i have a class which has a template by other purposes:
template<class t>
class MyClass {
public: //of course public...
t foo;
std::string text;
}
and i have another class which method get all kind of these class through the arguments, and want to store the pointer in an array. The class dont want to access the specific (tempalted) parts of the classes only the common attributes/methods.
class Container {
public: //of course public...
MyClass* array; //this is allocated with some magic.
void bar(MyClass& m) {
and want to store the class in a MyClass* array.
}
}
here is the error that argument list for template missing
how can i solve this?
The simplest method would be to make that function a template as well:
template <class t>
void bar(MyClass<t>& m) {
// ...
}
Note that that should probably be const MyClass<t>&, because you don't need to modify it.
Your new code is meaningless. There is no such that as an object of type MyClass, because MyClass is a template. If you want to operate on these classes irrespective of their template argument, then you need to factor out the non-template portions as a base class:
class MyClassBase
{
public:
// polymorphic base classes should always have virtual destructors
~MyClassBase() {}
virtual void some_function() = 0;
};
template <typename T>
class MyClass : public MyClassBase
{
public:
// implement abstract functions
void some_function()
{
// template argument T is available here
}
};
Then you can refer to that base, and when you call a virtual function it will dynamically dispatch:
class Container
{
public:
// no magic: use a std::vector for dynamic arrays
std::vector<MyClassBase*> array; // not by value! avoid slicing
void bar(MyClassBase& m)
{
array.push_back(&m);
}
void baz()
{
array[0]->some_function(); // for example
}
};
How about putting a common base class.
class MyClassCommon {
protected:
~MyClassCommon() { }
public:
std::string text;
};
template<class t>
class MyClass : public MyClassCommon {
public: // of course public...
t foo;
};
class Container {
public: // of course public...
MyClassCommon* array; // this is allocated with some magic.
void bar(MyClassCommon& m) {
/* ... */
}
};
If you want to create a "multi-template" array, you'd better use a non-template class as a base class of a template class. Or you can make a template array and store any objects in it.
the text variable in your class is private so unless you bar function is a method of the class you can't legally use it like that
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;
}
I have code like this:
class RetInterface {...}
class Ret1: public RetInterface {...}
class AInterface
{
public:
virtual boost::shared_ptr<RetInterface> get_r() const = 0;
...
};
class A1: public AInterface
{
public:
boost::shared_ptr<Ret1> get_r() const {...}
...
};
This code does not compile.
In visual studio it raises
C2555: overriding virtual function return type differs and is not
covariant
If I do not use boost::shared_ptr but return raw pointers, the code compiles (I understand this is due to covariant return types in C++). I can see the problem is because boost::shared_ptr of Ret1 is not derived from boost::shared_ptr of RetInterface. But I want to return boost::shared_ptr of Ret1 for use in other classes, else I must cast the returned value after the return.
Am I doing something wrong?
If not, why is the language like this - it should be extensible to handle conversion between smart pointers in this scenario? Is there a desirable workaround?
Firstly, this is indeed how it works in C++: the return type of a virtual function in a derived class must be the same as in the base class. There is the special exception that a function that returns a reference/pointer to some class X can be overridden by a function that returns a reference/pointer to a class that derives from X, but as you note this doesn't allow for smart pointers (such as shared_ptr), just for plain pointers.
If your interface RetInterface is sufficiently comprehensive, then you won't need to know the actual returned type in the calling code. In general it doesn't make sense anyway: the reason get_r is a virtual function in the first place is because you will be calling it through a pointer or reference to the base class AInterface, in which case you can't know what type the derived class would return. If you are calling this with an actual A1 reference, you can just create a separate get_r1 function in A1 that does what you need.
class A1: public AInterface
{
public:
boost::shared_ptr<RetInterface> get_r() const
{
return get_r1();
}
boost::shared_ptr<Ret1> get_r1() const {...}
...
};
Alternatively, you can use the visitor pattern or something like my Dynamic Double Dispatch technique to pass a callback in to the returned object which can then invoke the callback with the correct type.
There is a neat solution posted in this blog post (from Raoul Borges)
An excerpt of the bit prior to adding support for mulitple inheritance and abstract methods is:
template <typename Derived, typename Base>
class clone_inherit<Derived, Base> : public Base
{
public:
std::unique_ptr<Derived> clone() const
{
return std::unique_ptr<Derived>(static_cast<Derived *>(this->clone_impl()));
}
private:
virtual clone_inherit * clone_impl() const override
{
return new Derived(*this);
}
};
class concrete: public clone_inherit<concrete, cloneable>
{
};
int main()
{
std::unique_ptr<concrete> c = std::make_unique<concrete>();
std::unique_ptr<concrete> cc = c->clone();
cloneable * p = c.get();
std::unique_ptr<clonable> pp = p->clone();
}
I would encourage reading the full article. Its simply written and well explained.
You can't change return types (for non-pointer, non-reference return types) when overloading methods in C++. A1::get_r must return a boost::shared_ptr<RetInterface>.
Anthony Williams has a nice comprehensive answer.
What about this solution:
template<typename Derived, typename Base>
class SharedCovariant : public shared_ptr<Base>
{
public:
typedef Base BaseOf;
SharedCovariant(shared_ptr<Base> & container) :
shared_ptr<Base>(container)
{
}
shared_ptr<Derived> operator ->()
{
return boost::dynamic_pointer_cast<Derived>(*this);
}
};
e.g:
struct A {};
struct B : A {};
struct Test
{
shared_ptr<A> get() {return a_; }
shared_ptr<A> a_;
};
typedef SharedCovariant<B,A> SharedBFromA;
struct TestDerived : Test
{
SharedBFromA get() { return a_; }
};
Here is my attempt :
template<class T>
class Child : public T
{
public:
typedef T Parent;
};
template<typename _T>
class has_parent
{
private:
typedef char One;
typedef struct { char array[2]; } Two;
template<typename _C>
static One test(typename _C::Parent *);
template<typename _C>
static Two test(...);
public:
enum { value = (sizeof(test<_T>(nullptr)) == sizeof(One)) };
};
class A
{
public :
virtual void print() = 0;
};
class B : public Child<A>
{
public:
void print() override
{
printf("toto \n");
}
};
template<class T, bool hasParent = has_parent<T>::value>
class ICovariantSharedPtr;
template<class T>
class ICovariantSharedPtr<T, true> : public ICovariantSharedPtr<typename T::Parent>
{
public:
T * get() override = 0;
};
template<class T>
class ICovariantSharedPtr<T, false>
{
public:
virtual T * get() = 0;
};
template<class T>
class CovariantSharedPtr : public ICovariantSharedPtr<T>
{
public:
CovariantSharedPtr(){}
CovariantSharedPtr(std::shared_ptr<T> a_ptr) : m_ptr(std::move(a_ptr)){}
T * get() final
{
return m_ptr.get();
}
private:
std::shared_ptr<T> m_ptr;
};
And a little example :
class UseA
{
public:
virtual ICovariantSharedPtr<A> & GetPtr() = 0;
};
class UseB : public UseA
{
public:
CovariantSharedPtr<B> & GetPtr() final
{
return m_ptrB;
}
private:
CovariantSharedPtr<B> m_ptrB = std::make_shared<B>();
};
int _tmain(int argc, _TCHAR* argv[])
{
UseB b;
UseA & a = b;
a.GetPtr().get()->print();
}
Explanations :
This solution implies meta-progamming and to modify the classes used in covariant smart pointers.
The simple template struct Child is here to bind the type Parent and inheritance. Any class inheriting from Child<T> will inherit from T and define T as Parent. The classes used in covariant smart pointers needs this type to be defined.
The class has_parent is used to detect at compile time if a class defines the type Parent or not. This part is not mine, I used the same code as to detect if a method exists (see here)
As we want covariance with smart pointers, we want our smart pointers to mimic the existing class architecture. It's easier to explain how it works in the example.
When a CovariantSharedPtr<B> is defined, it inherits from ICovariantSharedPtr<B>, which is interpreted as ICovariantSharedPtr<B, has_parent<B>::value>. As B inherits from Child<A>, has_parent<B>::value is true, so ICovariantSharedPtr<B> is ICovariantSharedPtr<B, true> and inherits from ICovariantSharedPtr<B::Parent> which is ICovariantSharedPtr<A>. As A has no Parent defined, has_parent<A>::value is false, ICovariantSharedPtr<A> is ICovariantSharedPtr<A, false> and inherits from nothing.
The main point is as Binherits from A, we have ICovariantSharedPtr<B>inheriting from ICovariantSharedPtr<A>. So any method returning a pointer or a reference on ICovariantSharedPtr<A> can be overloaded by a method returning the same on ICovariantSharedPtr<B>.
Mr Fooz answered part 1 of your question. Part 2, it works this way because the compiler doesn't know if it will be calling AInterface::get_r or A1::get_r at compile time - it needs to know what return value it's going to get, so it insists on both methods returning the same type. This is part of the C++ specification.
For the workaround, if A1::get_r returns a pointer to RetInterface, the virtual methods in RetInterface will still work as expected, and the proper object will be deleted when the pointer is destroyed. There's no need for different return types.
maybe you could use an out parameter to get around "covariance with returned boost shared_ptrs.
void get_r_to(boost::shared_ptr<RetInterface>& ) ...
since I suspect a caller can drop in a more refined shared_ptr type as argument.