Why do we need virtual functions in C++?

Jake Wilson Source

I'm learning C++ and I'm just getting into virtual functions.

From what I've read (in the book and online), virtual functions are functions in the base class that you can override in derived classes.

But earlier in the book, when learning about basic inheritance, I was able to override base functions in derived classes without using virtual.

So what am I missing here? I know there is more to virtual functions, and it seems to be important so I want to be clear on what it is exactly. I just can't find a straight answer online.

c++virtual-functions

Answers

answered 9 years ago Henk Holterman #1

You need at least 1 level of inheritance and a downcast to demonstrate it. Here is a very simple example:

class Animal
{        
    public: 
      // turn the following virtual modifier on/off to see what happens
      //virtual   
      std::string Says() { return "?"; }  
};

class Dog: public Animal
{
    public: std::string Says() { return "Woof"; }
};

void test()
{
    Dog* d = new Dog();
    Animal* a = d;       // refer to Dog instance with Animal pointer

    cout << d->Says();   // always Woof
    cout << a->Says();   // Woof or ?, depends on virtual
}

answered 9 years ago h0b0 #2

You have to distinguish between overriding and overloading. Without the virtual keyword you only overload a method of a base class. This means nothing but hiding. Let's say you have a base class Base and a derived class Specialized which both implement void foo(). Now you have a pointer to Base pointing to an instance of Specialized. When you call foo() on it you can observe the difference that virtual makes: If the method is virtual, the implementation of Specialized will be used, if it is missing, the version from Base will be chosen. It is best practice to never overload methods from a base class. Making a method non-virtual is the way of its author to tell you that its extension in subclasses is not intended.

answered 9 years ago Alex Martelli #3

If the base class is Base, and a derived class is Der, you can have a Base *p pointer which actually points to an instance of Der. When you call p->foo();, if foo is not virtual, then Base's version of it executes, ignoring the fact that p actually points to a Der. If foo is virtual, p->foo() executes the "leafmost" override of foo, fully taking into account the actual class of the pointed-to item. So the difference between virtual and non-virtual is actually pretty crucial: the former allow runtime polymorphism, the core concept of OO programming, while the latter don't.

answered 9 years ago Steve314 #4

Without "virtual" you get "early binding". Which implementation of the method is used gets decided at compile time based on the type of the pointer that you call through.

With "virtual" you get "late binding". Which implementation of the method is used gets decided at run time based on the type of the pointed-to object - what it was originally constructed as. This is not necessarily what you'd think based on the type of the pointer that points to that object.

class Base
{
  public:
            void Method1 ()  {  std::cout << "Base::Method1" << std::endl;  }
    virtual void Method2 ()  {  std::cout << "Base::Method2" << std::endl;  }
};

class Derived : public Base
{
  public:
    void Method1 ()  {  std::cout << "Derived::Method1" << std::endl;  }
    void Method2 ()  {  std::cout << "Derived::Method2" << std::endl;  }
};

Base* obj = new Derived ();
  //  Note - constructed as Derived, but pointer stored as Base*

obj->Method1 ();  //  Prints "Base::Method1"
obj->Method2 ();  //  Prints "Derived::Method2"

EDIT - see this question.

Also - this tutorial covers early and late binding in C++.

answered 9 years ago M Perry #5

Here is how I understood not just what virtual functions are, but why they're required:

Let's say you have these two classes:

class Animal
{
    public:
        void eat() { std::cout << "I'm eating generic food."; }
};

class Cat : public Animal
{
    public:
        void eat() { std::cout << "I'm eating a rat."; }
};

In your main function:

Animal *animal = new Animal;
Cat *cat = new Cat;

animal->eat(); // Outputs: "I'm eating generic food."
cat->eat();    // Outputs: "I'm eating a rat."

So far so good, right? Animals eat generic food, cats eat rats, all without virtual.

Let's change it a little now so that eat() is called via an intermediate function (a trivial function just for this example):

// This can go at the top of the main.cpp file
void func(Animal *xyz) { xyz->eat(); }

Now our main function is:

Animal *animal = new Animal;
Cat *cat = new Cat;

func(animal); // Outputs: "I'm eating generic food."
func(cat);    // Outputs: "I'm eating generic food."

Uh oh... we passed a Cat into func(), but it won't eat rats. Should you overload func() so it takes a Cat*? If you have to derive more animals from Animal they would all need their own func().

The solution is to make eat() from the Animal class a virtual function:

class Animal
{
    public:
        virtual void eat() { std::cout << "I'm eating generic food."; }
};

class Cat : public Animal
{
    public:
        void eat() { std::cout << "I'm eating a rat."; }
};

Main:

func(animal); // Outputs: "I'm eating generic food."
func(cat);    // Outputs: "I'm eating a rat."

Done.

answered 7 years ago nitin_cherian #6

When you have a function in the base class, you can Redefine or Override it in the derived class.

Redefining a method : A new implementation for the method of base class is given in the derived class. Does not facilitate Dynamic binding.

Overriding a method: Redefining a virtual method of the base class in the derived class. Virtual method facilitates Dynamic Binding.

So when you said :

But earlier in the book, when learning about basic inheritance, I was able to override base methods in derived classes without using 'virtual'.

you were not overriding it as the method in the base class was not virtual, rather you were redefining it

answered 6 years ago Kev #7

It helps if you know the underlying mechanisms. C++ formalizes some coding techniques used by C programmers, "classes" replaced using "overlays" - structs with common header sections would be used to handle objects of different types but with some common data or operations. Normally the base struct of the overlay (the common part) has a pointer to a function table which points to a different set of routines for each object type. C++ does the same thing but hides the mechanisms i.e. the C++ ptr->func(...) where func is virtual as C would be (*ptr->func_table[func_num])(ptr,...), where what changes between derived classes is the func_table contents. [A non-virtual method ptr->func() just translates to mangled_func(ptr,..).]

The upshot of that is that you only need to understand the base class in order to call the methods of a derived class, i.e. if a routine understands class A, you can pass it a derived class B pointer then the virtual methods called will be those of B rather than A since you go through the function table B points at.

answered 4 years ago Cheers and hth. - Alf #8

You need virtual methods for safe downcasting, simplicity and conciseness.

That’s what virtual methods do: they downcast safely, with apparently simple and concise code, avoiding the unsafe manual casts in the more complex and verbose code that you otherwise would have.


Non-virtual method ⇒ static binding

The following code is intentionally “incorrect”. It doesn’t declare the value method as virtual, and therefore produces an unintended “wrong” result, namely 0:

#include <iostream>
using namespace std;

class Expression
{
public:
    auto value() const
        -> double
    { return 0.0; }         // This should never be invoked, really.
};

class Number
    : public Expression
{
private:
    double  number_;

public:
    auto value() const
        -> double
    { return number_; }     // This is OK.

    Number( double const number )
        : Expression()
        , number_( number )
    {}
};

class Sum
    : public Expression
{
private:
    Expression const*   a_;
    Expression const*   b_;

public:
    auto value() const
        -> double
    { return a_->value() + b_->value(); }       // Uhm, bad! Very bad!

    Sum( Expression const* const a, Expression const* const b )
        : Expression()
        , a_( a )
        , b_( b )
    {}
};

auto main() -> int
{
    Number const    a( 3.14 );
    Number const    b( 2.72 );
    Number const    c( 1.0 );

    Sum const       sum_ab( &a, &b );
    Sum const       sum( &sum_ab, &c );

    cout << sum.value() << endl;
}

In the line commented as “bad” the Expression::value method is called, because the statically known type (the type known at compile time) is Expression, and the value method is not virtual.


Virtual method ⇒ dynamic binding.

Declaring value as virtual in the statically known type Expression ensures that the each call will check what actual type of object this is, and call the relevant implementation of value for that dynamic type:

#include <iostream>
using namespace std;

class Expression
{
public:
    virtual
    auto value() const -> double
        = 0;
};

class Number
    : public Expression
{
private:
    double  number_;

public:
    auto value() const -> double
        override
    { return number_; }

    Number( double const number )
        : Expression()
        , number_( number )
    {}
};

class Sum
    : public Expression
{
private:
    Expression const*   a_;
    Expression const*   b_;

public:
    auto value() const -> double
        override
    { return a_->value() + b_->value(); }    // Dynamic binding, OK!

    Sum( Expression const* const a, Expression const* const b )
        : Expression()
        , a_( a )
        , b_( b )
    {}
};

auto main() -> int
{
    Number const    a( 3.14 );
    Number const    b( 2.72 );
    Number const    c( 1.0 );

    Sum const       sum_ab( &a, &b );
    Sum const       sum( &sum_ab, &c );

    cout << sum.value() << endl;
}

Here the output is 6.86 as it should be, since the virtual method is called virtually. This is also called dynamic binding of the calls. A little check is performed, finding the actual dynamic type of object, and the relevant method implementation for that dynamic type, is called.

The relevant implementation is the one in the most specific (most derived) class.

Note that method implementations in derived classes here are not marked virtual, but are instead marked override. They could be marked virtual but they’re automatically virtual. The override keyword ensures that if there is not such a virtual method in some base class, then you’ll get an error (which is desirable).


The ugliness of doing this without virtual methods

Without virtual one would have to implement some Do It Yourself version of the dynamic binding. It’s this that generally involves unsafe manual downcasting, complexity and verbosity.

For the case of a single function, as here, it suffices to store a function pointer in the object and call via that function pointer, but even so it involves some unsafe downcasts, complexity and verbosity, to wit:

#include <iostream>
using namespace std;

class Expression
{
protected:
    typedef auto Value_func( Expression const* ) -> double;

    Value_func* value_func_;

public:
    auto value() const
        -> double
    { return value_func_( this ); }

    Expression(): value_func_( nullptr ) {}     // Like a pure virtual.
};

class Number
    : public Expression
{
private:
    double  number_;

    static
    auto specific_value_func( Expression const* expr )
        -> double
    { return static_cast<Number const*>( expr )->number_; }

public:
    Number( double const number )
        : Expression()
        , number_( number )
    { value_func_ = &Number::specific_value_func; }
};

class Sum
    : public Expression
{
private:
    Expression const*   a_;
    Expression const*   b_;

    static
    auto specific_value_func( Expression const* expr )
        -> double
    {
        auto const p_self  = static_cast<Sum const*>( expr );
        return p_self->a_->value() + p_self->b_->value();
    }

public:
    Sum( Expression const* const a, Expression const* const b )
        : Expression()
        , a_( a )
        , b_( b )
    { value_func_ = &Sum::specific_value_func; }
};


auto main() -> int
{
    Number const    a( 3.14 );
    Number const    b( 2.72 );
    Number const    c( 1.0 );

    Sum const       sum_ab( &a, &b );
    Sum const       sum( &sum_ab, &c );

    cout << sum.value() << endl;
}

One positive way of looking at this is, if you encounter unsafe downcasting, complexity and verbosity as above, then often a virtual method or methods can really help.

answered 4 years ago Ajay GU #9

Need for Virtual Function explained [Easy to understand]

#include<iostream>

using namespace std;

class A{
public: 
        void show(){
        cout << " Hello from Class A";
    }
};

class B :public A{
public:
     void show(){
        cout << " Hello from Class B";
    }
};


int main(){

    A *a1 = new B; // Create a base class pointer and assign address of derived object.
    a1->show();

}

Output will be:

Hello from Class A.

But with virtual function:

#include<iostream>

using namespace std;

class A{
public:
    virtual void show(){
        cout << " Hello from Class A";
    }
};

class B :public A{
public:
    virtual void show(){
        cout << " Hello from Class B";
    }
};


int main(){

    A *a1 = new B;
    a1->show();

}

Output will be:

Hello from Class B.

Hence with virtual function you can achieve runtime polymorphism.

answered 4 years ago Duke #10

About efficiency, the virtual functions are slightly less efficient as the early-binding functions.

"This virtual call mechanism can be made almost as efficient as the "normal function call" mechanism (within 25%). Its space overhead is one pointer in each object of a class with virtual functions plus one vtbl for each such class" [A tour of C++ by Bjarne Stroustrup]

answered 4 years ago user2074102 #11

Virtual methods are used in interface design. For example in Windows there is an interface called IUnknown like below:

interface IUnknown {
  virtual HRESULT QueryInterface (REFIID riid, void **ppvObject) = 0;
  virtual ULONG   AddRef () = 0;
  virtual ULONG   Release () = 0;
};

These methods are left to the interface user to implement. They are essential for the creation and destruction of certain objects that must inherit IUnknown. In this case the run-time is aware of the three methods and expects them to be implemented when it calls them. So in a sense they act as a contract between the object itself and whatever uses that object.

answered 3 years ago rvkreddy #12

The keyword virtual tells the compiler it should not perform early binding. Instead, it should automatically install all the mechanisms necessary to perform late binding. To accomplish this, the typical compiler1 creates a single table (called the VTABLE) for each class that contains virtual functions.The compiler places the addresses of the virtual functions for that particular class in the VTABLE. In each class with virtual functions,it secretly places a pointer, called the vpointer (abbreviated as VPTR), which points to the VTABLE for that object. When you make a virtual function call through a base-class pointer the compiler quietly inserts code to fetch the VPTR and look up the function address in the VTABLE, thus calling the correct function and causing late binding to take place.

More details in this link http://cplusplusinterviews.blogspot.sg/2015/04/virtual-mechanism.html

answered 3 years ago Ziezi #13

Why do we need Virtual Methods in C++?

Quick Answer:

  1. It provides us with one of the needed "ingredients"1 for object oriented programming.

In Bjarne Stroustrup C++ Programming: Principles and Practice, (14.3):

The virtual function provides the ability to define a function in a base class and have a function of the same name and type in a derived class called when a user calls the base class function. That is often called run-time polymorphism, dynamic dispatch, or run-time dispatch because the function called is determined at run time based on the type of the object used.

  1. It is the fastest more efficient implementation if you need a virtual function call 2.

To handle a virtual call, one needs one or more pieces of data related to the derived object 3. The way that is usually done is to add the address of table of functions. This table is usually referred to as virtual table or virtual function table and its address is often called the virtual pointer. Each virtual function gets a slot in the virtual table. Depending of the caller's object (derived) type, the virtual function, in its turn, invokes the respective override.


1.The use of inheritance, run-time polymorphism, and encapsulation is the most common definition of object-oriented programming.

2. You can't code functionality to be any faster or to use less memory using other language features to select among alternatives at run time. Bjarne Stroustrup C++ Programming: Principles and Practice.(14.3.1).

3. Something to tell which function is really invoked when we call the base class containing the virtual function.

answered 2 years ago user6359267 #14

Virtual Functions are used to support Runtime Polymorphism.

That is, virtual keyword tells the compiler not to make the decision (of function binding) at compile time, rather postpone it for runtime".

  • You can make a function virtual by preceding the keyword virtual in its base class declaration. For example,

     class Base
     {
        virtual void func();
     }
    
  • When a Base Class has a virtual member function, any class that inherits from the Base Class can redefine the function with exactly the same prototype i.e. only functionality can be redefined, not the interface of the function.

     class Derive : public Base
     {
        void func();
     }
    
  • A Base class pointer can be used to point to Base class object as well as a Derived class object.

  • When the virtual function is called by using a Base class pointer, the compiler decides at run-time which version of the function - i.e. the Base class version or the overridden Derived class version - is to be called. This is called Runtime Polymorphism.

answered 2 years ago Aryaman Gupta #15

I would like to add another use of Virtual function though it uses the same concept as above stated answers but I guess its worth mentioning.

VIRTUAL DESTRUCTOR

Consider this program below, without declaring Base class destructor as virtual; memory for Cat may not be cleaned up.

class Animal {
    public:
    ~Animal() {
        cout << "Deleting an Animal" << endl;
    }
};
class Cat:public Animal {
    public:
    ~Cat() {
        cout << "Deleting an Animal name Cat" << endl;
    }
};

int main() {
    Animal *a = new Cat();
    delete a;
    return 0;
}

Output:

Deleting an Animal
class Animal {
    public:
    virtual ~Animal() {
        cout << "Deleting an Animal" << endl;
    }
};
class Cat:public Animal {
    public:
    ~Cat(){
        cout << "Deleting an Animal name Cat" << endl;
    }
};

int main() {
    Animal *a = new Cat();
    delete a;
    return 0;
}

Output:

Deleting an Animal name Cat
Deleting an Animal

answered 2 years ago rashedcs #16

We need virtual methods for supporting "Run time Polymorphism". When you refer to a derived class object using a pointer or a reference to the base class, you can call a virtual function for that object and execute the derived class's version of the function.

answered 1 year ago javaProgrammer #17

The virtual keyword forces the compiler to pick the method implementation defined in the object's class rather than in the pointer's class.

Shape *shape = new Triangle(); 
cout << shape->getName();

In the above example, Shape::getName will be called by default, unless the getName() is defined as virtual in the Base class Shape. This forces the compiler to look for the getName() implementation in the Triangle class rather than in the Shape class.

The virtual table is the mechanism in which the compiler keeps track of the various virtual-method implementations of the subclasses. This is also called dynamic dispatch, and there is some overhead associated with it.

Finally, why is virtual even needed in C++, why not make it the default behavior like in Java?

  1. C++ is based on the principles of "Zero Overhead" and "Pay for what you use". So it doesn't try to perform dynamic dispatch for you, unless you need it.
  2. To provide more control to the interface. By making a function non-virtual, the interface/abstract class can control the behavior in all its implementations.

answered 1 year ago akshaypmurgod #18

Why do we need virtual functions?

Virtual functions avoid unnecessary typecasting problem, and some of us can debate that why do we need virtual functions when we can use derived class pointer to call the function specific in derived class!the answer is - it nullifies the whole idea of inheritance in large system development, where having single pointer base class object is much desired.

Let's compare below two simple programs to understand the importance of virtual functions:

Program without virtual functions:

#include <iostream>
using namespace std;

class father
{
    public: void get_age() {cout << "Fathers age is 50 years" << endl;}
};

class son: public father
{
    public : void get_age() { cout << "son`s age is 26 years" << endl;}
};

int main(){
    father *p_father = new father;
    son *p_son = new son;

    p_father->get_age();
    p_father = p_son;
    p_father->get_age();
    p_son->get_age();
    return 0;
}

OUTPUT:

Fathers age is 50 years
Fathers age is 50 years
son`s age is 26 years

Program with virtual function:

#include <iostream>
using namespace std;

class father
{
    public:
        virtual void get_age() {cout << "Fathers age is 50 years" << endl;}
};

class son: public father
{
    public : void get_age() { cout << "son`s age is 26 years" << endl;}
};

int main(){
    father *p_father = new father;
    son *p_son = new son;

    p_father->get_age();
    p_father = p_son;
    p_father->get_age();
    p_son->get_age();
    return 0;
}

OUTPUT:

Fathers age is 50 years
son`s age is 26 years
son`s age is 26 years

By closely analyzing both the outputs one can understand the importance of virtual functions.

answered 1 year ago M-J #19

I've my answer in form of a conversation to be a better read:


Why do we need virtual functions?

Because of Polymorphism.

What is Polymorphism?

The fact that a base pointer can also point to derived type objects.

How does this definition of Polymorphism lead into the need for virtual functions?

Well, through early binding.

What is early binding?

Early binding(compile-time binding) in C++ means that a function call is fixed before the program is executed.

So...?

So if you use a base type as the parameter of a function, the compiler will only recognize the base interface, and if you call that function with any arguments from derived classes, it gets sliced off, which is not what you want to happen.

If it's not what we want to happen, why is this allowed?

Because we need Polymorphism!

What's the benefit of Polymorphism then?

You can use a base type pointer as the parameter of a single function, and then in the run-time of your program, you can access each of the derived type interfaces(e.g. their member functions) without any issues, using dereferencing of that single base pointer.

I still don't know what virtual functions are good for...! And this was my first question!

well, this is because you asked your question too soon!

Why do we need virtual functions?

Assume that you called a function with a base pointer, which had the address of an object from one of its derived classes. As we've talked about it above, in the run-time, this pointer gets dereferenced, so far so good, however, we expect a method(== a member function) "from our derived class" to be executed! However, a same method(one that has a same header) is already defined in the base class, so why should your program bother to choose the other method? In other words I mean, how can you tell this scenario off from what we used to see normally happen before?

The brief answer is "a Virtual member function in base", and a little longer answer is that, "at this step, if the program sees a virtual function in the base class, it knows(realizes) that you're trying to use polymorphism" and so goes to derived classes(using v-table, a form of late binding) to find that another method with the same header, but with -expectedly- a different implementation.

Why a different implementation?

You knuckle-head! Go read a good book!

OK, wait wait wait, why would one bother to use base pointers, when he/she could simply use derived type pointers? You be the judge, is all this headache worth it? Look at these two snippets:

//1:

Parent* p1 = &boy;
p1 -> task();
Parent* p2 = &girl;
p2 -> task();

//2:

Boy* p1 = &boy;
p1 -> task();
Girl* p2 = &girl;
p2 -> task();

OK, although I think that 1 is still better than 2, you could write 1 like this either:

//1:

Parent* p1 = &boy;
p1 -> task();
p1 = &girl;
p1 -> task();

and moreover, you should be aware that this is yet just a contrived use of all the things I've explained to you so far. Instead of this, assume for example a situation in which you had a function in your program that used the methods from each of the derived classes respectively(getMonthBenefit()):

double totalMonthBenefit = 0;    
std::vector<CentralShop*> mainShop = { &shop1, &shop2, &shop3, &shop4, &shop5, &shop6};
for(CentralShop* x : mainShop){
     totalMonthBenefit += x -> getMonthBenefit();
}

Now, try to re-write this, without any headaches!

double totalMonthBenefit=0;
Shop1* branch1 = &shop1;
Shop2* branch2 = &shop2;
Shop3* branch3 = &shop3;
Shop4* branch4 = &shop4;
Shop5* branch5 = &shop5;
Shop6* branch6 = &shop6;
totalMonthBenefit += branch1 -> getMonthBenefit();
totalMonthBenefit += branch2 -> getMonthBenefit();
totalMonthBenefit += branch3 -> getMonthBenefit();
totalMonthBenefit += branch4 -> getMonthBenefit();
totalMonthBenefit += branch5 -> getMonthBenefit();
totalMonthBenefit += branch6 -> getMonthBenefit();

And actually, this might be yet a contrived example either!

answered 8 months ago user3371350 #20

Here is complete example that illustrates why virtual method is used.

#include <iostream>

using namespace std;

class Basic
{
    public:
    virtual void Test1()
    {
        cout << "Test1 from Basic." << endl;
    }
    virtual ~Basic(){};
};
class VariantA : public Basic
{
    public:
    void Test1()
    {
        cout << "Test1 from VariantA." << endl;
    }
};
class VariantB : public Basic
{
    public:
    void Test1()
    {
        cout << "Test1 from VariantB." << endl;
    }
};

int main()
{
    Basic *object;
    VariantA *vobjectA = new VariantA();
    VariantB *vobjectB = new VariantB();

    object=(Basic *) vobjectA;
    object->Test1();

    object=(Basic *) vobjectB;
    object->Test1();

    delete vobjectA;
    delete vobjectB;
    return 0;
}

answered 2 months ago edwinc #21

i think you are referring to the fact once a method is declared virtual you don't need to use the 'virtual' keyword in overrides.

class Base { virtual void foo(); };

class Derived : Base 
{ 
  void foo(); // this is overriding Base::foo
};

If you don't use 'virtual' in Base's foo declaration then Derived's foo would just be shadowing it.

answered 2 weeks ago Leon Chang #22

Here is a merged version of the C++ code for the first two answers.

#include        <iostream>
#include        <string>

using   namespace       std;

class   Animal
{
        public:
#ifdef  VIRTUAL
                virtual string  says()  {       return  "??";   }
#else
                string  says()  {       return  "??";   }
#endif
};

class   Dog:    public Animal
{
        public:
                string  says()  {       return  "woof"; }
};

string  func(Animal *a)
{
        return  a->says();
}

int     main()
{
        Animal  *a = new Animal();
        Dog     *d = new Dog();
        Animal  *ad = d;

        cout << "Animal a says\t\t" << a->says() << endl;
        cout << "Dog d says\t\t" << d->says() << endl;
        cout << "Animal dog ad says\t" << ad->says() << endl;

        cout << "func(a) :\t\t" <<      func(a) <<      endl;
        cout << "func(d) :\t\t" <<      func(d) <<      endl;
        cout << "func(ad):\t\t" <<      func(ad)<<      endl;
}

Two different results are:

Without #define virtual, it binds at compile time. Animal *ad and func(Animal *) all point to the Animal's says() method.

$ g++ virtual.cpp -o virtual
$ ./virtual 
Animal a says       ??
Dog d says      woof
Animal dog ad says  ??
func(a) :       ??
func(d) :       ??
func(ad):       ??

With #define virtual, it binds at run time. Dog *d, Animal *ad and func(Animal *) point/refer to the Dog's says() method as Dog is their object type. Unless [Dog's says() "woof"] method is not defined, it will be the one searched first in the class tree, i.e. derived classes may override methods of their base classes [Animal's says()].

$ g++ virtual.cpp -D VIRTUAL -o virtual
$ ./virtual 
Animal a says       ??
Dog d says      woof
Animal dog ad says  woof
func(a) :       ??
func(d) :       woof
func(ad):       woof

It is interesting to note that all class attributes (data and methods) in Python are effectively virtual. Since all objects are dynamically created at runtime, there is no type declaration or a need for keyword virtual. Below is Python's version of code:

class   Animal:
        def     says(self):
                return  "??"

class   Dog(Animal):
        def     says(self):
                return  "woof"

def     func(a):
        return  a.says()

if      __name__ == "__main__":

        a = Animal()
        d = Dog()
        ad = d  #       dynamic typing by assignment

        print("Animal a says\t\t{}".format(a.says()))
        print("Dog d says\t\t{}".format(d.says()))
        print("Animal dog ad says\t{}".format(ad.says()))

        print("func(a) :\t\t{}".format(func(a)))
        print("func(d) :\t\t{}".format(func(d)))
        print("func(ad):\t\t{}".format(func(ad)))

The output is:

Animal a says       ??
Dog d says      woof
Animal dog ad says  woof
func(a) :       ??
func(d) :       woof
func(ad):       woof

which is identical to C++'s virtual define. Note that d and ad are two different pointer variables referring/pointing to the same Dog instance. The expression (ad is d) returns True and their values are the same <main.Dog object at 0xb79f72cc>.

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