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javaobjectconstructorsuperclass

invoking constructor in java

class A {

    A() {
        System.out.print("A");
    }
}

class B extends A {
     B() {
        System.out.print("B");
    }
}

class C extends B {
 C() {
        System.out.print("C");
    }
}

public class My extends C {
My(){
super();
}
    public static void main(String[] args) {
        My m = new My();
    }
}

Question starts from one Interview Question (what happens when an object is created in Java?)

and answer is...

The constructor for the most derived class is invoked. The first thing a constructor does is call the consctructor for its superclasses. This process continues until the constrcutor for java.lang.Object is called, as java.lang.Object is the base class for all objects in java. Before the body of the constructor is executed, all instance variable initializers and initialization blocks are executed. Then the body of the constructor is executed. Thus, the constructor for the base class completes first and constructor for the most derived class completes last.

So, according to above statement, answer should be ABCC, but it showing only ABC. Although, when i'm commenting the super() in derived constructor. Then, output is ABC. Please, help me to figure out, did i misunderstand the above paragraph. ?

Solution

  • No, the answer is ABC

    My m = new My();

    The above first invokes My class, then a super call is made to its super class i.e., C class, then a super call to B class is made, then a super call to A Class, then a Super call to java.lang.Object as all objects extend java.lang.Object.

    Thus the answer is ABC.

    You don't really need to explicitly call super() in your My class as it'd be included by the compiler unless you call an overloaded constructor of that class like this(something).