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

No, this is not possible, because static member functions lack a this pointer. And static members (both functions and variables) are not really class members per-se. They just happen to be invoked by ClassName::member, and adhere to the class access specifiers. Their storage is defined somewhere outside the class; storage is not created each time you instantiated an object of the class. Pointers to class members are special in semantics and syntax. A pointer to a static member is a normal pointer in all regards.

virtual functions in a class needs the this pointer, and is very coupled to the class, hence they can t be static.

It s not possible, but that s just because an omission. It isn t something that "doesn t make sense" as a lot of people seem to claim. To be clear, I m talking about something like this:

struct Base {
  static virtual void sayMyName() {
    cout << "Base
";
  }
};

struct Derived : public Base {
  static void sayMyName() override {
    cout << "Derived
";
  }
};

void foo(Base *b) {
  b->sayMyName();
  Derived::sayMyName(); // Also would work.
}

This is 100% something that could be implemented (it just hasn t), and I d argue something that is useful.

Consider how normal virtual functions work. Remove the statics and add in some other stuff and we have:

struct Base {
  virtual void sayMyName() {
    cout << "Base
";
  }
  virtual void foo() {
  }
  int somedata;
};

struct Derived : public Base {
  void sayMyName() override {
    cout << "Derived
";
  }
};

void foo(Base *b) {
  b->sayMyName();
}

This works fine and basically what happens is the compiler makes two tables, called VTables, and assigns indices to the virtual functions like this

enum Base_Virtual_Functions {
  sayMyName = 0;
  foo = 1;
};

using VTable = void*[];

const VTable Base_VTable = {
  &Base::sayMyName,
  &Base::foo
};

const VTable Derived_VTable = {
  &Derived::sayMyName,
  &Base::foo
};

Next each class with virtual functions is augmented with another field that points to its VTable, so the compiler basically changes them to be like this:

struct Base {
  VTable* vtable;
  virtual void sayMyName() {
    cout << "Base
";
  }
  virtual void foo() {
  }
  int somedata;
};

struct Derived : public Base {
  VTable* vtable;
  void sayMyName() override {
    cout << "Derived
";
  }
};

Then what actually happens when you call b->sayMyName()? Basically this:

b->vtable[Base_Virtual_Functions::sayMyName](b);

(The first parameter becomes this.)

Ok fine, so how would it work with static virtual functions? Well what s the difference between static and non-static member functions? The only difference is that the latter get a this pointer.

We can do exactly the same with static virtual functions - just remove the this pointer.

b->vtable[Base_Virtual_Functions::sayMyName]();

This could then support both syntaxes:

b->sayMyName(); // Prints "Base" or "Derived"...
Base::sayMyName(); // Always prints "Base".

So ignore all the naysayers. It does make sense. Why isn t it supported then? I think it s because it has very little benefit and could even be a little confusing.

The only technical advantage over a normal virtual function is that you don t need to pass this to the function but I don t think that would make any measurable difference to performance.

It does mean you don t have a separate static and non-static function for cases when you have an instance, and when you don t have an instance, but also it might be confusing that it s only really "virtual" when you use the instance call.

Well , quite a late answer but it is possible using the curiously recurring template pattern. This wikipedia article has the info you need and also the example under static polymorphism is what you are asked for.

This question is over a decade old, but it looks like it gets a good amount of traffic, so I wanted to post an alternative using modern C++ features that I haven t seen anywhere else.

This solution uses CRTP and SFINAE to perform static dispatching. That, in itself, is nothing new, but all such implementations I ve found lack strict signature checking for "overrides." This implementation requires that the "overriding" method signature exactly matches that of the "overridden" method. This behavior more closely resembles that of virtual functions, while also allowing us to effectively overload and "override" a static method.

Note that I put override in quotes because, strictly speaking, we re not technically overriding anything. Instead, we re calling a dispatch method X with signature Y that forwards all of its arguments to T::X, where T is to the first type among a list of types such that T::X exists with signature Y. This list of types considered for dispatching can be anything, but generally would include a default implementation class and the derived class.

Implementation

#include <experimental/type_traits>

template <template <class...> class Op, class... Types>
struct dispatcher;

template <template <class...> class Op, class T>
struct dispatcher<Op, T> : std::experimental::detected_t<Op, T> {};

template <template <class...> class Op, class T, class... Types>
struct dispatcher<Op, T, Types...>
  : std::experimental::detected_or_t<
    typename dispatcher<Op, Types...>::type, Op, T> {};


// Helper to convert a signature to a function pointer
template <class Signature> struct function_ptr;

template <class R, class... Args> struct function_ptr<R(Args...)> {
    using type = R (*)(Args...);
};


// Macro to simplify creation of the dispatcher
// NOTE: This macro isn t smart enough to handle creating an overloaded
//       dispatcher because both dispatchers will try to use the same
//       integral_constant type alias name. If you want to overload, do it
//       manually or make a smarter macro that can somehow put the signature in
//       the integral_constant type alias name.
#define virtual_static_method(name, signature, ...)                            
    template <class VSM_T>                                                     
    using vsm_##name##_type = std::integral_constant<                          
        function_ptr<signature>::type, &VSM_T::name>;                          
                                                                               
    template <class... VSM_Args>                                               
    static auto name(VSM_Args&&... args)                                       
    {                                                                          
        return dispatcher<vsm_##name##_type, __VA_ARGS__>::value(              
            std::forward<VSM_Args>(args)...);                                  
    }

Example Usage

#include <iostream>

template <class T>
struct Base {
    // Define the default implementations
    struct defaults {
        static std::string alpha() { return "Base::alpha"; };
        static std::string bravo(int) { return "Base::bravo"; }
    };

    // Create the dispatchers
    virtual_static_method(alpha, std::string(void), T, defaults);
    virtual_static_method(bravo, std::string(int), T, defaults);
    
    static void where_are_the_turtles() {
        std::cout << alpha() << std::endl;  // Derived::alpha
        std::cout << bravo(1) << std::endl; // Base::bravo
    }
};

struct Derived : Base<Derived> {
    // Overrides Base::alpha
    static std::string alpha(){ return "Derived::alpha"; }

    // Does not override Base::bravo because signatures differ (even though
    // int is implicitly convertible to bool)
    static std::string bravo(bool){ return "Derived::bravo"; }
};

int main() {
    Derived::where_are_the_turtles();
}

I think what you re trying to do can be done through templates. I m trying to read between the lines here. What you re trying to do is to call a method from some code, where it calls a derived version but the caller doesn t specify which class. Example:

class Foo {
public:
    void M() {...}
};

class Bar : public Foo {
public:
    void M() {...}
};

void Try()
{
    xxx::M();
}

int main()
{
    Try();
}

You want Try() to call the Bar version of M without specifying Bar. The way you do that for statics is to use a template. So change it like so:

class Foo {
public:
    void M() {...}
};

class Bar : public Foo {
public:
    void M() {...}
};

template <class T>
void Try()
{
    T::M();
}

int main()
{
    Try<Bar>();
}

No, Static member function can t be virtual .since virtual concept is resolved at run time with the help of vptr, and vptr is non static member of a class.due to that static member function can t acess vptr so static member can t be virtual.

No, its not possible, since static members are bound at compile time, while virtual members are bound at runtime.

If your desired use for a virtual static is to be able to define an interface over the static section of a class then there is a solution to your problem using C++20 concept s.

class ExBase { //object properties
  public: virtual int do(int) = 0;
};

template <typename T> //type properties
concept ExReq = std::derived_from<T, ExBase> && requires(int i) { //~constexpr bool
  {
    T::do_static(i) //checks that this compiles
  } -> std::same_as<int> //checks the expression type is int
};

class ExImpl : virtual public ExBase { //satisfies ExReq
  public: int do(int i) override {return i;} //overrides do in ExBase
  public: static int do_static(int i) {return i;} //satisfies ExReq
};

//...

void some_func(ExReq auto o) {o.do(0); decltype(o)::do_static(0);}

(this works the same way on members aswell!)

For more on how concepts work: https://en.cppreference.com/w/cpp/language/constraints

For the standard concepts added in C++20: https://en.cppreference.com/w/cpp/concepts

First, the replies are correct that what the OP is requesting is a contradiction in terms: virtual methods depend on the run-time type of an instance; static functions specifically don t depend on an instance -- just on a type. That said, it makes sense to have static functions return something specific to a type. For example, I had a family of MouseTool classes for the State pattern and I started having each one have a static function returning the keyboard modifier that went with it; I used those static functions in the factory function that made the correct MouseTool instance. That function checked the mouse state against MouseToolA::keyboardModifier(), MouseToolB::keyboardModifier(), etc. and then instantiated the appropriate one. Of course later I wanted to check if the state was right so I wanted write something like "if (keyboardModifier == dynamic_type(*state)::keyboardModifier())" (not real C++ syntax), which is what this question is asking.

So, if you find yourself wanting this, you may want to rething your solution. Still, I understand the desire to have static methods and then call them dynamically based on the dynamic type of an instance. I think the Visitor Pattern can give you what you want. It gives you what you want. It s a bit of extra code, but it could be useful for other visitors.

See: http://en.wikipedia.org/wiki/Visitor_pattern for background.

struct ObjectVisitor;

struct Object
{
     struct TypeInformation;

     static TypeInformation GetTypeInformation();
     virtual void accept(ObjectVisitor& v);
};

struct SomeObject : public Object
{
     static TypeInformation GetTypeInformation();
     virtual void accept(ObjectVisitor& v) const;
};

struct AnotherObject : public Object
{
     static TypeInformation GetTypeInformation();
     virtual void accept(ObjectVisitor& v) const;
};

Then for each concrete Object:

void SomeObject::accept(ObjectVisitor& v) const {
    v.visit(*this); // The compiler statically picks the visit method based on *this being a const SomeObject&.
}
void AnotherObject::accept(ObjectVisitor& v) const {
    v.visit(*this); // Here *this is a const AnotherObject& at compile time.
}

and then define the base visitor:

struct ObjectVisitor {
    virtual ~ObjectVisitor() {}
    virtual void visit(const SomeObject& o) {} // Or = 0, depending what you feel like.
    virtual void visit(const AnotherObject& o) {} // Or = 0, depending what you feel like.
    // More virtual void visit() methods for each Object class.
};

Then the concrete visitor that selects the appropriate static function:

struct ObjectVisitorGetTypeInfo {
    Object::TypeInformation result;
    virtual void visit(const SomeObject& o) {
        result = SomeObject::GetTypeInformation();
    }
    virtual void visit(const AnotherObject& o) {
        result = AnotherObject::GetTypeInformation();
    }
    // Again, an implementation for each concrete Object.
};

finally, use it:

void printInfo(Object& o) {
    ObjectVisitorGetTypeInfo getTypeInfo;
    Object::TypeInformation info = o.accept(getTypeInfo).result;
    std::cout << info << std::endl;
}

Notes:

• Constness left as an exercise.
• You returned a reference from a static. Unless you have a singleton, that s questionable.

If you want to avoid copy-paste errors where one of your visit methods calls the wrong static function, you could use a templated helper function (which can t itself be virtual) t your visitor with a template like this:

struct ObjectVisitorGetTypeInfo {
    Object::TypeInformation result;
    virtual void visit(const SomeObject& o) { doVisit(o); }
    virtual void visit(const AnotherObject& o) { doVisit(o); }
    // Again, an implementation for each concrete Object.

  private:
    template <typename T>
    void doVisit(const T& o) {
        result = T::GetTypeInformation();
    }
};

With c++ you can use static inheritance with the crt method. For the example, it is used widely on window template atl & wtl.

See https://en.wikipedia.org/wiki/Curiously_recurring_template_pattern

To be simple, you have a class that is templated from itself like class myclass : public myancestor. From this point the myancestor class can now call your static T::YourImpl function.

I had a browse through the other answers and none of them seem to mention virtual function tables (vtable), which explains why this is not possible.

A static function inside a C++ class compiles to something which is effectively the same as any other function in a regular namespace.

In other words, when you declare a function static you are using the class name as a namespace rather than an object (which has an instance, with some associated data).

Let s quickly look at this...

// This example is the same as the example below
class ExampleClass
{
    static void exampleFunction();

    int someData;
};

// This example is the same as the example above
namespace ExampleClass
{
    void exampleFunction();

    // Doesn t work quite the same. Each instance of a class
    // has independent data. Here the data is global.
    int someData;
}

With that out of the way, and an understanding of what a static member function really is, we can now consider vtables.

If you declare any virtual function in a class, then the compiler creates a block of data which (usually) precedes other data members. This block of data contains runtime information which tells the program at runtime where in memory it needs to jump to in order to execute the correct (virtual) function for each instance of a class which might be created during runtime.

The important point here is "block of data". In order for that block of data to exist, it has to be stored as part of an instance of an object (class). If your function is static, then we already said it uses the name of the class as a namespace. There is no object associated with that function call.

To add slightly more detail: A static function does not have an implicit this pointer, which points to the memory where the object lives. Because it doesn t have that, you can t jump to a place in memory and find the vtable for that object. So you can t do virtual function dispatch.

I m not an expert in compiler engineering by any means, but understanding things at least to this level of detail is helpful, and (hopefully?) makes it easy to understand why (at least in C++) static virtual does not make sense, and cannot be translated into something sensible by the compiler.

Maybe you can try my solution below:

class Base {
public:
    Base(void);
    virtual ~Base(void);

public:
    virtual void MyVirtualFun(void) = 0;
    static void  MyStaticFun(void) { assert( mSelf != NULL); mSelf->MyVirtualFun(); }
private:
    static Base* mSelf;
};

Base::mSelf = NULL;

Base::Base(void) {
    mSelf = this;
}

Base::~Base(void) {
    // please never delete mSelf or reset the Value of mSelf in any deconstructors
}

class DerivedClass : public Base {
public:
    DerivedClass(void) : Base() {}
    ~DerivedClass(void){}

public:
    virtual void MyVirtualFun(void) { cout<<"Hello, it is DerivedClass!"<<endl; }
};

int main() {
    DerivedClass testCls;
    testCls.MyStaticFun(); //correct way to invoke this kind of static fun
    DerivedClass::MyStaticFun(); //wrong way
    return 0;
}