i have a factory class that i've used for quite some time, which i have tried to do a little template magic to, and it simply will not work. I don't think you'll need to see that class, as it's quite large, but if you really think it's related then i will edit it in. Anyway, i have a setup like this which i've done forever.
class Base :
public Factory<Base>
{
};
class Derived :
public Base
{
static Factory<Base>::DerivedRegister<Derived> m_Reg;
};
Factory<Base>::DerivedRegister<Derived> Derived::m_Reg;
the factory gives it the static method createInstance(const std::string& name) which will return an instance of the registered object. The DerivedRegister object derives from the factory object and when declared statically, must be constructed during static initialization. This means i have access to the factories static internals within it's constructor, and so i utilize it's constructor to have all registered object available to the factory during static initialization. This works and has for awhile, but then i tried something like this so that derived classes could avoid needing to explicitly make that static derivedregister object:
class Base :
public Factory<Base>
{
};
template<class TDerived>
class BaseReg :
public Base
{
static Factory<Base>::DerivedRegister<TDerived> m_Reg;
};
template<class TDerived>
Factory<Base>::DerivedRegister<TDerived> BaseReg<TDerived>::m_Reg;
class Derived :
public BaseReg<Derived>
{
};
When i run the program and try to create an instance of the Derived class, it doesn't exist in the factory, and after a little doodling, i find that the static DerivedRegister is never being constructed in derived, as in the constructor is never called. I find this odd. How could it not be forced to statically initialize if the Derived class isn't a template? When i make a dummy static object as seen below, and use a method from m_Reg in it's static construction, suddenly the DerivedRegister constructs correctly and the factory can instantiate it, no sweat.
class Base :
public Factory<Base>
{
};
template<class TDerived>
class BaseReg :
public Base
{
protected:
static Factory<Base>::DerivedRegister<TDerived> m_Reg;
};
template<class TDerived>
Factory<Base>::DerivedRegister<TDerived> BaseReg<TDerived>::m_Reg;
class Derived :
public BaseReg<Derived>
{
class RandomClass
{
public:
RandomClass(std::string meh) {}
};
private:
static RandomClass m_Obj;
};
Derived::RandomClass Derived::m_Obj(m_Reg.func());
So what am i not getting about this template class's static member initialization here? doesn't it have to statically initialize the object like any other non-template object?
EDIT: mkay, to give a little insight, i'll post the Factory object. Just beware the WALLS of text. ignore the extra includes, and the superfluous GetNames, and func functions, they are just voodoo code.
#ifndef FACTORY_H
#define FACTORY_H
// library tools
#include <map>
#include <string>
#include <typeinfo>
#include <cstdlib>
#include <vector>
#include <iostream>
#include <cxxabi.h>
const std::string demangle(const char* name);
template<class base>
class Factory
{
protected:
template<class T>
static base * createT() { return new T;}
typedef std::map<std::string, base*(*)()> map_type;
static map_type& GetMap()
{
static map_type map;
return map;
}
public:
virtual ~Factory(){}
static base * createInstance(const std::string & s)
{
if(!GetMap().count(s))
return nullptr;
typename map_type::iterator it = GetMap().find(s);
return it->second();
}
template <class TDerived>
struct DerivedRegister :
public Factory<base>
{
DerivedRegister()
{
std::string name = demangle(typeid(TDerived).name());
GetMap().insert(std::pair<std::string, base*(*)()>(name, &createT<TDerived>));
}
DerivedRegister(const std::string& name)
{
GetMap().insert(std::pair<std::string, base*(*)()>(name, &createT<TDerived>));
}
std::string func() {return "meh";}
};
static void GetNames(std::vector<std::string>& names)
{
names.clear();
for(auto it = GetMap().begin(); it != GetMap().end(); ++it)
names.push_back(it->first);
}
};
#endif
What you might be running up against here is the woes of unpredictable static initialisation order, assuming that you are constructing the factory itself statically, or that it is constructed as part of a chain of other static constructions.
You can get around this by wrapping and using the static in its own function:
template<class TDerived>
class BaseReg :
public Base
{
protected:
static Factory<Base>::DerivedRegister<TDerived> & Reg()
{
static Factory<Base>::DerivedRegister<TDerived> m_Reg;
return m_Reg;
}
};
Now, the only way to access m_Reg is to call Reg(), after which time you can be certain that m_Reg is constructed. That is, it's constructed at whatever time it is first used.
I have successfully fixed some pretty hairy problems with the above approach, and now I use it as a matter of course, to avoid having to debug strange crashes. I almost never have a need for a static member variable anymore.
I need several C++ classes to have a static method "register", however the implementation of register varies between those classes.
It should be static because my idea is to "register" all those classes with Lua (only once of course).
Obviously I can't declare an interface with a static pure virtual function. What do you guys suggest me to do ? Simplicity is welcome, but I think some kind of template could work.
Example of what I would like to achieve
class registerInterface
{
public:
static virtual void register() = 0; //obviously illegal
};
class someClass: public registerInterface
{
static virtual void register()
{
//I register myself with Lua
}
}
class someOtherClass: public registerInterface
{
static virtual void register()
{
//I register myself with Lua in a different way
}
}
int main()
{
someClass::register();
someOtherClass::register();
return 0;
}
Based on how you've described the problem, it's unclear to me why you even need the 'virtual static method' on the classes. This should be perfectly legal.
class SomeClass {
static void register(void) {
...
}
}
class SomeOtherClass {
static void register(void) {
...
}
}
int main(int argc, char* argv[]) {
SomeClass::register();
SomeOtherClass::register();
return 0;
}
Drop the RegisterInterface, I don't think you need it.
If it helps, you could take Hitesh's answer, and add:
struct luaRegisterManager {
template <typename T>
void registrate() {
T::registrate();
// do something else to record the fact that we've registered -
// perhaps "registrate" should be returning some object to help with that
}
};
Then:
int main() {
luaRegisterManager lrm;
lrm.registrate<someClass>();
lrm.registrate<someOtherClass>();
}
More generally, if you want to introduce any dynamic polymorphism in C++, then you need an object, not just a class. So again, perhaps the various register functions should be returning objects, with some common interface base class registeredClass, or classRegistrationInfo, or something along those lines.
Could provide an example of what you feel it is that you need dynamic polymorphism for? Hitesh's code precisely matches your one example, as far as I can see, so that example must not cover all of your anticipated use cases. If you write the code that would be using it, perhaps it will become clear to you how to implement it, or perhaps someone can advise.
Something else that might help:
#include <iostream>
#include <string>
#include <vector>
struct Registered {
virtual std::string name() = 0;
virtual ~Registered() {}
Registered() {
all.push_back(this);
}
static std::vector<Registered*> all;
};
std::vector<Registered*> Registered::all;
typedef std::vector<Registered*>::iterator Iter;
template <typename T>
struct RegisteredT : Registered {
std::string n;
RegisteredT(const std::string &name) : n(name) { T::registrate(); }
std::string name() { return n; }
// other functions here could be implemented in terms of calls to static
// functions of T.
};
struct someClass {
static Registered *r;
static void registrate() { std::cout << "registering someClass\n"; }
};
Registered *someClass::r = new RegisteredT<someClass>("someClass");
struct someOtherClass {
static Registered *r;
static void registrate() { std::cout << "registering someOtherClass\n"; }
};
Registered *someOtherClass::r = new RegisteredT<someOtherClass>("someOtherClass");
int main() {
for (Iter it = Registered::all.begin(); it < Registered::all.end(); ++it) {
std::cout << (*it)->name() << "\n";
}
}
There are all sorts of problems with this code if you try to split it across multiple compilation units. Furthermore, this kind of thing leads to spurious reports from memory leak detectors unless you also write some code to tear everything down at the end, or use a vector of shared_ptr, Boost pointer vector, etc. But you see the general idea that a class can "register itself", and that you need an object to make virtual calls.
In C++ you usually try to avoid static initialisation, though, in favour of some sort of setup / dependency injection at the start of your program. So normally you would just list all the classes you care about (calling a function on each one) rather than try to do this automatically.
Your intentions are noble, but your solution is inkling towards "overengineering" (unless I am missing an obvious solution).
Here is one possibility: You can use the Virtual Friend function idiom For example,
class RegisterInterface{
friend void register(RegisterInterface* x){x->do_real_register();}
protected:
virtual void do_real_register();
}
class Foo : public RegisterInterface{
protected:
virtual void do_real_register(){}
};
class Bar : public RegisterInterface{
protected:
virtual void do_real_register(){}
};
int main(int argc, char* argv[]) {
BOOST_FOREACH(RegisterInterface* ri, registered_interfaces)
{
register(ri);
}
return 0;
}
I know you've already accepted an answer, but I figured I would write this up anyway. You can have self-registering classes if you use some static initialization and the CRTP:
#include <vector>
#include <iostream>
using namespace std;
class RegisterableRoot // Holds the list of functions to call, doesn't actually need
// need to be a class, could just be a collection of globals
{
public:
typedef void (*registration_func)();
protected:
static std::vector<registration_func> s_registery;
public:
static void do_registration()
{
for(int i = 0; i < s_registery.size(); ++i)
s_registery[i]();
}
static bool add_func(registration_func func) // returns something so we can use it in
// in an initializer
{
s_registery.push_back(func);
return true;
}
};
template<typename RegisterableType> // Doesn't really need to inherit from
class Registerable : public RegisterableRoot // RegisterableRoot
{
protected:
static const bool s_effect;
};
class A : public Registerable<A> // Honestly, neither does A need to inherit from
// Registerable<T>
{
public:
static void Register()
{
cout << "A" << endl;
}
};
class B : public Registerable<B>
{
public:
static void Register()
{
cout << "B" << endl;
}
};
int main()
{
RegisterableRoot::do_registration();
return 0;
}
std::vector<RegisterableRoot::registration_func> RegisterableRoot::s_registery;
template <typename RegisterableType> // This is the "cute" part, we initialize the
// static s_effect so we build the list "magically"
const bool Registerable<RegisterableType>::s_effect = add_func(&RegisterableType::Register);
template class Registerable<A>; // Explicitly instantiate the template
// causes the equivalent of
// s_registery.push_back(&A::Register) to
// be executed
template class Registerable<B>;
This outputs
A
B
although I wouldn't rely on this order if I were you. Note that the template class Registerable<X> need not be in the same translation unit as the call to do_registration, you can put it with the rest of your definition of Foo. If you inherit from Registerable<> and you don't write a static void Register() function for your class you'll get a (admittedly probably cryptic) compiler error much like you might expect if there really was such a thing as "static virtuals". The "magic" merely adds the class specific function to the list to be called, this avoids several of the pitfalls of doing the actual registration in a static initializer. You still have to call do_registration for anything to happen.
How about this way? Define an interface class:
// IFoobar.h
class IFoobar{
public:
virtual void Register(void) = 0;
}
Then define the class that handles the register..
// RegisterFoobar.h
class RegisterFoobar{
public:
// Constructors etc...
IFoobar* fooBar;
static void RegisterFoobar(IFoobar& fubar){
foobar = &fubar;
}
private:
void Raise(void){ foobar->Register(); }
}
Now, then define another class like this
// MyFuBar.h
class MyFuBar : IFoobar{
public:
// Constructors etc...
void Register(void);
private:
RegisterFoobar* _regFoobar;
}
Call the code like this:
//MyFuBar.cpp
MyFuBar::MyFuBar(){
_regFoobar = new Foobar();
_regFoobar->RegisterFoobar(this);
}
void MyFuBar::Register(void){
// Raised here...
}
Maybe I have misunderstood your requirements...
I have an abstract base class like so:
class AbstractBaseClass
{};
a templated concrete class that derives from it:
template<class T>
class ConcreteClass : public AbstractBaseClass
{
public:
ConcreteClass(T input) : data(input) {}
private:
T data;
};
AndI have a factory class that creates AbstractBaseClasses
class MyFactory
{
public:
boost::shared_ptr<AbstractBaseClass> CreateBlah();
boost::shared_ptr<AbstractBaseClass> CreateFoo();
template<class T>
boost::shared_ptr<AbstractBaseClass> Create(T input)
{
return boost::shared_ptr<AbstractBaseClass>(new ConcreteClass<T>(input));
}
};
The problem with this, is that now EVERYTHING that uses MyFactory has to include the entire implementation to ConcreteClass. Ideally, I want nothing but MyFactory to know about ConcreteClass.
Is there any way to architect this to achieve this goal? (Besides manually making a new Create function in MyFactory for every type I want instead of templating it).
you'll need to put the factory implementation into the implementation file (which you mentioned you'd like to avoid, but it is the lesser evil unless the interfaces are small, and/or your projects are small).
of course, there are a few other ways you could approach this, such as putting the implementation into base classes, and making derived bases factories, or using some other really weird template syntax to reduce instantiations in dependent translations. this really comes down to convenience and scale for your project. if you are working on one or more large projects, then full abstraction wrt instantiation will serve your needs best in the long run (assuming you need dynamic polymorphism and memory).
you may also try other approaches (such as overloading) to reduce errors by using type-safety.
the short answer is that you'll really need to abstract the interfaces/instantiation into one or multiple implementation files to remove header dependencies - very common idiom, and many ways to tackle it. you can over course further divide and use polymorphism for your factories as well.
you may also use template forward declarations to minimize the sets to the compilation unit. provided:
/** in MyIntermediateFactory.hpp */
class MyIntermediateFactory {
public:
static template<class T> boost::shared_ptr<T> Create(float);
};
/** in Concrete.hpp */
template<Concrete>
boost::shared_ptr<T> MyIntermediateFactory::Create<Concrete>(float arg) {
/* … */
}
using this you can select portions of programs/interfaces which you need in the library, then wrap it all up in a real Factory (for the build at hand). the linker/instantiation should fail along the way if you actually attempt to request a creation which is not visible.
there a lot of options, really - you need to figure out how big your scale is in order to determine what to abstract (or not). instantiation requires interface, to remove header dependencies, you'll have to abstract the instantiation someplace.
My approach to the same problem in the past was the creation of a set of concrete factories (one per type) that get registered in the global factory (for illustration purposes, indexing by object name):
class AbstractBaseClass;
class ConcreteFactory
{
public:
AbstractBaseClass * create();
};
class AbstractFactory
{
public:
void registerFactory( std::string const & name, std::shared_ptr<ConcreteFactory> const & f )
{
factory[ name ] = f; // check for collisions, complain if so ...
}
AbstractBaseClass * create( std::string const & name )
{
return factory[name]->create(); // check for existence before dereferencing...
}
private:
std::map<std::string, std::shared_ptr<ConcreteFactory> > factory;
};
I used this in a piece of code that was heavily templated to reduce compilation time. Each concrete factory and the class that it creates need only be in a single translation unit that registers the concrete factory. The rest of the code only need to use the common interface to AbstractBaseClass.
I realize I am answering this five years later. Maybe the language has grown a tad since then. I'd like to offer something that seems right, if I understand the question properly, if for no other point than to help others who might find this question and wonder what they could do.
factory.hpp
#include "base.hpp"
namespace tvr
{
namespace test
{
class factory
{
public:
typedef base::ptr Ptr;
enum eSpecial
{
eDerived
};
template<typename Type>
Ptr create()
{
Ptr result;
result.reset(new Type());
return result;
}
template<typename Type, typename DataType>
Ptr create(const DataType& data)
{
Ptr result;
result.reset(new Type(data));
return result;
}
template<typename Type, typename DataType>
Ptr create(const DataType& data, eSpecial tag)
{
Ptr result;
result.reset(new Type());
static_cast<Type*>(result.get())->set_item(data);
return result;
}
};
}
}
base.hpp
#include <memory>
namespace tvr
{
namespace test
{
class base
{
public:
typedef std::shared_ptr<base> ptr;
public:
base() {}
virtual ~base() {}
virtual void do_something() = 0;
};
}
}
some_class.hpp
#include <ostream>
namespace tvr
{
namespace test
{
struct some_class
{
};
}
}
std::ostream& operator<<(std::ostream& out, const tvr::test::some_class& item)
{
out << "This is just some class.";
return out;
}
template_derived.hpp
#include <iostream>
#include "base.hpp"
namespace tvr
{
namespace test
{
template<typename Type>
class template_derived : public base
{
public:
template_derived(){}
virtual ~template_derived(){}
virtual void do_something()
{
std::cout << "Doing something, like printing _item as \"" << _item << "\"." << std::endl;
}
void set_item(const Type data)
{
_item = data;
}
private:
Type _item;
};
}
}
and, finally, main.cpp
#include <vector>
#include "base.hpp"
#include "factory.hpp"
namespace tvr
{
namespace test
{
typedef std::vector<tvr::test::base::ptr> ptr_collection;
struct iterate_collection
{
void operator()(const ptr_collection& col)
{
for (ptr_collection::const_iterator iter = col.begin();
iter != col.end();
++iter)
{
iter->get()->do_something();
}
}
};
}
}
#include "template_derived.hpp"
#include "some_class.hpp"
namespace tvr
{
namespace test
{
inline int test()
{
ptr_collection items;
tvr::test::factory Factory;
typedef template_derived<unsigned int> UIntConcrete;
typedef template_derived<double> DoubleConcrete;
typedef template_derived<std::string> StringConcrete;
typedef template_derived<some_class> SomeClassConcrete;
items.push_back(Factory.create<SomeClassConcrete>(some_class(), tvr::test::factory::eDerived));
for (unsigned int i = 5; i < 7; ++i)
{
items.push_back(Factory.create<UIntConcrete>(i, tvr::test::factory::eDerived));
}
items.push_back(Factory.create<DoubleConcrete>(4.5, tvr::test::factory::eDerived));
items.push_back(Factory.create<StringConcrete>(std::string("Hi there!"), tvr::test::factory::eDerived));
iterate_collection DoThem;
DoThem(items);
return 0;
}
}
}
int main(int argc, const char* argv[])
{
tvr::test::test();
}
output
Doing something, like printing _item as "This is just some class.".
Doing something, like printing _item as "5".
Doing something, like printing _item as "6".
Doing something, like printing _item as "4.5".
Doing something, like printing _item as "Hi there!".
This uses a combination of templates, function overloading, and tagging through enums to help create a flexible factory class that doesn't require knowing much about the individual classes it instantiates, to include templated concrete classes as the OP asked about.
The 'eDerived' tag (in the form of an enum) tells the compiler to use the version of the factory's create function that takes a class like the template_derived class, which has a function that allows me to assign data to one of its members. As you can tell from the way I ordered the headers in main.cpp, the factory doesn't know anything about template_derived. Neither does the function calling the base class's virtual function (do_something). I think this is what the OP wanted, but without having to add a various create functions within every class that this factory might generate.
I also showed how one doesn't have to explicitly create functions for each class the factory should create. The factory's overloaded create functions can create anything derived from the base class that matches the appropriate signature.
I didn't do an extensive performance analysis on this code, but I did enough to see that the majority of the work happens in the streaming operator. This compiles in about 1 second on my 3.30Ghz quad core machine. You might need to experiment with more robust code to see how badly it might bog down the compiler, if much at all.
I've tested this code in VC++ 2015, although it probably works in other compilers pretty easily. If you want to copy this, you'll need to add your own guard headers. In any event, I hope this is useful.
You could use explicit template instanciation. Trying to call the factory method with a template parameter not explicit instanciated will give you a linker error. Note the explicit template instanciation in MyFactory.cpp
template AbstractBaseClass* MyFactory::Create(int input);
All put together looks like this (I removed shared_ptr for the sake of simplicity):
Main.cpp:
#include "AbstractBaseClass.h"
#include "MyFactory.h"
//we do not need to know nothing about concreteclass (neither MyFactory.h includes it)
int main()
{
MyFactory f;
AbstractBaseClass* ab = f.Create(10);
ab = f.Create(10.0f);
return 0;
}
MyFactory.h:
#include "AbstractBaseClass.h"
class MyFactory
{
public:
template<class T>
AbstractBaseClass* Create(T input);
};
MyFactory.cpp:
#include "MyFactory.h"
#include "ConcreteClass.h"
template<class T>
AbstractBaseClass* MyFactory::Create(T input) {
return new ConcreteClass<T>(input);
}
//explicit template instanciation
template AbstractBaseClass* MyFactory::Create(int input);
//you could use as well specialisation for certain types
template<>
AbstractBaseClass* MyFactory::Create(float input) {
return new ConcreteClass<float>(input);
}
AbstractBaseClass.h:
class AbstractBaseClass{};
ConcreteClass.h:
#include "AbstractBaseClass.h"
template<class T>
class ConcreteClass : public AbstractBaseClass
{
public:
ConcreteClass(T input) : data(input) {}
private:
T data;
};
You're looking for the "PIMPL" idiom. There's a good explanation at Herb Sutter's GOTW site
It can't be done because ConcreteClass is a template, that means you need the complete implementation available at compile time. Same reason why you can't precompile templates and have to write them all in header files instead.
Is it possible in C++ to have a member function that is both static and virtual? Apparently, there isn't a straightforward way to do it (static virtual member(); is a compile error), but is there at least a way to achieve the same effect?
I.E:
struct Object
{
struct TypeInformation;
static virtual const TypeInformation &GetTypeInformation() const;
};
struct SomeObject : public Object
{
static virtual const TypeInformation &GetTypeInformation() const;
};
It makes sense to use GetTypeInformation() both on an instance (object->GetTypeInformation()) and on a class (SomeObject::GetTypeInformation()), which can be useful for comparisons and vital for templates.
The only ways I can think of involves writing two functions / a function and a constant, per class, or use macros.
Any other solutions?
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:
less duplication of code (but we
have to call
OVERRIDE_STATIC_FUNCTIONS in every
constructor)
Disadvantages:
OVERRIDE_STATIC_FUNCTIONS in every
constructor
memory and performance
overhead
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\n";
}
};
struct Derived : public Base {
static void sayMyName() override {
cout << "Derived\n";
}
};
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\n";
}
virtual void foo() {
}
int somedata;
};
struct Derived : public Base {
void sayMyName() override {
cout << "Derived\n";
}
};
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\n";
}
virtual void foo() {
}
int somedata;
};
struct Derived : public Base {
VTable* vtable;
void sayMyName() override {
cout << "Derived\n";
}
};
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;
}
Like others have said, there are 2 important pieces of information:
there is no this pointer when making a static function call and
the this pointer points to the structure where the virtual table, or thunk, are used to look up which runtime method to call.
A static function is determined at compile time.
I showed this code example in C++ static members in class; it shows that you can call a static method given a null pointer:
struct Foo
{
static int boo() { return 2; }
};
int _tmain(int argc, _TCHAR* argv[])
{
Foo* pFoo = NULL;
int b = pFoo->boo(); // b will now have the value 2
return 0;
}