Virtual functions in C++

Introduction

Virtual functions let a program choose the right function to run when using a base class pointer or reference. This helps when you want different behaviors for different types of objects.

When you have a base class pointer pointing to different derived class objects and want to call the correct function for each.

When you want to write flexible code that can work with many types of objects sharing a common base.

When you want to change or extend behavior in derived classes without changing the base class code.

Syntax

C++

class Base {
public:
    virtual void functionName();
};

The keyword virtual before a function means it can be overridden in derived classes.

Virtual functions must be members of a class.

Examples

This base class has a virtual function sound that can be changed in derived classes.

C++

class Animal {
public:
    virtual void sound() {
        std::cout << "Some sound" << std::endl;
    }
};

The derived class Dog overrides the virtual function to provide its own behavior.

C++

class Dog : public Animal {
public:
    void sound() override {
        std::cout << "Bark" << std::endl;
    }
};

Using a base class pointer to a derived object calls the Dog version of sound because it is virtual.

C++

#include <iostream>

int main() {
    Animal* a = new Dog();
    a->sound();
    delete a;
    return 0;
}

Sample Program

This program shows how virtual functions let the program call the right sound function depending on the actual object type, even when using base class pointers.

C++

#include <iostream>

class Animal {
public:
    virtual void sound() {
        std::cout << "Some sound" << std::endl;
    }
    virtual ~Animal() = default;
};

class Dog : public Animal {
public:
    void sound() override {
        std::cout << "Bark" << std::endl;
    }
};

class Cat : public Animal {
public:
    void sound() override {
        std::cout << "Meow" << std::endl;
    }
};

int main() {
    Animal* animal1 = new Dog();
    Animal* animal2 = new Cat();

    animal1->sound();  // Calls Dog's sound
    animal2->sound();  // Calls Cat's sound

    delete animal1;
    delete animal2;
    return 0;
}

Important Notes

If a function is not virtual, the base class version is called even if the pointer points to a derived object.

Use override keyword in derived classes to help catch mistakes.

Remember to make destructors virtual if you use virtual functions to avoid memory leaks.

Summary

Virtual functions allow dynamic choice of function at runtime based on object type.

They enable polymorphism, making code flexible and reusable.

Always use virtual in base class and override in derived classes for clarity.

Practice

1. What is the main purpose of declaring a function as virtual in a C++ base class?

easy

A. To allow derived classes to override the function and enable runtime polymorphism

B. To make the function inline for faster execution

C. To prevent the function from being overridden in derived classes

D. To make the function static and shared among all objects

Solution

1. Step 1: Understand virtual function role

   A virtual function in a base class allows derived classes to provide their own implementation that will be called at runtime.

2. Step 2: Recognize runtime polymorphism

   This mechanism enables dynamic dispatch, meaning the program decides which function to call based on the actual object type, not the pointer type.

3. Final Answer:

   To allow derived classes to override the function and enable runtime polymorphism -> Option A

4. Quick Check:

   Virtual function = runtime polymorphism [OK]

Hint: Virtual means choose function at runtime, not compile time [OK]

Common Mistakes:

• Confusing virtual with inline functions
• Thinking virtual prevents overriding
• Assuming virtual makes function static

2. Which of the following is the correct syntax to declare a virtual function named display in a base class?

easy

A. virtual void display();

B. void virtual display();

C. virtual display void();

D. void display() virtual;

Solution

1. Step 1: Recall correct virtual function syntax

   The keyword virtual must come before the return type and function name.

2. Step 2: Check each option

   virtual void display(); correctly places virtual before void display();. Others have incorrect order or placement.

3. Final Answer:

   virtual void display(); -> Option A

4. Quick Check:

   virtual before return type = correct syntax [OK]

Hint: Put 'virtual' before return type in function declaration [OK]

Common Mistakes:

• Placing virtual after function name
• Mixing order of keywords
• Adding virtual at the end of declaration

3. Consider the following code:

class Base {
public:
    virtual void show() { std::cout << "Base"; }
};
class Derived : public Base {
public:
    void show() override { std::cout << "Derived"; }
};

int main() {
    Base* ptr = new Derived();
    ptr->show();
    delete ptr;
    return 0;
}

What is the output when this program runs?

medium

A. Base

B. Runtime error

C. Derived

D. Compilation error

Solution

1. Step 1: Analyze pointer and object types

   The pointer is of type Base* but points to a Derived object.

2. Step 2: Understand virtual function call

   Because show() is virtual, the Derived version is called at runtime, printing "Derived".

3. Final Answer:

   Derived -> Option C

4. Quick Check:

   Virtual function calls derived version [OK]

Hint: Virtual calls use actual object type, not pointer type [OK]

Common Mistakes:

• Assuming base class function runs
• Confusing override keyword effect
• Expecting compilation or runtime errors

4. Identify the error in the following code snippet:

class Base {
public:
    void display() { std::cout << "Base"; }
};
class Derived : public Base {
public:
    void display() override { std::cout << "Derived"; }
};

medium

A. Derived class function cannot have the same name as base class function

B. No error, code is correct

C. Override keyword is not allowed in derived class

D. Base class function must be declared virtual to allow override

Solution

1. Step 1: Check base class function declaration

   The base class function display() is not declared virtual.

2. Step 2: Understand override keyword usage

   The override keyword requires the base function to be virtual; otherwise, it causes a compilation error.

3. Final Answer:

   Base class function must be declared virtual to allow override -> Option D

4. Quick Check:

   Override needs virtual base function [OK]

Hint: Base function must be virtual to use override in derived [OK]

Common Mistakes:

• Ignoring missing virtual in base
• Thinking override is optional
• Assuming same name always means override

5. Given the classes below, what will be the output of the program?

class Animal {
public:
    virtual void sound() { std::cout << "Animal sound "; }
};
class Dog : public Animal {
public:
    void sound() override { std::cout << "Bark "; }
};
class Cat : public Animal {
public:
    void sound() override { std::cout << "Meow "; }
};

int main() {
    Animal* animals[3];
    animals[0] = new Animal();
    animals[1] = new Dog();
    animals[2] = new Cat();

    for (int i = 0; i < 3; i++) {
        animals[i]->sound();
    }

    for (int i = 0; i < 3; i++) {
        delete animals[i];
    }
    return 0;
}

hard

A. Animal sound Animal sound Animal sound

B. Animal sound Bark Meow

C. Bark Meow Animal sound

D. Compilation error due to array of pointers

Solution

1. Step 1: Analyze array of Animal pointers

   The array holds pointers to objects of types Animal, Dog, and Cat respectively.

2. Step 2: Understand virtual function calls in loop

   Calling sound() on each pointer calls the overridden version based on actual object type, printing "Animal sound ", "Bark ", and "Meow ".

3. Final Answer:

   Animal sound Bark Meow -> Option B

4. Quick Check:

   Virtual calls print actual object sounds [OK]

Hint: Virtual functions call derived versions in arrays of base pointers [OK]

Common Mistakes:

• Expecting base class sound for all
• Confusing array pointer syntax
• Assuming compilation error with polymorphism