so that the chain of constructor remains connected if we need to create an object for the superclass. At the time of object creation, only one constructor can be called. Through super, we can call the other constructor from within the current constructor when needed.
If you are thinking why it's there for a class that is not extending any other class, then just remember every class follows object class by default. So it's a good practice to keep super in your constructor.
Note: Even if you don't have super() in your first statement, the compiler will add it for you!
We can Access SuperClass members using super keyword
If your method overrides one of its superclass's methods, you can invoke the overridden method through the use of the keyword super. You can also use super to refer to a hidden field (although hiding fields is discouraged). Consider this class, Superclass:
public class Superclass {
public void printMethod() {
System.out.println("Printed in Superclass.");
}
}
// Here is a subclass, called Subclass, that overrides printMethod():
public class Subclass extends Superclass {
// overrides printMethod in Superclass
public void printMethod() {
super.printMethod();
System.out.println("Printed in Subclass");
}
public static void main(String[] args) {
Subclass s = new Subclass();
s.printMethod();
}
}
Within Subclass, the simple name printMethod() refers to the one declared in Subclass, which overrides the one in Superclass. So, to refer to printMethod() inherited from Superclass, Subclass must use a qualified name, using super as shown. Compiling and executing Subclass prints the following:
Printed in Superclass.
Printed in Subclass
as constructor is not a part of class,
so while calling it cannot be implemented,
by using SUPER() we can call the members and memberfunctions in constructor.
I have parent-child classes in two different packages. I am overriding a method of protected type. I want to access super class protected method in subclass.
Consider below code:
package package1;
public class Super
{
protected void demoMethod()
{
System.out.println("In super method");
}
}
package package2;
import package1.Super;
public class Sub extends Super
{
#Override
protected void demoMethod()
{
System.out.println("In sub method");
}
public static void main(String[] args)
{
//code for accessing superclass demoMethod to print "In super method"
}
}
In the main method of sub class, I want to access super class demoMethod which prints "In super method". I know demoMethod won't be visible from sub class using super class object reference to call demoMethod.
Is it possible or not? If yes, how?
Consider me new to Java and provide answers replacing comment in main method.
In the child class use super.demoMethod() or just remove it altogether from the child class
Your main() method cannot access the superclass implementation of demoMethod() -- because it's overridden in the child class.
Your main() method can access demoMethod(), through a reference of the subclass type, even though it's protected, because it's in the same package as your subclass. But it will call the subclass implementation.
But if you're "using superclass object reference to call demoMethod", the method will not be accessible, and your code will not compile. Your superclass is in a different package. Methods marked protected can only be accessed by subclasses and by code in the same package.
If you made the method public in both subclass and superclass, calling demoMethod() would call the subclass implementation, regardless whether the reference was of the super or subclass type.
An instance of the subclass can call super.demoMethod() as part of the implementation of its methods. But the main() method cannot.
I'm currently learning about class inheritance in my Java course and I don't understand when to use the super() call?
Edit:
I found this example of code where super.variable is used:
class A
{
int k = 10;
}
class Test extends A
{
public void m() {
System.out.println(super.k);
}
}
So I understand that here, you must use super to access the k variable in the super-class. However, in any other case, what does super(); do? On its own?
Calling exactly super() is always redundant. It's explicitly doing what would be implicitly done otherwise. That's because if you omit a call to the super constructor, the no-argument super constructor will be invoked automatically anyway. Not to say that it's bad style; some people like being explicit.
However, where it becomes useful is when the super constructor takes arguments that you want to pass in from the subclass.
public class Animal {
private final String noise;
protected Animal(String noise) {
this.noise = noise;
}
public void makeNoise() {
System.out.println(noise);
}
}
public class Pig extends Animal {
public Pig() {
super("Oink");
}
}
super is used to call the constructor, methods and properties of parent class.
You may also use the super keyword in the sub class when you want to invoke a method from the parent class when you have overridden it in the subclass.
Example:
public class CellPhone {
public void print() {
System.out.println("I'm a cellphone");
}
}
public class TouchPhone extends CellPhone {
#Override
public void print() {
super.print();
System.out.println("I'm a touch screen cellphone");
}
public static void main (strings[] args) {
TouchPhone p = new TouchPhone();
p.print();
}
}
Here, the line super.print() invokes the print() method of the superclass CellPhone. The output will be:
I'm a cellphone
I'm a touch screen cellphone
You would use it as the first line of a subclass constructor to call the constructor of its parent class.
For example:
public class TheSuper{
public TheSuper(){
eatCake();
}
}
public class TheSub extends TheSuper{
public TheSub(){
super();
eatMoreCake();
}
}
Constructing an instance of TheSub would call both eatCake() and eatMoreCake()
When you want the super class constructor to be called - to initialize the fields within it. Take a look at this article for an understanding of when to use it:
http://download.oracle.com/javase/tutorial/java/IandI/super.html
You could use it to call a superclass's method (such as when you are overriding such a method, super.foo() etc) -- this would allow you to keep that functionality and add on to it with whatever else you have in the overriden method.
Super will call your parent method. See: http://leepoint.net/notes-java/oop/constructors/constructor-super.html
You call super() to specifically run a constructor of your superclass. Given that a class can have multiple constructors, you can either call a specific constructor using super() or super(param,param) oder you can let Java handle that and call the standard constructor. Remember that classes that follow a class hierarchy follow the "is-a" relationship.
The first line of your subclass' constructor must be a call to super() to ensure that the constructor of the superclass is called.
I just tried it, commenting super(); does the same thing without commenting it as #Mark Peters said
package javaapplication6;
/**
*
* #author sborusu
*/
public class Super_Test {
Super_Test(){
System.out.println("This is super class, no object is created");
}
}
class Super_sub extends Super_Test{
Super_sub(){
super();
System.out.println("This is sub class, object is created");
}
public static void main(String args[]){
new Super_sub();
}
}
From oracle documentation page:
If your method overrides one of its superclass's methods, you can invoke the overridden method through the use of the keyword super.
You can also use super to refer to a hidden field (although hiding fields is discouraged).
Use of super in constructor of subclasses:
Invocation of a superclass constructor must be the first line in the subclass constructor.
The syntax for calling a superclass constructor is
super();
or:
super(parameter list);
With super(), the superclass no-argument constructor is called. With super(parameter list), the superclass constructor with a matching parameter list is called.
Note: If a constructor does not explicitly invoke a superclass constructor, the Java compiler automatically inserts a call to the no-argument constructor of the superclass. If the super class does not have a no-argument constructor, you will get a compile-time error.
Related post:
Polymorphism vs Overriding vs Overloading
I'm dealing with a class which extends JFrame.
It's not my code and it makes a call to super before it begins constructing the GUI. I'm wondering why this is done since I've always just accessed the methods of the superclass without having to call super();
There is an implicit call to super() with no arguments for all classes that have a parent - which is every user defined class in Java - so calling it explicitly is usually not required. However, you may use the call to super() with arguments if the parent's constructor takes parameters, and you wish to specify them. Moreover, if the parent's constructor takes parameters, and it has no default parameter-less constructor, you will need to call super() with argument(s).
An example, where the explicit call to super() gives you some extra control over the title of the frame:
class MyFrame extends JFrame
{
public MyFrame() {
super("My Window Title");
...
}
}
A call to your parent class's empty constructor super() is done automatically when you don't do it yourself. That's the reason you've never had to do it in your code. It was done for you.
When your superclass doesn't have a no-arg constructor, the compiler will require you to call super with the appropriate arguments. The compiler will make sure that you instantiate the class correctly. So this is not something you have to worry about too much.
Whether you call super() in your constructor or not, it doesn't affect your ability to call the methods of your parent class.
As a side note, some say that it's generally best to make that call manually for reasons of clarity.
None of the above answers answer the 'why'.
Found a good explanation here:
A subclass can have its own private data members, so a subclass can
also have its own constructors.
The constructors of the subclass can initialize only the instance
variables of the subclass. Thus, when a subclass object is
instantiated the subclass object must also automatically execute one
of the constructors of the superclass.
You might also want to read everything about the super keyword here or watch everything about the super keyword here.
We can access super class elements by using super keyword
Consider we have two classes, Parent class and Child class, with different implementations of method foo. Now in child class if we want to call the method foo of parent class, we can do so by super.foo(); we can also access parent elements by super keyword.
class parent {
String str="I am parent";
//method of parent Class
public void foo() {
System.out.println("Hello World " + str);
}
}
class child extends parent {
String str="I am child";
// different foo implementation in child Class
public void foo() {
System.out.println("Hello World "+str);
}
// calling the foo method of parent class
public void parentClassFoo(){
super.foo();
}
// changing the value of str in parent class and calling the foo method of parent class
public void parentClassFooStr(){
super.str="parent string changed";
super.foo();
}
}
public class Main{
public static void main(String args[]) {
child obj = new child();
obj.foo();
obj.parentClassFoo();
obj.parentClassFooStr();
}
}
It simply calls the default constructor of the superclass.
We use super keyword to call the members of the Superclass.
As a subclass inherits all the members (fields, methods, nested classes) from its parent and since Constructors are NOT members (They don't belong to objects. They are responsible for creating objects), they are NOT inherited by subclasses.
So we have to explicitly give the call for parent constructor so that the chain of constructor remains connected if we need to create an object for the superclass. At the time of object creation, only one constructor can be called. Through super, we can call the other constructor from within the current constructor when needed.
If you are thinking why it's there for a class that is not extending any other class, then just remember every class follows object class by default. So it's a good practice to keep super in your constructor.
Note: Even if you don't have super() in your first statement, the compiler will add it for you!
We can Access SuperClass members using super keyword
If your method overrides one of its superclass's methods, you can invoke the overridden method through the use of the keyword super. You can also use super to refer to a hidden field (although hiding fields is discouraged). Consider this class, Superclass:
public class Superclass {
public void printMethod() {
System.out.println("Printed in Superclass.");
}
}
// Here is a subclass, called Subclass, that overrides printMethod():
public class Subclass extends Superclass {
// overrides printMethod in Superclass
public void printMethod() {
super.printMethod();
System.out.println("Printed in Subclass");
}
public static void main(String[] args) {
Subclass s = new Subclass();
s.printMethod();
}
}
Within Subclass, the simple name printMethod() refers to the one declared in Subclass, which overrides the one in Superclass. So, to refer to printMethod() inherited from Superclass, Subclass must use a qualified name, using super as shown. Compiling and executing Subclass prints the following:
Printed in Superclass.
Printed in Subclass
as constructor is not a part of class,
so while calling it cannot be implemented,
by using SUPER() we can call the members and memberfunctions in constructor.