Super keyword in Java

Introduction

The super keyword helps you access methods and variables from a parent class. It is useful when you want to use or extend the behavior of the parent class.

When you want to call a parent class constructor from a child class.

When you want to access a parent class method that is overridden in the child class.

When you want to access a parent class variable that is hidden by a child class variable.

Syntax

Java

super.methodName();
super.variableName;
super();  // calls parent class constructor

super() must be the first statement in a child class constructor if used.

You can use super to avoid confusion when child and parent have variables or methods with the same name.

Examples

This example shows how super calls the parent method and accesses the parent variable.

Java

class Parent {
    int x = 10;
    void show() {
        System.out.println("Parent show");
    }
}

class Child extends Parent {
    int x = 20;
    void show() {
        System.out.println("Child show");
        super.show();  // calls Parent's show
        System.out.println(super.x);  // accesses Parent's x
    }
}

This example shows how super() calls the parent constructor from the child constructor.

Java

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

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

Sample Program

This program shows how the child class uses super to access the parent class variable and method even when they are overridden or hidden.

Java

class Parent {
    int number = 100;
    void display() {
        System.out.println("Parent display: " + number);
    }
}

class Child extends Parent {
    int number = 200;
    void display() {
        System.out.println("Child display: " + number);
        System.out.println("Parent number using super: " + super.number);
        super.display();
    }
}

public class Main {
    public static void main(String[] args) {
        Child c = new Child();
        c.display();
    }
}

Important Notes

If you do not use super() in a child constructor, Java automatically calls the parent's no-argument constructor.

You cannot use super in static methods because super refers to instance members.

Summary

super helps access parent class methods and variables.

Use super() to call the parent constructor from a child constructor.

It is useful when child class overrides or hides parent class members.

Practice

1. What does the super keyword do in Java?

easy

A. It defines a static method in the class.

B. It creates a new object of the child class.

C. It accesses methods and variables from the parent class.

D. It terminates the program execution.

Solution

1. Step 1: Understand the role of super

   The super keyword is used to refer to the parent class's members (methods or variables) from a child class.

2. Step 2: Compare options with definition

   Only It accesses methods and variables from the parent class. correctly describes this behavior. Other options describe unrelated actions.

3. Final Answer:

   It accesses methods and variables from the parent class. -> Option C

4. Quick Check:

   super accesses parent members = A [OK]

Hint: Remember: super means parent class access [OK]

Common Mistakes:

• Thinking super creates new objects
• Confusing super with this keyword
• Assuming super ends program

2. Which of the following is the correct way to call a parent class constructor in Java?

easy

A. super();

B. this();

C. parent();

D. base();

Solution

1. Step 1: Recall syntax for parent constructor call

   In Java, super() is used inside a child constructor to call the parent class constructor.

2. Step 2: Evaluate options

   Only super(); uses the correct keyword super(). Others are invalid or refer to different concepts.

3. Final Answer:

   super(); -> Option A

4. Quick Check:

   Parent constructor call = super() [OK]

Hint: Use super() to call parent constructor [OK]

Common Mistakes:

• Using this() instead of super()
• Trying to call parent() which is invalid
• Confusing base() with super()

3. What will be the output of the following code?

class Parent {
  int x = 10;
}
class Child extends Parent {
  int x = 20;
  void printX() {
    System.out.println(super.x);
  }
}
public class Test {
  public static void main(String[] args) {
    Child c = new Child();
    c.printX();
  }
}

medium

A. 20

B. Compilation error

C. 0

D. 10

Solution

1. Step 1: Understand variable hiding and super usage

   The child class has its own x = 20, but super.x accesses the parent's x which is 10.

2. Step 2: Trace the print statement

   The method printX() prints super.x, so it prints 10.

3. Final Answer:

   10 -> Option D

4. Quick Check:

   super.x accesses parent variable = 10 [OK]

Hint: super.variable accesses parent class variable [OK]

Common Mistakes:

• Printing child variable instead of parent
• Confusing super.x with this.x
• Expecting compilation error

4. Identify the error in this code snippet:

class Parent {
  void show() {
    System.out.println("Parent show");
  }
}
class Child extends Parent {
  void show() {
    super.show();
    System.out.println("Child show");
  }
  void display() {
    super();
  }
}

medium

A. super.show() is invalid inside child class

B. super() cannot be called like a method in display()

C. Child class cannot override show() method

D. No error, code is correct

Solution

1. Step 1: Check usage of super in methods

   Calling super.show() inside overridden method is valid to call parent method.

2. Step 2: Analyze super() call in display()

   super() can only be used to call parent constructor inside child constructor, not as a method call elsewhere.

3. Final Answer:

   super() cannot be called like a method in display() -> Option B

4. Quick Check:

   super() only in constructor = D [OK]

Hint: super() only calls parent constructor inside child constructor [OK]

Common Mistakes:

• Using super() outside constructor
• Thinking super.show() is invalid
• Believing overriding is not allowed

5. Given these classes:

class Animal {
  String name;
  Animal(String name) {
    this.name = name;
  }
  void sound() {
    System.out.println("Animal sound");
  }
}
class Dog extends Animal {
  Dog() {
    super("Dog");
  }
  void sound() {
    super.sound();
    System.out.println("Bark");
  }
}
public class Test {
  public static void main(String[] args) {
    Dog d = new Dog();
    d.sound();
    System.out.println(d.name);
  }
}

What is the output when running Test.main()?

hard

A. Animal sound
Bark
Dog

B. Bark
Animal sound
Dog

C. Animal sound
Dog
Bark

D. Compilation error due to super()

Solution

1. Step 1: Understand constructor chaining

   The Dog constructor calls super("Dog"), setting name to "Dog" in Animal.

2. Step 2: Trace the sound() method call

   Dog.sound() calls super.sound() which prints "Animal sound", then prints "Bark".

3. Step 3: Print the name field

   Printing d.name outputs "Dog".

4. Final Answer:

   Animal sound Bark Dog -> Option A

5. Quick Check:

   super() sets name, super.sound() prints parent sound = A [OK]

Hint: super() sets parent state; super.method() calls parent method [OK]

Common Mistakes:

• Expecting Dog before Animal sound
• Confusing order of prints
• Thinking super() causes error