Super keyword in Java - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of super keyword in constructor call

What is the output of this Java program that uses super to call the parent class constructor?

Java

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

class Child extends Parent {
    Child() {
        super();
        System.out.println("Child constructor");
    }
}

public class Test {
    public static void main(String[] args) {
        new Child();
    }
}

A Child constructor

B Child constructor\nParent constructor

C Parent constructor

D Parent constructor\nChild constructor

💡 Hint

Remember that super() calls the parent constructor before the child constructor body runs.

The super() call in the Child constructor invokes the Parent constructor first, printing "Parent constructor". Then the Child constructor prints "Child constructor".

❓ Predict Output

intermediate

Using super to access parent method

What is the output of this Java program that uses super to call a parent class method?

Java

class Animal {
    void sound() {
        System.out.println("Animal sound");
    }
}

class Dog extends Animal {
    void sound() {
        System.out.println("Dog barks");
    }
    void printSounds() {
        sound();
        super.sound();
    }
}

public class Test {
    public static void main(String[] args) {
        Dog d = new Dog();
        d.printSounds();
    }
}

A Animal sound\nDog barks

B Dog barks\nAnimal sound

C Dog barks\nDog barks

D Animal sound\nAnimal sound

💡 Hint

The super keyword calls the parent class method, while calling the method normally calls the child class version.

Calling sound() inside printSounds() calls the overridden method in Dog, printing "Dog barks". Calling super.sound() calls the Animal class method, printing "Animal sound".

❓ Predict Output

advanced

Output when super keyword used with variables

What is the output of this Java program that uses super to access a parent class variable?

Java

class Base {
    int x = 10;
}

class Derived extends Base {
    int x = 20;
    void printX() {
        System.out.println(x);
        System.out.println(super.x);
    }
}

public class Test {
    public static void main(String[] args) {
        Derived d = new Derived();
        d.printX();
    }
}

A 20\n10

B 10\n20

C 10\n10

D 20\n20

💡 Hint

The variable x in Derived hides the one in Base. Use super.x to access the parent's variable.

The x in Derived is 20, so System.out.println(x) prints 20. The super.x accesses the Base class variable, which is 10.

❓ Predict Output

advanced

Output of super in multi-level inheritance

What is the output of this Java program using super in a multi-level inheritance chain?

Java

class A {
    void show() {
        System.out.println("A show");
    }
}

class B extends A {
    void show() {
        System.out.println("B show");
    }
}

class C extends B {
    void show() {
        super.show();
        System.out.println("C show");
    }
}

public class Test {
    public static void main(String[] args) {
        C c = new C();
        c.show();
    }
}

A B show\nC show

B A show\nC show

C C show\nB show

D A show\nB show\nC show

💡 Hint

In class C, super.show() calls the immediate parent class B's show() method.

The super.show() in C calls B's show(), which prints "B show". Then C prints "C show".

🧠 Conceptual

expert

Error caused by incorrect use of super

Which option will cause a compile-time error due to incorrect use of super in Java?

A Using <code>super</code> to call a parent class method from a child instance method

B Using <code>super</code> to access a parent class variable when it is hidden by child variable

C Using <code>super()</code> inside a static method

D Using <code>super()</code> as the first statement in a child constructor

💡 Hint

Remember that super relates to instance context, not static context.

Using super() inside a static method causes a compile-time error because static methods do not belong to an instance, so super cannot be used there.