This is best described in pseudo-code:
class Thing {};
interface ThingGetter<T extends Thing> {
T getThing();
}
class Product extends Thing {};
class ProductGetter<Product> {
Product getThing() {
// Some product code
}
}
class SpecialProductGetter extends ProductGetter {
Product getThing() {
p = ProductGetter::getThing();
// Do some special stuff;
return p;
}
}
class NeatProductGetter extends ProductGetter {
Product getThing() {
p = ProductGetter::getThing();
// Do some neat stuff;
return p;
}
}
I will have other "things" with other getters for those as well.
I've tried codifying this in C++ but it doesn't like :
template <> class ThingGetter <Product> {
nor:
class NeatProductGetter : public ThingGetter <Product> {
Can you model this in C++ without making getThing return a Thing and then having to cast it like crazy?
If so, how? If not, what is the best way to do this?
Thanks!
Since the question comes from someone with a Java background, I will interpret it in Java terms. You want to define a generic interface that will return an object derived from T:
template<typename T>
struct getter
{
virtual T* get() = 0; // just for the signature, no implementation
};
Note the changes: the function is declared virtual so that it will behave polimorphically in derived objects. The return value is a pointer instead of an object. If you keep an object in your interface, the compiler will slice (cut the non-base part of the returned object) the return. With some metaprogramming magic (or resorting to boost) you can have the compiler test the constraint that T derives from a given type.
Now the question is why you would like to do so... If you define a non-generic interface (abstract class) that returns a Thing by pointer you can just get the same semantics more easily:
struct Thing {};
struct AnotherThing : public Thing {};
struct getter
{
virtual Thing* get() = 0;
};
struct AnotherGetter : public getter
{
virtual AnotherThing* get() { return 0; }
};
struct Error : public getter
{
int get() { return 0; } // error conflicting return type for get()
};
struct AnotherError : public getter
{
};
int main() {
AnotherError e; // error, AnotherError is abstract (get is still pure virtual at this level)
}
The compiler will require all instantiable classes derived from getter to implement a get() method that returns a Thing by pointer (or a covariant return type: pointer to a class derived from Thing). If the derived class tries to return another type the compiler will flag it as an error.
The problem, as you post it is that if you use the interface, then the objects can only be handled as pointers to the base Thing class. Now, whether that is a problem or not is another question... In general, if you have correctly designed your hierarchy, you should be able to use the returned objects polimorphically without having to resort to down casting.
Note that if you are using the different getters at the most derived level in the hierarchy, each get() method returns the most derived element from the Thing hierarchy, at that point you will not need to down cast at all. Only if you use the different getters through the base getter interface you will get the same return type for all (Thing*)
It is important to note that templates are resolved completely at compile time. That means that trying to use templates to solve the need (do you really need it?) to downcast will not help you if you use the different getters through the interface (reference/pointers to the basic getter).
Maybe posting a little more details on your specific domain, where and how you intend to use this code, can help in providing more useful responses.
In C++, this would look like this:
(I've changed "class" to "struct" to avoid a bunch of "public:" declarations. It's the same thing, though.)
class Thing { };
template <typename T>
struct ThingGetter {
T getThing();
};
struct Product : public Thing { };
struct ProductGetter : public ThingGetter<Product>{
Product getThing() { return Product(); }
};
struct SpecialProductGetter : public ProductGetter {
Product getThing() {
Product p = this->ProductGetter::getThing();
return p;
}
};
struct NeatProductGetter : public ProductGetter {
Product getThing() {
Product p = ProductGetter::getThing();
return p;
}
};
...
SpecialProductGetter spg;
spg.getThing();
NeatProductGetter npg;
npg.getThing();
Another solution is to use template specialization:
struct Thing { };
struct Widget { };
template <typename T>
struct Getter {
T getIt() = 0;
};
template <>
struct Getter<Thing> {
Thing getIt() { return Thing(); }
};
template <>
struct Getter<Widget> {
Widget getIt() { return Widget(); }
};
typedef Getter<Widget> WidgetGetter;
struct SpecialWidgetGetter : public WidgetGetter {
Widget getIt() { return this->WidgetGetter::getIt(); }
};
struct FizzledWidgetGetter : public WidgetGetter {
Widget getFizzledWidget() { return getIt().fizzle(); }
};
...
Getter<Thing> tg;
tg.getIt();
WidgetGetter wg;
wg.getIt();
Related
I have the following hierarchy pattern in various places in the codebase:
enum DerivedType {
A, B, C };
class Base {
public:
static Base* Create(DerivedType t);
};
template <DerivedType T>
class Derived : public Base {
};
The Create method returns a new object of class Derived<A>, Derived<B>, or Derived<C>, depending on its argument:
Base* Base::Create(DerivedType t) {
switch (t) {
case A: return new Derived<A>;
case B: return new Derived<B>;
case C: return new Derived<C>;
default: return NULL;
}
}
The problem is that there are many such Base -> Derived hierarchies, with essentially the same implementation of Create() copy-pasted all over the place. Is there an elegant, yet easy-to-understand way to avoid duplication here?
We can abstract away the Factory details, and rely on the client to provide us the mapping of enum to class type:
template<typename ENUM, typename T>
struct Factory
{
typedef std::map<ENUM, T*(*)()> map_type;
static map_type factoryMapping_;
static T* Create(ENUM c)
{
return factoryMapping_[c]();
}
static void Init(map_type _mapping)
{
factoryMapping_ = _mapping;
}
};
template<typename ENUM, typename T>
typename Factory<ENUM, T>::map_type Factory<ENUM,T>::factoryMapping_;
Now it's the client's job to provide us methods for creating a Base* given some enum value.
If you're willing and able to abstract away the creation of a derived class with a template, then you can save a fair bit of typing.
What I mean is, let's create a templated function to create a derived class (without any real checking for correctness):
template<typename Base, typename Derived>
Base* CreateDerived()
{
return new Derived();
}
Now I can define an enum and associated class hierarchy:
enum ClassType {A, B};
struct Foo
{
virtual void PrintName() const
{
std::cout << "Foo\n";
}
};
typedef Factory<ClassType, Foo> FooFactory ;
struct DerivedOne : public Foo
{
virtual void PrintName() const
{
std::cout << "DerivedOne\n";
}
};
struct DerivedTwo : public Foo
{
virtual void PrintName() const
{
std::cout << "DerivedTwo\n";
}
};
And then use it like so:
// set up factory
std::map<ClassType, Foo*(*)()> mapping;
mapping[A] = &CreateDerived<Foo, DerivedOne>;
mapping[B] = &CreateDerived<Foo, DerivedTwo>;
FooFactory::Init(mapping);
// Use the factory
Foo* f = FooFactory::Create(A);
f->PrintName();
Live Demo
Of course this simplifies your problem a bit, namely moving the factory details out of the base class and ignoring for a minute that the children themselves are templated. Depending on how hard it is in your domain to create a good CreateDerived function for each type, you may not end up saving a ton of typing.
EDIT: We can modify our Create function to return NULL by taking advantage of std::map::find. I omitted it for brevity. If you concerned about performance, then yes, an O(log n) search is slower asymptotically than a simple switch, but I strongly doubt this will wind up being the hot path.
You could make it a template
template<typename Base, typename Derive, template<Derive> class Derived>
Base* Base::Create(Derive t) {
switch (t) {
case Derive::A: return new Derived<Derive::A>;
case Derive::B: return new Derived<Derive::B>;
case Derive::C: return new Derived<Derive::C>;
default: return nullptr;
}
}
but that assumes that there are only ever A, B, and C in struct enum Derive.
Instead of using a C++ style enum you could use a Java style enum where each value is a singleton of a class derived from a common base.
Then define a set of pure virtual functions on the base class that create the desired flavour of derived class, and implement them appropriately in each of the singletons.
Then instead of switch (t) ... you can use t->createDerived().
Or to put it more succinctly: replace switch with polymorphism.
I want to implement a class hierarchy for object dispatching. Different classes dispatch different elements, and each class can dispatch its element represented as different data types.
It is better understood through a (faulty) example. This is what I would like to have if virtual function templating was allowed:
class Dispatcher {
template <class ReturnType>
virtual ReturnType getStuffAs();
};
So that I can implement subclasses as:
class CakeDispatcher : public Dispatcher {
template <>
virtual Recipe getStuffAs(){ ... }
template <>
virtual Baked getStuffAs(){ ... }
};
class DonutDispatcher : public Dispatcher {
template <>
virtual Frozen getStuffAs(){ ... }
template <>
virtual Baked getStuffAs(){ ... }
}
So that I can do the following later on:
void function( Dispatcher * disp ) {
// Works for Donut and Cake, but result will be a different Baked object
Baked b = disp->getStuffAs<Baked>();
// works if disp points to a DonutDispatcher
// fails if it is a CakeDispatcher
// can be compiling/linking time error or runtime error. I don't care
Frozen f = disp->getStuffAs<Frozen>();
}
Requirements/constraints:
All possible return types are not known beforehand. That's why I "need" templates.
Each class can provide just some return types.
Classes must have a common ancestor, so that I can store objects through a pointer to parent class and invoke functions through this pointer.
EDIT: I CAN'T use C++11 features, but I CAN use boost library.
Things I've thought about, but are not a solution:
Obviously, virtual template functions
Curiously Recurring Template Pattern: breaks the condition of common ancestor
Using some kind of traits class containing the functionality of children classes, but it does not work because a non-virtual implementation in the parent class does not have access to this information
I could maybe store some typeid info in the parent class, passed by children on construction. This makes possible for the non-virtual parent dispatching method to dynamic-cast itself to the children type... but it appears to be ugly as hell, and I don't know if this can cause some kind cycle-referencing problem.
class Dispatcher {
private:
typeid(?) childType;
public:
Dispatcher(typeid childT) : childType(childT) {}
// NOT VIRTUAL
template <class ReturnType>
ReturnType getStuffAs()
{
// or something equivalent to this cast, which I doubt is a correct expression
return dynamic_cast<childType *>(this)->childGetStuffAs<ReturnType>();
}
};
Then child classes would implement childGetStuffAs functions, which are not virtual too.
I've read like 5-10 related questions, but none of the provided solutions seems to fit this problem.
Can any of you come up with a better solution?
Is there a standard pattern/technique for solving this problem?
EDIT: The real problem
In the real problem, I have physical models with properties that can be represented in multiple ways: functions, matrices, probability distributions, polynomials, and some others (for example, a non-linear system can be represented as a function but not as a a matrix, while a linear system can be transformed to both).
There are also algorithms which can use those models indistinctly, but they could require specific representations for some model features. That's the reason for the "getStuffAs" function. The whole think is a bit complicated --too much to explain it here properly--, but I can guarantee that in this context the interface is well defined: input, computation and output.
My intention was to make this possible assuming that the number of possible representations is fully defined beforehand, and making it possible to transform the products to already existing types/classes that cannot be modified.
However, i'm starting to realize that this is, indeed, not possible in a simple way --I don't want to write a library just for this problem.
#include <cstdio>
// as a type identifier
struct stuff {
virtual void foo() {}
};
template <typename T>
struct stuff_inh : stuff {
};
struct Dispatcher {
template <typename T>
T* getStuffAs() {
return (T*)((getStuffAsImpl( new stuff_inh<T>() )));
}
virtual void* getStuffAsImpl(void*) = 0;
virtual void type() {printf("type::dispatcher\n");}
};
struct Cake : public Dispatcher {
void* getStuffAsImpl(void* p) {
stuff* s = static_cast<stuff*>(p);
printf("cake impl\n");
if (dynamic_cast<stuff_inh<Cake>*>(s) == NULL) {
throw "bad cast";
}
return (void*)(new Cake());
}
virtual void type() {printf("type::Cake\n");}
};
struct Rabbit : public Dispatcher {
void* getStuffAsImpl(void* p) {
stuff* s = static_cast<stuff*>(p);
printf("rabbit impl\n");
if (dynamic_cast<stuff_inh<Rabbit>*>(s) != NULL) {
return (void*)(new Rabbit());
}
else if (dynamic_cast<stuff_inh<Cake>*>(s) != NULL) {
return (void*)(new Cake());
}
else {
throw "bad cast";
}
}
virtual void type() {printf("type::Rabbit\n");}
};
void foo(Dispatcher* d) {
d->getStuffAs<Cake>()->type();
d->getStuffAs<Rabbit>()->type();
}
int main() {
Rabbit* r = new Rabbit;
foo(r);
Cake* c = new Cake;
foo(c);
}
I an not sure about the correctness of this ugly solution, may it be helpful for you. >_<
deletion of resource is not coded for a clearer look.
My solution is a combination of recurring template and diamond inheritance.
At least it's working. :)
#include <iostream>
class Dispatcher
{
public:
template<class T>
T getStuff()
{
return T();
}
};
template<class T>
class Stuffer : public Dispatcher
{
public:
template<class TT=T>
TT getStuff(){
return reinterpret_cast<TT>(this);
}
};
class Cake{
public:
Cake(){}
void print()
{
std::cout << "Cake" << std::endl;
}
};
class Recipe
{
public:
Recipe(){}
void print()
{
std::cout << "Recipe" << std::endl;
}
};
class CakeRecipe : public Stuffer<Cake>, public Stuffer< Recipe >
{
public:
};
int main()
{
Dispatcher* cr = reinterpret_cast<Dispatcher*>(new CakeRecipe());
cr->getStuff<Cake>().print();
cr->getStuff<Recipe>().print();
getchar();
return 1;
}
I'm working with a simple object model in which objects can implement interfaces to provide optional functionality. At it's heart, an object has to implement a getInterface method which is given a (unique) interface ID. The method then returns a pointer to an interface - or null, in case the object doesn't implement the requested interface. Here's a code sketch to illustrate this:
struct Interface { };
struct FooInterface : public Interface { enum { Id = 1 }; virtual void doFoo() = 0; };
struct BarInterface : public Interface { enum { Id = 2 }; virtual void doBar() = 0; };
struct YoyoInterface : public Interface { enum { Id = 3 }; virtual void doYoyo() = 0; };
struct Object {
virtual Interface *getInterface( int id ) { return 0; }
};
To make things easier for clients who work in this framework, I'm using a little template which automatically generates the 'getInterface' implementation so that clients just have to implement the actual functions required by the interfaces. The idea is to derive a concrete type from Object as well as all the interfaces and then let getInterface just return pointers to this (casted to the right type). Here's the template and a demo usage:
struct NullType { };
template <class T, class U>
struct TypeList {
typedef T Head;
typedef U Tail;
};
template <class Base, class IfaceList>
class ObjectWithIface :
public ObjectWithIface<Base, typename IfaceList::Tail>,
public IfaceList::Head
{
public:
virtual Interface *getInterface( int id ) {
if ( id == IfaceList::Head::Id ) {
return static_cast<IfaceList::Head *>( this );
}
return ObjectWithIface<Base, IfaceList::Tail>::getInterface( id );
}
};
template <class Base>
class ObjectWithIface<Base, NullType> : public Base
{
public:
virtual Interface *getInterface( int id ) {
return Base::getInterface( id );
}
};
class MyObjectWithFooAndBar : public ObjectWithIface< Object, TypeList<FooInterface, TypeList<BarInterface, NullType> > >
{
public:
// We get the getInterface() implementation for free from ObjectWithIface
virtual void doFoo() { }
virtual void doBar() { }
};
This works quite well, but there are two problems which are ugly:
A blocker for me is that this doesn't work with MSVC6 (which has poor support for templates, but unfortunately I need to support it). MSVC6 yields a C1202 error when compiling this.
A whole range of classes (a linear hierarchy) is generated by the recursive ObjectWithIface template. This is not a problem for me per se, but unfortunately I can't just do a single switch statement to map an interface ID to a pointer in getInterface. Instead, each step in the hierarchy checks for a single interface and then forwards the request to the base class.
Does anybody have suggestions how to improve this situation? Either by fixing the above two problems with the ObjectWithIface template, or by suggesting alternatives which would make the Object/Interface framework easier to use.
dynamic_cast exists within the language to solve this exact problem.
Example usage:
class Interface {
virtual ~Interface() {}
}; // Must have at least one virtual function
class X : public Interface {};
class Y : public Interface {};
void func(Interface* ptr) {
if (Y* yptr = dynamic_cast<Y*>(ptr)) {
// Returns a valid Y* if ptr is a Y, null otherwise
}
if (X* xptr = dynamic_cast<X*>(ptr)) {
// same for X
}
}
dynamic_cast will also seamlessly handle things like multiple and virtual inheritance, which you may well struggle with.
Edit:
You could check COM's QueryInterface for this- they use a similar design with a compiler extension. I've never seen COM code implemented, only used the headers, but you could search for it.
What about something like that ?
struct Interface
{
virtual ~Interface() {}
virtual std::type_info const& type() = 0;
};
template <typename T>
class InterfaceImplementer : public virtual Interface
{
std::type_info const& type() { return typeid(T); }
};
struct FooInterface : InterfaceImplementer<FooInterface>
{
virtual void foo();
};
struct BarInterface : InterfaceImplementer<BarInterface>
{
virtual void bar();
};
struct InterfaceNotFound : std::exception {};
struct Object
{
void addInterface(Interface *i)
{
// Add error handling if interface exists
interfaces.insert(&i->type(), i);
}
template <typename I>
I* queryInterface()
{
typedef std::map<std::type_info const*, Interface*>::iterator Iter;
Iter i = interfaces.find(&typeid(I));
if (i == interfaces.end())
throw InterfaceNotFound();
else return static_cast<I*>(i->second);
}
private:
std::map<std::type_info const*, Interface*> interfaces;
};
You may want something more elaborate than type_info const* if you want to do this across dynamic libraries boundaries. Something like std::string and type_info::name() will work fine (albeit a little slow, but this kind of extreme dispatch will likely need something slow). You can also manufacture numeric IDs, but this is maybe harder to maintain.
Storing hashes of type_infos is another option:
template <typename T>
struct InterfaceImplementer<T>
{
std::string const& type(); // This returns a unique hash
static std::string hash(); // This memoizes a unique hash
};
and use FooInterface::hash() when you add the interface, and the virtual Interface::type() when you query.
I have something like this:
class Base
{
public:
static int Lolz()
{
return 0;
}
};
class Child : public Base
{
public:
int nothing;
};
template <typename T>
int Produce()
{
return T::Lolz();
}
and
Produce<Base>();
Produce<Child>();
both return 0, which is of course correct, but unwanted. Is there anyway to enforce the explicit declaration of the Lolz() method in the second class, or maybe throwing an compile-time error when using Produce<Child>()?
Or is it bad OO design and I should do something completely different?
EDIT:
What I am basically trying to do, is to make something like this work:
Manager manager;
manager.RegisterProducer(&Woot::Produce, "Woot");
manager.RegisterProducer(&Goop::Produce, "Goop");
Object obj = manager.Produce("Woot");
or, more generally, an external abstract factory that doesn't know the types of objects it is producing, so that new types can be added without writing more code.
There are two ways to avoid it. Actually, it depends on what you want to say.
(1) Making Produce() as an interface of Base class.
template <typename T>
int Produce()
{
return T::Lolz();
}
class Base
{
friend int Produce<Base>();
protected:
static int Lolz()
{
return 0;
}
};
class Child : public Base
{
public:
int nothing;
};
int main(void)
{
Produce<Base>(); // Ok.
Produce<Child>(); // error :'Base::Lolz' : cannot access protected member declared in class 'Base'
}
(2) Using template specialization.
template <typename T>
int Produce()
{
return T::Lolz();
}
class Base
{
public:
static int Lolz()
{
return 0;
}
};
class Child : public Base
{
public:
int nothing;
};
template<>
int Produce<Child>()
{
throw std::bad_exception("oops!");
return 0;
}
int main(void)
{
Produce<Base>(); // Ok.
Produce<Child>(); // it will throw an exception!
}
There is no way to override a static method in a subclass, you can only hide it. Nor is there anything analogous to an abstract method that would force a subclass to provide a definition. If you really need different behaviour in different subclasses, then you should make Lolz() an instance method and override it as normal.
I suspect that you are treading close to a design problem here. One of the principals of object-oriented design is the substitution principal. It basically says that if B is a subclass of A, then it must be valid to use a B wherever you could use an A.
C++ doesn't support virtual static functions. Think about what the vtable would have to look like to support that and you'll realize its a no-go.
or maybe throwing a compile-time error when using Produce<Child>()
The modern-day solution for this is to use delete:
class Child : public Base
{
public:
int nothing;
static int Lolz() = delete;
};
It helps avoid a lot of boilerplate and express your intentions clearly.
As far as I understand your question, you want to disable static method from the parent class. You can do something like this in the derived class:
class Child : public Base
{
public:
int nothing;
private:
using Base::Lolz;
};
Now Child::Lolz becomes private.
But, of course, it's much better to fix the design :)
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.