A constructors puzzle.

[Dmitry Zhuravlev]

Ladys and Gents, please help with the following:

I have an example code:

 class SuperClass {
    SuperClass() {
        System.out.println("SUPER");
    }

    SuperClass(int i) {
        this();
        System.out.println("SUPER " + i);
    }
}

class SubClass extends SuperClass {
    SubClass() {
        this(77);
        System.out.println("SUB");

    }
    SubClass(int i) {
        super(i);
        System.out.println("SUB" + i);
    }
}

public class Main {
    public static void main(String[] args)  {
        new SubClass();
}
}

Output is

SUPER
SUPER 77
SUB77
SUB

I dont understand why! Its seems that SUPER should be printed twice: first when SubClass() is called in the beginning and then because of this() in the SuperClass(). But one of these two options doesn't work. Why? Which one?

[Seetharaman Venkatasamy]

*this()* calls the overloaded constructor of class itself ; and *super()* calls the super class constructor *according to the arguments* .

do you have any IDE like eclipse ? if yes, please debug the flow.

hth

[Frits Walraven]

Hi Dmitry,

This is what happens:

constructor SubClass() is called
constructor SubClass(77) is called
constructor SuperClass(77) is called
constructor SuperClass() is called

from there going back
System.out.println("SUPER");
System.out.println("SUPER " + 77);
System.out.println("SUB" + 77);
System.out.println("SUB");

Regards,
Frits

[bhanu chowdary]

You can have either this or super in a constructor. When you call

 new SubClass()   

it will go to the no-arg constructor SubClass(), but since

this(77)

in this constructor it will go into the args constructor of SubClass and not no-args constructor of SuperClass.
Hope this helps.