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.

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

void A::f() {
    cout<<"Test"<<endl;
}

Is code above OK?

What's the purpose to make it a pure virtual function with an implementation?

Answer 1

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 2

Yes this is correct. In your example, classes that derive from A inherit both the interface f() and a default implementation. But you force derived classes to implement the method f() (even if it is only to call the default implementation provided by A).

Scott Meyers discusses this in Effective C++ (2nd Edition) Item #36 Differentiate between inheritance of interface and inheritance of implementation. The item number may have changed in the latest edition.

Answer 3

Pure virtual functions with or without a body simply mean that the derived types must provide their own implementation.

Pure virtual function bodies in the base class are useful if your derived classes wants to call your base class implementation.

Answer 4

The advantage of it is that it forces derived types to still override the method but also provides a default or additive implementation.

Answer 5

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 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)