Saturday 18 August 2018

Understanding Virtual Functions in C++

Until I had to explain it to someone, I never appreciated how confusing the virtual keyword can be. After all, in certain situations there seems to be no functional difference between virtual and non-virtual functions except that your IDE or editor might complain at you.

This article is aimed at programmers who are new to C++ or an initiate to programming in general.


The term conjures up a vision of something that is non-tangible. That is, something that doesn't exist in the physical world. In C++, this meaning is extended to describe functions which may not exist. By that, we mean that it may not be defined.

A virtual function, therefore, is a function which may not be real. While a class may call a function it defines, a virtual function could also exist somewhere else in it's object hierarchy. The class doesn't know what function it's actually calling until it is compiled or possibly until runtime.

Essentially, the virtual keyword signals to a developer that the function is intended to be able to be implemented or overridden in a derived class.

The Base Class (Non-Virtual)

To demonstrate, start by creating a simple base class with two public functions: Foo() and Bar(). Inside the Foo() function, call Bar(). It may be helpful to add some output so you know which function is being called.

Something like this:

class Base {
    void Foo() { cout << "Base::Foo" << endl; Bar(); }
    void Bar() { cout << "Base::Bar" << endl; };};
In a main function, instantiate a new Base class and call Foo(). You should get ouput similar to this:
This is pretty standard fair and works as expected.

The Derived Class

Next, create a derived function that inherits from Base. In it, create a function with the same name and signature as Bar(). If printing out a statement, make sure you update it to report the new class name.
class Derived : public Base {
    void Bar() { cout << "Derived::Bar" << endl; }
Call Foo() on this new class and you'll see that you get the same output. The function Bar() is only called on the Base class.

The Base Class (Virtual)

Add the virtual keyword to the function Bar() in the Base class then try running the code again. The code should look like this:
class Base {
    void Foo() { cout << "Base::Foo" << endl; Bar(); }
    virtual void Bar() { cout << "Base::Bar" << endl; };};
You should get output like this:
This time, Bar() was called in the Derived class instead of the Base class. Why?

Virtual Explained

The virtual keyword signaled a change in visibility. A non-virtual function tells the compiler not to bother looking for another function with the same name, just to look for the function in the class where it is called. A virtual function tells the compiler to see if a function with a matching signature exists in any derived classes first, before calling the one in the same class.


Virtual functions are those whom are intended to exist higher up in the class hierarchy in derived classes and should be called if they are defined.

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