CODEWITHSUNDEEP
javapolymorphism
user1104775
user1104775

Reputation: 259

Polymorphism and Constructors

I am an AP Java Student and I am practicing for my exam. I came across this question and I don't understand the answer:

Consider the following classes:

public class A
{
  public A() { methodOne(); }

  public void methodOne() { System.out.print("A"); }
}

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

  public void methodOne() { System.out.print("B"); }
}

What is the output when the following code is executed:

A obj = new B();

The correct answer is B*. Can someone please explain to me the sequence of method calls?

Upvotes: 19

Views: 14970

Answers (2)

Pubby
Pubby

Reputation: 53037

The base class must be constructed before the derived class.

First A() is called which calls methodOne() which prints B.

Next, B() is called which prints *.

Upvotes: 2

JB Nizet
JB Nizet

Reputation: 691685

The B constructor is called. The first implicit instruction of the B constructor is super() (call the default constructor of super class). So A's constructor is called. A's constructor calls super(), which invokes the java.lang.Object constructor, which doesn't print anything. Then methodOne() is called. Since the object is of type B, the B's version of methodOne is called, and B is printed. Then the B constructor continues executing, and * is printed.

It must be noted that calling an overridable method from a constructor (like A's constructor does) is very bad practice: it calls a method on an object which is not constructed yet.

Upvotes: 40

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