Why call super in a constructor

Question

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

It simply calls the default constructor of the superclass

[java] Why call super() in a constructor?

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