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

public void foo() {
System.out.println("Class A");
}
}

class B extends A {
public B(){}

public void foo() {
System.out.println("Class B");
}
}

class main {
public static void main(String[] args) {

A a = new B();
}
}

What does it print? Class A or Class B?

Question

class A {
    public A() {
          foo();
    }
    
    public void foo() {
            System.out.println("Class A");
    }
}

class B extends A {
    public B(){}
    
    public void foo() {
            System.out.println("Class B");
    }
}

class main {
     public static void main(String[] args) {
          
          A a = new B();
     }
}

What does it print? Class A or Class B?

tommy · Accepted Answer

it's absolutely Java code, no C++!! It will print ClassB, because default contructor of A will be automatically called from contructor of B (it isn't mandatory to call super default contructor from a derived default contructor). The method with the same signature overrides the base class method.
If you "translate" the code to C++ without declaring as virtual the A's foo method, then it will print class A. Otherwise it will print Class B.

Usee · Answer

If this is C++, then it prints out A. But this looks like java and it printed B

YI · Answer

thats why you are rejected... it will definitely prints out Class A, if the fun is defined as virtual, it would print out class B

Sigiloso · Answer

Class B

Ghareeb · Answer

I don't think it'll print anything. There's no super(), so A's constructor will never be called.

Sigiloso · Answer

Class B is correct

Mahigupta · Answer

Its going to print Class B. What is going to happen that when we say new B() its going to call constructor of Class B, but since Class B inherit class A, so first constructor of Class A will be called ,hence function foo will be executed. Now function foo is overwritten in Class B  ,so function foo defined in Class B will be executed. Hence it will print ClassB.

Hj0755 · Answer

Guys, Guys!! Has anyone of you tested the code in VC? Come on it is absolutely Class A.

YY · Answer

It prints B

Sigiloso · Answer

Class B. First it's java style code. We can see it uses "System.out.println". If it's c++, it should be written as cout. Second the knowledge point it's inheritance. Variable a is actually defined as class B. That's why it will print Class B.

raat · Answer

it will "class B" twice. its a Java code due to extends keyword & it is tested.
The reason is: it will go to constructor of A, but then it will call foo() of B, then again constructor of B with foo() of B.
I know its weird, but that's what I found!

cq · Answer

This is not C++, Read item 9 Effective C++/Meyers. Never call virtual functions during construction or destruction. Moreover the syntax  A a = new B(); won't work in C++. No viable conversion possible.

Sigiloso · Answer

I have tried the problem and it prints Class B in JAVA atleast thats because before a class object calls its own constructor it makes sure to call the super's default constructor if any.

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