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Does a super class need a default constructor?


By : DaveM86
Date : October 17 2020, 06:10 PM
it fixes the issue No, that is not a correct assumption.
You just have an error. Your derived class constructor must call the one and only Base constructor you provide which is a one parameter constructor.
code :
    Child(int a)
    : Base(a)
    {
        cout << "Child" << endl;
    }


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Implicit super constructor Num() is undefined for default constructor. Must define an explicit constructor, what is the


By : TOstojich
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further When you define your own constructor,the compiler does not provide a no-argument constructor for you. When you define a class with no constructor,the compiler inserts a no-arg constructor for you with a call to super().
code :
class Example{
}
class Example{

Example(){
super();   // an accessible no-arg constructor must be present for the class to compile.
}
class Num 
{
    Num(double x) 
    { 
        System.out.println( x ) ; 
    }
}

class Number extends Num 
{ 


 Number(double x){
 super(x);
 }

 public static void main(String[] args)
    { 
        Num num = new Num(2) ; 
    } 
} 

why does the default constructor in MyClass calls super i.e of Object class


By : user3508402
Date : March 29 2020, 07:55 AM
Does that help Every constructor must call either a different constructor of the same class or a constructor of its direct super class. The call to the super class constructor is added implicitly if you don't call it explicitly.
Since an instance of a class inherits the state of its ancestors, it must initialize it by calling the constructors of its ancestors.

How does default constructor used to instantiate object if it's super class no argument constructor has empty body?


By : Michael Jones
Date : March 29 2020, 07:55 AM
hop of those help? default constructor invoke no_argument constructor in super class reference and this last constructor used to instantiate object from this class,How ? i mean what does superclass constructor body exactly do ? ,
The syntax for calling a superclass constructor is
code :
super();  
super(parameter list);
Fragment.class.getConstructor().newInstance();
Child a = new Child() ;
a.getClass().getSuperclass().getConstructor().newInstance();

Why am I getting unreported exception when the super class default constructor has a 'throws' clause?


By : Ionut Ghica
Date : March 29 2020, 07:55 AM
seems to work fine You have not defined a default constructor for Sub, and Super is reporting a checked exception. To fix it add this constructor to Sub:

Why does the subclass in this Java code have a default constructor if the super-class does not have one?


By : Solag
Date : March 29 2020, 07:55 AM
it should still fix some issue This my friend would produce a compiler error no suitable constructor found for BinaryTree().
A default no-arg constructor would be created for a Class only if you don't provide any constructor for the class. If you provide a constructor to the class, then you have to create a no-arg constructor on your own.
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