Pure virtual function with implementation

Question

My basic understanding is that there is no implementation for a pure virtual function, however, I was told there might be implementation for pure virtual function.

Answer 1

One important use-case of having a pure virtual method with an implementation body, is when you want to have an abstract class, but you do not have any proper methods in the class to make it pure virtual. In this case, you can make the destructor of the class pure virtual and put your desired implementation (even an empty body) for that. As an example:

class Foo
{
   virtual ~Foo() = 0;
   void bar1() {}
   void bar2(int x) {}
   // other methods
};

Foo::~Foo()
{
}

This technique, makes the Foo class abstract and as a result impossible to instantiate the class directly. At the same time you have not added an additional pure virtual method to make the Foo class abstract.

Answer 2

The 'virtual void foo() =0;' syntax does not mean you can't implement foo() in current class, you can. It also does not mean you must implement it in derived classes. Before you slap me, let's observe the Diamond Problem: (Implicit code, mind you).

class A
{
public: 
    virtual void foo()=0;
    virtual void bar();
}

class B : public virtual A
{
public:
    void foo() { bar(); }
}

class C : public virtual A
{
public:
    void bar();
}

class D : public B, public C
{}

int main(int argc, const char* argv[])
{
    A* obj = new D();
    **obj->foo();**
    return 0;
}

Now, the obj->foo() invocation will result in B::foo() and then C::bar().

You see... pure virtual methods do not have to be implemented in derived classes (foo() has no implementation in class C - compiler will compile) In C++ there are a lot of loopholes.

Hope I could help :-)

Answer 3

If you have code that should be executed by the deriving class, but you don't want it to be executed directly -- and you want to force it to be overriden.

Your code is correct, although all in all this isn't an often used feature, and usually only seen when trying to define a pure virtual destructor -- in that case you must provide an implementation. The funny thing is that once you derive from that class you don't need to override the destructor.

Hence the one sensible usage of pure virtual functions is specifying a pure virtual destructor as a "non-final" keyword.

The following code is surprisingly correct:

class Base {
public:
  virtual ~Base() = 0;
};

Base::~Base() {}

class Derived : public Base {};

int main() { 
  // Base b; -- compile error
  Derived d; 
}

Answer 4

A pure virtual function must be implemented in a derived type that will be directly instantiated, however the base type can still define an implementation. A derived class can explicitly call the base class implementation (if access permissions allow it) by using a fully-scoped name (by calling A::f() in your example - if A::f() were public or protected). Something like:

class B : public A {

    virtual void f() {
        // class B doesn't have anything special to do for f()
        //  so we'll call A's

        // note that A's declaration of f() would have to be public 
        //  or protected to avoid a compile time problem

        A::f();
    }

};

The use case I can think of off the top of my head is when there's a more-or-less reasonable default behavior, but the class designer wants that sort-of-default behavior be invoked only explicitly. It can also be the case what you want derived classes to always perform their own work but also be able to call a common set of functionality.

Note that even though it's permitted by the language, it's not something that I see commonly used (and the fact that it can be done seems to surprise most C++ programmers, even experienced ones).

Answer 5

To be clear, you are misunderstanding what = 0; after a virtual function means.

= 0 means derived classes must provide an implementation, not that the base class can not provide an implementation.

In practice, when you mark a virtual function as pure (=0), there is very little point in providing a definition, because it will never be called unless someone explicitly does so via Base::Function(...) or if the Base class constructor calls the virtual function in question.

Answer 6

You'd have to give a body to a pure virtual destructor, for example :)

Read: http://cplusplus.co.il/2009/08/22/pure-virtual-destructor/

(Link broken, use archive)