Why default constructor is required in a parent class if it has an argument-ed constructor?

Question

Why default constructor is required(explicitly) in a parent class if it has an argumented constructor

class A {    
  A(int i){    
  }
}

class B extends          


        

Answer 1

There are a few things to be noted when using constructors and how you should declare them in your base class and super class. This can get somewhat confusing solely because there can be many possibilities of the availability or existence of constructors in the super class or base class. I will try to delve into all the possibilities:

• If you explicitly define constructors in any class(base class/super class), the Java compiler will not create any other constructor for you in that respective class.

• If you don't explicitly define constructors in any class(base class/super class), the Java compiler will create a no-argument constructor for you in that respective class.

• If your class is a base class inheriting from a super class and you do not explicitly define constructors in that base class, not only will a no-argument constructor be created for you (like the above point) by the compiler, but it will also implicitly call the no-argument constructor from the super class.

  class A 
  {
     A() 
   {
    super();
   }
  }
  

• Now if you do not explicity type super(), (or super(parameters)), the compiler will put in the super() for you in your code.

• If super() is being called (explicitly or implicitly by the compiler) , the compiler will expect your superclass to have a constructor without parameters. If it does not find any constructor in your superclass without parameters, it will give you a compiler error.

• Similary if super(parameters) is called, the compiler will expect your superclass to have a constructor with parameters(number and type of parameters should match). If it does not find such a constructor in your superclass, it will give you a compiler error. ( Super(parameters) can never be called implicitly by the compiler. It has to be explicitly put in your code if one is required.)

We can summarize a few things from the above rules

• If your superclass only has a constructor with parameters and has no no-argument constructor, you must have an explicit super(parameters) statement in your constructor. This is because if you do not do that a super() statement will be implicitly put in your code and since your superclass does not have a no-argument constructor, it will show a compiler error.
• If your superclass has a constructor with parameters and another no-argument constructor, it is not necessary to have an explicit super(parameters) statement in your constructor. This is because a super() statement will be implicitly put in your code by the compiler and since your superclass has a no-argument constructor, it will work fine.
• If your superclass only has a no-argument constructor you can refer to the point above as it is the same thing.

Another thing to be noted is if your superclass has a private constructor, that will create an error when you compile your subclass. That is because if you don't write a constructor in your subclass it will call the superclass constructor and the implicit super() will try to look for a no-argument constructor in the superclass but will not find one.