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What is the purpose of calling super class's method while overriding a method?

While overriding methods like onPostExecute from AsyncTask we can avoid or remove super.onPostExecute(result); from it. Can you tell me why and when do we need to use this call?
as I here implemented AsyncTask as a inner class in my Main Activity
private class CustomAsyncTask extends AsyncTask<String, Void, Event> {
#Override
protected Event doInBackground(String... strings) {
Event result = Utils.fetchEarthquakeData(strings[0]);
return result;
}
#Override
protected void onPostExecute(Event result) {
super.onPostExecute(result);// what is the effect of calling this? If it was possible to skip or remove it.
updateUi(result);
}
}

When you override a method, you redefine the method i.e. if this method is called on the instance of the child class (in which you have overridden the method), this new version of the definition will be called.
When you override a method (i.e. redefine or in other words, replace the superclass definition with the child's version of the definition), you have two choices:
Do not use anything from the superclass definition and rewrite it in a completely new way.
Reuse the superclass definition at some point (in the beginning/middle/end) in the new definition. It's like putting a cherry on the ice cream cone where ice cream cone is the superclass definition of the method and cherry is added in the new definition.
Note that as already pointed out by Andy Turner, a requirement to call super is considered an antipattern.

when we override some method then it means:
we have interface in our project.
In that interface One Class is Parent and Other one is Child.
Now Overriding means having same method in both Parent and Child.
Se below example:
class Parent
{
public Parent()
{
"Inside Parent Constructor";
}
public void meth()
{
"Inside Parent Method";
}
}
class Child
{
public Child()
{
"Inside Child Constructor";
}
public void meth()
{
super();
"Inside Parent Method";
}
}
Due to super keyword in Child Class it will go to Constructor of Parent. and Print "Inside Parent Constructor".
So main thing is that many of the time Andoid OS will want to access Parent Class from which the method is overrided and for that we write super().

What is the purpose of calling super class's method while overriding a method?
The purpose is to run the method in the class from which the current class inherits from to perform some work that class is responsible, as it could be some setup of the super class' state.
For your particular case, there is no need to call super.onPostExcecute() as stated in the question below:
Should I call super.onPostExecute(result) in Android AsyncTask?

Java: Is there a way to use a method defined in a superclass, along with added functionality in the subclass? [duplicate]

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

Invoke a method of a class which is implemented in her supperclass

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.

What does it mean to #override the method of a super class?

Let's say I have a method called mymethod()
and this method overrides the method of the super class method.
What does it mean to override a method?
Does that mean mymethod() ignores everything that is in the method of the superclass, or does that means mymethod() also includes everything in the superclass method?
When overriding a method, can I only override the methods of the same name, or I can override methods of any name?
thanks.

An example:
public class Base {
public void saySomething() {
System.out.println("Hi, I'm a base class");
}
}
public class Child extends Base {
#Override
public void saySomething() {
System.out.println("Hi, I'm a child class");
}
}
Now assume we have a main function somewhere...
public static void main(String [] args) {
Base obj = new Child();
obj.saySomething();
}
When this runs, it will call Child's version of saySomething, because you overrode the parent's version by giving a new version of the function in Child.
The #Override annotation allows other developers (and you, when you forget) to know that this method overrides something in a base class/interface, and it also allows the compiler to yell at you if you're not actually overriding anything in a base class. For example, if you got the number of arguments wrong for a function, the compiler will give you an error saying your #Override is incorrect.
For example:
public class Child extends Base {
#Override
public void saySomething(int x) {
System.out.println("I'm a child and x is: " + x);
}
}
The compiler will yell at you because this version of saySomething takes an argument, but the parent's version doesn't have an argument, so you're #Override-ing something that's not in the parent.
On super
The Child version of saySomething will not invoke the Base version, you have to do it yourself with super.method().
For example:
public class Child extends Base {
#Override
public void saySomething() {
super.saySomething();
System.out.println("I'm also a child");
}
}
If you ran the main and used this Child class, it would print out I'm a base and I'm also a child.

Overriding means that when you call a method on your object, your object's method is called instead of the super class. The #Override annotation is something you use to make sure that you are overriding the correct method of the superclass. If you annotate a method that does not exist in the superclass, the Java compiler will give you an error. This way you can be sure that you are overriding the correct methods. This is especially useful in cases like this:
public class MyClass {
...
public boolean equals(MyClass myClass) {
...
}
}
There is a logic-bug in the code above. You haven't actually overridden the Object class's equals method. If you add the #Override annotation:
public class MyClass {
...
#Override
public boolean equals(MyClass myClass) {
...
}
}
The Java compiler will now complain because there is no corresponding method in the parent class. You'll then know that the correct solution is:
public class MyClass {
...
#Override
public boolean equals(Object o) {
...
}
}
To call the parent class's method, you can call super.overriddenMethod() where overriddenMethod is the name of the method you have overridden. So if you want to do something in addition to what the parent class already does, you can do something like this:
public class MyClass {
...
#Override
public void overriddenMethod() {
super.overriddenMethod();
/* whatever additional stuff you want to do */
}
}

If an inheriting class has on override method of the same name as the parent class it will be called instead of the one in the parent class. This only works if the names are the same, and of course if the signature of the method matches your call to the method.

What does it mean to override a method?
It means you replace the super class definition of the method with your own definition.
does that mean mymethod() ignores everything that is in the method of the super class?
or does that means mymethod() also includes everything in the superclass method?
You can choose whether to include the super class definition within your definition. To include it, you need to call super.mymethod() within mymethod().
and when overriding a method, can I only override the methods of the same name, or I can override methods of any name?
To override a method, you must supply a method in the sub class with the same signature (which means the same name, parameters and return type).
As a side note, the #Override annotation in your question does not actually cause your method to override another method. It causes a compile-time error if a method annotated with it does not have a signature matching a public or protected method of a super class (or interface as of 1.6).

I once had a student come to ask me why his code wasn't working. He had spent several days wondering why he could put something into a collection but was not able to find it. His code was something like:
public int hashcode()
instead of:
public int hashCode()
So the hashCode method never got called.
Adding #Overrides to a method makes it clear that you are overriding the method AND make sure that you really are overriding a method.

When you override a method of the super class, calling that method on your object calls its method instead of that of the super class.
You can call the super class's method (despite having overridden it) using super.methodName(). A common reason for this is when the overridden method would otherwise reimplement the super class method and add additional code specific to the extending class (public void methodName() { super.methodName(); /* more code */ }).
#Override annotation allows you to cause warning at compile time if the method isn't actually overriding anything. It isn't necessary, but these warning are a hit to you that you might have got the signature wrong in the extending class, forgot to implement the method at all in the super class, or some other silly mistake.

When do I use super()?

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