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12.2 — Virtual functions and polymorphism

In the previous lesson on pointers and references to the base class of derived objects, we took a look at a number of examples where using pointers or references to a base class had the potential to simplify code. However, in every case, we ran up against the problem that the base pointer or reference was only able to call the base version of a function, not a derived version.

Here’s a simple example of this behavior:

This example prints the result:

rBase is a Base

Because rBase is a Base pointer, it calls Base::GetName(), even though it’s actually pointing to the Base portion of a Derived object.

In this lesson, we will address this issue using virtual functions.

Virtual functions and polymorphism

A virtual function is a special type of function that resolves to the most-derived version of the function with the same signature. This capability is known as polymorphism.

To make a function virtual, simply place the “virtual” keyword before the function declaration.

Note that virtual functions and virtual base classes are two entirely different concepts, even though they share the same keyword.

Here’s the above example with a virtual function:

This example prints the result:

rBase is a Derived

Because rBase is a pointer to the Base portion of a Derived object, when rBase.GetName() is evaluated, it would normally resolve to Base::GetName(). However, Base::GetName() is virtual, which tells the program to go look and see if there are any more-derived versions of the function available. Because the Base object that rBase is pointing to is actually part of a Derived object, the program will check every inherited class between Base and Derived and use the most-derived version of the function that it finds. In this case, that is Derived::GetName()!

Let’s take a look at a slightly more complex example:

What do you think this program will output?

Let’s look at how this works. First, we instantiate a C class object. rBase is an A pointer, which we set to point to the A portion of the C object. Finally, we call rBase.GetName(). rBase.GetName() evaluates to A::GetName(). However, A::GetName() is virtual, so the compiler will check all the classes between A and C to see if it can find a more-derived match. First, it checks B::GetName(), and finds a match. Then it checks C::GetName() and finds a better match. It does not check D::GetName() because our original object was a C, not a D. Consequently, rBase.GetName() resolves to C::GetName().

As a result, our program outputs:

rBase is a C

A more complex example

Let’s take another look at the Animal example we were working with in the previous lesson. Here’s the original class:

And here’s the class with virtual functions:

Note that we didn’t make Animal::GetName() virtual. This is because GetName() is never overridden in any of the derived classes, therefore there is no need.

Now, using the virtual Speak() function, the following function should work correctly:

Indeed, this program produces the result:

Fred says Meow
Garbo says Woof

When cAnimal.Speak() is evaluated, the program notes that it is a virtual function. In the case where rAnimal is pointing to the Animal portion of a Cat object, the program looks at all the classes between Animal and Cat to see if it can find a more derived function. In that case, it finds Cat::Speak(). In the case where rAnimal points to the Animal portion of a Dog object, the program resolves the function call to Dog::Speak().

Similarly, the following array example now works as expected:

Which produces the result:

Fred says Meow
Garbo says Woof
Pooky says Woof
Truffle says Woof
Tyson says Meow
Zeke says Meow

Even though these two examples only use Cat and Dog, any other classes we derive from Animal would also work with our Report() function and Animal array without further modification! This is perhaps the biggest benefit of virtual functions -- the ability to structure your code in such a way that newly derived classes will automatically work with the old code without modification!

A word of warning: the signature of the derived class function must exactly match the signature of the base class virtual function in order for the derived class function to be used. If the derived class function has different parameter types, the program will likely still compile fine, but the virtual function will not resolve as intended.

Use of the virtual keyword

Technically, the virtual keyword is not needed in derived class. For example:

prints

rBase is a Derived

Exactly the same as if Derived::GetName() was explicitly tagged as virtual. Only the most base class function needs to be tagged as virtual for all of the derived functions to work virtually. However, having the keyword virtual on the derived functions does not hurt, and it serves as a useful reminder that the function is a virtual function rather than a normal one. Consequently, it’s generally a good idea to use the virtual keyword for virtualized functions in derived classes even though it’s not strictly necessary.

Return types of virtual functions

Under normal circumstances, the return type of a virtual function and it’s override must match. Thus, the following will not work:

However, there is one special case in which this is not true. If the return type of a virtual function is a pointer or a reference to a class, override functions can return a pointer or a reference to a derived class. These are called covariant return types. Here is an example:

Note that some older compilers (eg. Visual Studio 6) do not support covariant return types.

12.3 -- Virtual destructors, virtual assignment, and overriding virtualization
Index
12.1 -- Pointers and references to the base class of derived objects

41 comments to 12.2 — Virtual functions and polymorphism

  • eldeaca

    although it doesn't change the instance, it lets change in the future

  • eldeaca

    const on the left doesn't allow return any reference to the class itself

  • meet_999

    Hi Alex,
    Why i am getting compile time errors in following program:
    #include
    class Base {
    public:
    virtual Base& fun() const
    {
    std::cout<<"fun() in Base";
    return *this;
    }
    };
    class Derived : public Base
    {
    public:
    Derived& fun() const
    {
    std::cout<<"fun() in Derivedn";
    }
    };
    int main()
    {
    Base* p=new Derived();
    p->fun();
    delete p;
    p=0;
    return 0;
    }
    Errors:
    [Error] invalid initialization of reference of type 'Base&' from expression of type 'const Base'
    [Error] invalid initialization of reference of type 'Derived&' from expression of type 'const Derived'

    Please help me. How can i fix these errors?

  • Base &rBase = &cDerived; { can't work like dat.. )

    but

    Base &rBase = cDerived; (it works well like dat)

    WILL IT GIVES DE SAME RESULT OR DIFFERENT RESULT ?

    THANKS ALOT

  • vipulcr

    "Because rBase is a pointer to the Base portion of a Derived object, when rBase.GetName() is evaluated, it would normally resolve to Base::GetName()."

    It's a reference bro, not pointer and you haven't fixed that mistake in example with virtual functions. second example still consists it. Hope you fix it.

  • tanekim77

    Typo: Base &rBase = &cDerived;

    Please fix.

  • BonnieB

    Granted I'm new to C++, but I'm reading through Stroustrup's book, and his explanation of how virtual functions work is completely different than what's presented here. The whole thing about how rBase.GetName() will check A::GetName then B::GetName then C::GetName..... Actually the way it seems to work is that each object will have a hidden field for a virtual functions table (a vtbl). A vtbl is a lookup table of pointers to functions. If we create a B object, then its vtbl would have pointers to B functions, and if we create a C object, its vtbl would have pointers to C functions. So when we have a variable to hold a reference to an A object, and we assign it a C object, then that A variable doesn't actually know it's holding a C, nor does it need to know. It just looks at the vtbl and follows the pointer to a function.

  • M Asif

    I think everything is fine but have only two confusion

    1. first you did'nt use namespce in any case just becouse of this it will show error of predefined objects like cout (not defined in this scope).

    2. we can't do

    Base &rbase = &cDrived;
    although we can do

    Base *rBase = &cDrived;

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