C++ static virtual members?

Question

No, there's no way to do it, since what would happen when you called Object::GetTypeInformation()? It can't know which derived class version to call since there's no object associated with it.

You'll have to make it a non-static virtual function to work properly; if you also want to be able to call a specific derived class's version non-virtually without an object instance, you'll have to provide a second redunduant static non-virtual version as well.

Many say it is not possible, I would go one step further and say it is not meaningfull.

A static member is something that does not relate to any instance, only to the class.

A virtual member is something that does not relate directly to any class, only to an instance.

So a static virtual member would be something that does not relate to any instance or any class.

I ran into this problem the other day: I had some classes full of static methods but I wanted to use inheritance and virtual methods and reduce code repetition. My solution was:

Instead of using static methods, use a singleton with virtual methods.

In other words, each class should contain a static method that you call to get a pointer to a single, shared instance of the class. You can make the true constructors private or protected so that outside code can't misuse it by creating additional instances.

In practice, using a singleton is a lot like using static methods except that you can take advantage of inheritance and virtual methods.

While Alsk has already given a pretty detailed answer, I'd like to add an alternative, since I think his enhanced implementation is overcomplicated.

We start with an abstract base class, that provides the interface for all the object types:

class Object
{
public:
    virtual char* GetClassName() = 0;
};

Now we need an actual implementation. But to avoid having to write both the static and the virtual methods, we will have our actual object classes inherit the virtual methods. This does obviously only work, if the base class knows how to access the static member function. So we need to use a template and pass the actual objects class name to it:

template<class ObjectType>
class ObjectImpl : public Object
{
public:
    virtual char* GetClassName()
    {
        return ObjectType::GetClassNameStatic();
    }
};

Finally we need to implement our real object(s). Here we only need to implement the static member function, the virtual member functions will be inherited from the ObjectImpl template class, instantiated with the name of the derived class, so it will access it's static members.

class MyObject : public ObjectImpl<MyObject>
{
public:
    static char* GetClassNameStatic()
    {
        return "MyObject";
    }
};

class YourObject : public ObjectImpl<YourObject>
{
public:
    static char* GetClassNameStatic()
    {
        return "YourObject";
    }
};

Let's add some code to test:

char* GetObjectClassName(Object* object)
{
    return object->GetClassName();
}

int main()
{
    MyObject myObject;
    YourObject yourObject;

    printf("%s\n", MyObject::GetClassNameStatic());
    printf("%s\n", myObject.GetClassName());
    printf("%s\n", GetObjectClassName(&myObject));
    printf("%s\n", YourObject::GetClassNameStatic());
    printf("%s\n", yourObject.GetClassName());
    printf("%s\n", GetObjectClassName(&yourObject));

    return 0;
}

Addendum (Jan 12th 2019):

Instead of using the GetClassNameStatic() function, you can also define the the class name as a static member, even "inline", which IIRC works since C++11 (don't get scared by all the modifiers :)):

class MyObject : public ObjectImpl<MyObject>
{
public:
    // Access this from the template class as `ObjectType::s_ClassName` 
    static inline const char* const s_ClassName = "MyObject";

    // ...
};

It is possible!

But what exactly is possible, let's narrow down. People often want some kind of "static virtual function" because of duplication of code needed for being able to call the same function through static call "SomeDerivedClass::myfunction()" and polymorphic call "base_class_pointer->myfunction()". "Legal" method for allowing such functionality is duplication of function definitions:

class Object
{
public:
    static string getTypeInformationStatic() { return "base class";}
    virtual string getTypeInformation() { return getTypeInformationStatic(); }
}; 
class Foo: public Object
{
public:
    static string getTypeInformationStatic() { return "derived class";}
    virtual string getTypeInformation() { return getTypeInformationStatic(); }
};

What if base class has a great number of static functions and derived class has to override every of them and one forgot to provide a duplicating definition for virtual function. Right, we'll get some strange error during runtime which is hard to track down. Cause duplication of code is a bad thing. The following tries to resolve this problem (and I want to tell beforehand that it is completely type-safe and doesn't contain any black magic like typeid's or dynamic_cast's :)

So, we want to provide only one definition of getTypeInformation() per derived class and it is obvious that it has to be a definition of static function because it is not possible to call "SomeDerivedClass::getTypeInformation()" if getTypeInformation() is virtual. How can we call static function of derived class through pointer to base class? It is not possible with vtable because vtable stores pointers only to virtual functions and since we decided not to use virtual functions, we cannot modify vtable for our benefit. Then, to be able to access static function for derived class through pointer to base class we have to store somehow the type of an object within its base class. One approach is to make base class templatized using "curiously recurring template pattern" but it is not appropriate here and we'll use a technique called "type erasure":

class TypeKeeper
{
public:
    virtual string getTypeInformation() = 0;
};
template<class T>
class TypeKeeperImpl: public TypeKeeper
{
public:
    virtual string getTypeInformation() { return T::getTypeInformationStatic(); }
};

Now we can store the type of an object within base class "Object" with a variable "keeper":

class Object
{
public:
    Object(){}
    boost::scoped_ptr<TypeKeeper> keeper;

    //not virtual
    string getTypeInformation() const 
    { return keeper? keeper->getTypeInformation(): string("base class"); }

};

In a derived class keeper must be initialized during construction:

class Foo: public Object
{
public:
    Foo() { keeper.reset(new TypeKeeperImpl<Foo>()); }
    //note the name of the function
    static string getTypeInformationStatic() 
    { return "class for proving static virtual functions concept"; }
};

Let's add syntactic sugar:

template<class T>
void override_static_functions(T* t)
{ t->keeper.reset(new TypeKeeperImpl<T>()); }
#define OVERRIDE_STATIC_FUNCTIONS override_static_functions(this)

Now declarations of descendants look like:

class Foo: public Object
{
public:
    Foo() { OVERRIDE_STATIC_FUNCTIONS; }
    static string getTypeInformationStatic() 
    { return "class for proving static virtual functions concept"; }
};

class Bar: public Foo
{
public:
    Bar() { OVERRIDE_STATIC_FUNCTIONS; }
    static string getTypeInformationStatic() 
    { return "another class for the same reason"; }
};

usage:

Object* obj = new Foo();
cout << obj->getTypeInformation() << endl;  //calls Foo::getTypeInformationStatic()
obj = new Bar();
cout << obj->getTypeInformation() << endl;  //calls Bar::getTypeInformationStatic()
Foo* foo = new Bar();
cout << foo->getTypeInformation() << endl; //calls Bar::getTypeInformationStatic()
Foo::getTypeInformation(); //compile-time error
Foo::getTypeInformationStatic(); //calls Foo::getTypeInformationStatic()
Bar::getTypeInformationStatic(); //calls Bar::getTypeInformationStatic()

Advantages:

1. less duplication of code (but we have to call OVERRIDE_STATIC_FUNCTIONS in every constructor)

Disadvantages:

1. OVERRIDE_STATIC_FUNCTIONS in every constructor
2. memory and performance overhead
3. increased complexity

Open issues:

1) there are different names for static and virtual functions how to solve ambiguity here?

class Foo
{
public:
    static void f(bool f=true) { cout << "static";}
    virtual void f() { cout << "virtual";}
};
//somewhere
Foo::f(); //calls static f(), no ambiguity
ptr_to_foo->f(); //ambiguity

2) how to implicitly call OVERRIDE_STATIC_FUNCTIONS inside every constructor?

It is possible. Make two functions: static and virtual

struct Object{     
  struct TypeInformation;
  static  const TypeInformation &GetTypeInformationStatic() const 
  { 
      return GetTypeInformationMain1();
  }
  virtual const TypeInformation &GetTypeInformation() const
  { 
      return GetTypeInformationMain1();
  }
protected:
  static const TypeInformation &GetTypeInformationMain1(); // Main function
};

struct SomeObject : public Object {     
  static  const TypeInformation &GetTypeInformationStatic() const 
  { 
      return GetTypeInformationMain2();
  }
  virtual const TypeInformation &GetTypeInformation() const
  { 
      return GetTypeInformationMain2();
  }
protected:
  static const TypeInformation &GetTypeInformationMain2(); // Main function
};