when is super() called in the below code

Question

Trying to understand when does the super() method called. In the below code, Child class has a no-argument-constructor with this(), so compiler cannot insert super(). Then how is parent constructor called.

public class Parent
{
    public Parent()
    {
        System.out.println("In parent constructor");
    }
 }


public class Child extends Parent
{
private int age;

public Child()
{   
    this(10);
    System.out.println("In child constructor with no argument");
}

public Child(int age)
{
    this.age = age;
    System.out.println("In child constructor with argument");
}

public static void main(String[] args)
{
    System.out.println("In main method");
    Child child = new Child();
}

}

Output :

In main method

In parent constructor

In child constructor with argument

In child constructor with no argument

Answer 1

Here is what happens:

public class Parent
{
    public Parent()
    {
        System.out.println("In parent constructor"); // 4 <------
    }
}


public class Child extends Parent
{
    private int age;

    public Child()
    {
        this(10); // 2 <------
        System.out.println("In child constructor with no argument"); // 6 <------
    }

    public Child(int age)
    {
        // 3 -- implicit call to super()  <------
        this.age = age;
        System.out.println("In child constructor with argument"); // 5 <------
    }

    public static void main(String[] args)
    {
        System.out.println("In main method"); // 1 <------
        Child child = new Child();
    }
}

Answer 2

super() is called implicitly before the first line of any constructor, unless it explicitly calls super() or an overload itself, or the class is java.lang.Object.