I'm trying to create an abstract factory template for multiple abstract factories in C++ and came up with this.
#define _CRTDBG_MAP_ALLOC
#include <crtdbg.h>
#include <map>
#include <stdio.h>
class Base
{
public:
virtual ~Base() {}
virtual bool Get() = 0;
};
class DerivedA : public Base
{
public:
bool Get()
{
return true;
}
};
class DerivedB : public Base
{
public:
bool Get()
{
return false;
}
};
template <class T>
class Creator
{
public:
virtual ~Creator(){}
virtual T* Create() = 0;
};
template <class T>
class DerivedCreator : public Creator<T>
{
public:
T* Create()
{
return new T;
}
};
template <class T, class Key>
class Factory
{
public:
void Register(Key Id, Creator<T>* Fn)
{
FunctionMap[Id] = Fn;
}
T* Create(Key Id)
{
return FunctionMap[Id]->Create();
}
~Factory()
{
std::map<Key, Creator<T>*>::iterator i = FunctionMap.begin();
while (i != FunctionMap.end())
{
delete (*i).second;
++i;
}
}
private:
std::map<Key, Creator<T>*> FunctionMap;
};
int main(int argc, char** argv[])
{
_CrtSetDbgFlag(_CrtSetDbgFlag(_CRTDBG_REPORT_FLAG) | _CRTDBG_LEAK_CHECK_DF);
//Register
Factory<Base, char*> temp;
temp.Register("DA", (Creator<Base>*)new DerivedCreator<DerivedA>);
temp.Register("DB", (Creator<Base>*)new DerivedCreator<DerivedB>);
//Pointer to base interface
Base* pBase = 0;
//Create and call
pBase = temp.Create("DA");
printf("DerivedA %u\n", pBase->Get());
delete pBase;
//Create and call
pBase = temp.Create("DB");
printf("DerivedB %u\n", pBase->Get());
delete pBase;
return 0;
}
It compiles and runs fine with no memory leaks (win32 crtdbg) but I don't know if this is really the correct way to do an abstract factory template.
temp.Register("DA", (Creator<Base>*)new DerivedCreator<DerivedA>);
I'm also wondering about the line above. I'm confused why I have to cast. I don't understand templates very well but I'd assume that it should work fine considering that both the template class and the actual class are derived.
That code actually works fine as shown above and even deletes fine with no memory leaks. I just don't feel entirely comfortable with it.
I haven't been able to find any real examples of template classes except for this from MaNGOS (wow emulator) - https://mangos.svn.sourceforge.net/svnroot/mangos/trunk/src/framework/Dynamic/ObjectRegistry.h
But I don't think I can use that method in my project because I plan on using DLLs at some point in my project and it uses CRTP which is against my requirement for runtime polymorphism.
The class DerivedCreator<DerivedA> is a Creator<DerivedA> not a Creator<Base>.
You need to tell the derived template what the base type so it can implement the interface of Creator<Base> by creating an instance of the derived type:
// DerivedCreator is Creator<BaseType> which creates a
// DerivedType, not a Creator<DerivedType>
template <class DerivedType, class BaseType>
class DerivedCreator : public Creator<BaseType>
{
public:
BaseType* Create()
{
return new DerivedType;
}
};
// Register
Factory<Base, std::string> temp;
temp.Register("DA", new DerivedCreator<DerivedA, Base>);
temp.Register("DB", new DerivedCreator<DerivedB, Base>);
// or if you want to create lots with the same base:
template <class DerivedType>
class DerivedBaseCreator : public DerivedCreator<DerivedType, Base> {};
//Register
Factory<Base, std::string> temp;
temp.Register("DA", new DerivedBaseCreator<DerivedA>);
temp.Register("DB", new DerivedBaseCreator<DerivedB>);
Small remarks to improve the design :
1) Use shared_ptr instead of raw pointers
2) use std::string instead of char*
You have to cast, because types Creator, Creator and Creator< DerivedB > are completely different types. The way to fix that is to remove casts :
//Register
Factory<Base, char*> temp;
temp.Register("DA", new DerivedCreator<Base>);
temp.Register("DB", new DerivedCreator<Base>);
You can make Factory::Register a template method and use boost mpl assert inside
#include <boost/mpl/assert.hpp>
#include <boost/type_traits.hpp>
template <class T, class Key>
class Factory
{
public:
///////////////////////////////////
template <typename _Base>
void Register(Key Id, Creator<_Base> * Fn)
{
BOOST_MPL_ASSERT((boost::is_base_of<T, _Base>));
FunctionMap[Id] = reinterpret_cast<Creator<T>*>(Fn);
}
///////////////////////////////////
//...
};
Related
Is it possible to declare some type of base class with template methods which i can override in derived classes? Following example:
#include <iostream>
#include <stdexcept>
#include <string>
class Base
{
public:
template<typename T>
std::string method() { return "Base"; }
};
class Derived : public Base
{
public:
template<typename T>
std::string method() override { return "Derived"; }
};
int main()
{
Base *b = new Derived();
std::cout << b->method<bool>() << std::endl;
return 0;
}
I would expect Derived as the output but it is Base. I assume it is necessary to make a templated wrapper class which receives the implementing class as the template parameter. But i want to make sure.
1) Your functions, in order to be polymorphic, should be marked with virtual
2) Templated functions are instantiated at the POI and can't be virtual (what is the signature??How many vtable entries do you reserve?). Templated functions are a compile-time mechanism, virtual functions a runtime one.
Some possible solutions involve:
Change design (recommended)
Follow another approach e.g. multimethod by Andrei Alexandrescu (http://www.icodeguru.com/CPP/ModernCppDesign/0201704315_ch11.html)
Template methods cannot be virtual. One solution is to use static polymorphism to simulate the behavior of "template virtual" methods:
#include <iostream>
#include <stdexcept>
#include <string>
template<typename D>
class Base
{
template<typename T>
std::string _method() { return "Base"; }
public:
template<typename T>
std::string method()
{
return static_cast<D&>(*this).template _method<T>();
}
};
class Derived : public Base<Derived>
{
friend class Base<Derived>;
template<typename T>
std::string _method() { return "Derived"; }
public:
//...
};
int main()
{
Base<Derived> *b = new Derived();
std::cout << b->method<bool>() << std::endl;
return 0;
}
where method is the interface and _method is the implementation. To simulate a pure virtual method, _method would absent from Base.
Unfortunately, this way Base changes to Base<Derived> so you can no longer e.g. have a container of Base*.
Also note that for a const method, static_cast<D&> changes to static_cast<const D&>. Similarly, for an rvalue-reference (&&) method, it changes to static_cast<D&&>.
Another possible aproach to make your example work as you expect is to use std::function:
class Base {
public:
Base() {
virtualFunction = [] () -> string { return {"Base"}; };
}
template <class T> string do_smth() { return virtualFunction(); }
function<string()> virtualFunction;
};
class Derived : public Base {
public:
Derived() {
virtualFunction = [] () -> string { return {"Derived"}; };
}
};
int main() {
auto ptr = unique_ptr<Base>(new Derived);
cout << ptr->do_smth<bool>() << endl;
}
This outputs "Derived". I'm not sure that this is what you realy want, but I hope it will help you..
I had the same problem, but I actually came up with a working solution. The best way to show the solution is by an example:
What we want(doesn't work, since you can't have virtual templates):
class Base
{
template <class T>
virtual T func(T a, T b) {};
}
class Derived
{
template <class T>
T func(T a, T b) { return a + b; };
}
int main()
{
Base* obj = new Derived();
std::cout << obj->func(1, 2) << obj->func(std::string("Hello"), std::string("World")) << obj->func(0.2, 0.1);
return 0;
}
The solution(prints 3HelloWorld0.3):
class BaseType
{
public:
virtual BaseType* add(BaseType* b) { return {}; };
};
template <class T>
class Type : public BaseType
{
public:
Type(T t) : value(t) {};
BaseType* add(BaseType* b)
{
Type<T>* a = new Type<T>(value + ((Type<T>*)b)->value);
return a;
};
T getValue() { return value; };
private:
T value;
};
class Base
{
public:
virtual BaseType* function(BaseType* a, BaseType* b) { return {}; };
template <class T>
T func(T a, T b)
{
BaseType* argA = new Type<T>(a);
BaseType* argB = new Type<T>(b);
BaseType* value = this->function(argA, argB);
T result = ((Type<T>*)value)->getValue();
delete argA;
delete argB;
delete value;
return result;
};
};
class Derived : public Base
{
public:
BaseType* function(BaseType* a, BaseType* b)
{
return a->add(b);
};
};
int main()
{
Base* obj = new Derived();
std::cout << obj->func(1, 2) << obj->func(std::string("Hello"), std::string("World")) << obj->func(0.2, 0.1);
return 0;
}
We use the BaseType class to represent any datatype or class you would usually use in a template. The members(and possibly operators) you would use in a template are described here with the virtual tag. Note that the pointers are necessary in order to get the polymorphism to work.
Type is a template class that extends Derived. This actually represents a specific type, for example Type<int>. This class is very important, since it allows us to convert any type into the BaseType. The definition of the members we described described in BaseType are implemented here.
function is the function we want to override. Instead of using a real template we use pointers to BaseType to represent a typename. The actual template function is in the Base class defined as func. It basically just calls function and converts T to Type<T>. If we now extend from Base and override function, the new overridden function gets called for the derived class.
I have a sprite class, which has a templatised data member. It holds an object, which has a pointer to this specialised sprite template class.
That object requires a forward declaration of my sprite class, but since sprite is a template class, I need to include the full header. Therefore I get a cyclic dependancy which I am unable to figure out
Sprite.h
#include "myclass.h"
template<typename SpriteType, typename = typename std::enable_if_t<std::is_base_of_v<sf::Transformable, SpriteType> && std::is_base_of_v<sf::Drawable, SpriteType>>>
class Sprite {
public:
SpriteType s;
myclass<SpriteType>;
Sprite() {
}
auto foo() {
return s;
}
private:
};
myclass.h
#include "Sprite.h"
//a sprite of type T, is going to create a myclass<Sprite<T>>, a pointer of the Sprite<T> is held in myclass.
template<typename T>
class myclass
{
public:
std::shared_ptr<Sprite<T>> ptr;
myclass() {
}
private:
};
How could I solve this cyclic dependency?
So in summary:
-Sprite is a template class.
-Sprite holds an object to another class. This other class holds a pointer to my this templated sprite class.
-This gives me a cyclic dependency, since both classes are now templates, and need to have their implementations written in their header files.
Simplified decoupling, based on #Taekahns solution.
template<typename T>
class myclass
{
public:
std::shared_ptr<T> ptr;
myclass() {
}
private:
};
template<typename SpriteType, typename = typename std::enable_if_t<std::is_base_of_v<sf::Transformable, SpriteType> && std::is_base_of_v<sf::Drawable, SpriteType>>>
class Sprite {
public:
SpriteType s;
// DO NOT PASS SpriteType here, put the whole Sprite<SpriteType>
myclass<Sprite<SpriteType>> t;
Sprite() {
}
auto foo() {
return s;
}
private:
};
One of the great thing about templates is breaking type dependencies.
You could do something like this. Simplified for readability.
template<typename T>
class myclass
{
public:
std::shared_ptr<T> ptr;
myclass() {
}
private:
};
template<typename SpriteType, typename = std::enable_if_t<std::is_base_of_v<base_class, SpriteType>>>
class Sprite {
public:
SpriteType s;
myclass<Sprite<SpriteType>> t;
Sprite() {
}
auto foo() {
return s;
}
private:
};
That is one of many options.
Another option is to use an interface. i.e. a pure virtual base class that isn't a template.
Example:
I think something like this should do it. Starting to get a hard to follow though.
class base_sprite
{
public:
virtual ~base_sprite(){};
virtual int foo() = 0;
};
template<typename T>
class myclass
{
public:
std::shared_ptr<base_sprite> ptr;
myclass() : ptr(std::make_shared<T>())
{
};
};
template<typename SpriteType>
class Sprite : public base_sprite{
public:
myclass<Sprite<SpriteType>> l;
int foo() override {return 0;};
};
I have the following classes:
class Box{...};
class MyBox : public Box{...};
And the template:
template <type T>
class ObjectManager{...};
Which I use in some other class:
class Engine{
ObjectManager<Box> * manager = nullptr;
...
};
Then I extend (implement) the Engine interface:
class MyEngine : public Engine{...}
And in that implementation (not earlier!) I know that manager should be like that:
MyEngine::MyEngine(){
manager = new ObjectManager<MyBox>();
}
But this gives me an error because of types conflict (conversion between ObjectManager<Box> and ObjectManager<MyBox>), even when MyBox inherits from Box.
Is there any way around that problem? I don't want to modify the Box, MyBox, ObjectManager and Engine classes.
Templatize Engine and then inherit MyEngine with Engine carrying the template instance of Box that you desire. Something like this: (http://codepad.org/SZMSbCRB)
#include <iostream>
using namespace std;
class Box{
};
class MyBox : public Box{
};
template <typename T>
class ObjectManager{
};
template <typename T>
class Engine{
public:
ObjectManager<T*> * manager;
};
class MyEngine : public Engine<MyBox>{
public:
MyEngine(){
manager = new ObjectManager<MyBox*>();
cout<<"myEngine created"<<endl;
}
};
int main() {
MyEngine eng = MyEngine();
return 0;
}
The benefit here is, if tomorrow you create a new Box namely MyBox2 and want to create a custom engine MyEngine2 for that as well, simply inherit MyEngine : public Engine <MyBox2>. Just make sure to take care of type conversions.
As far as I remember, in Java you can declare Engine as something like Engine<extends T> which lets you instantiate Engine with any subtype of the template T provided. That is a safer and better way to do it, but I do not know if C++ provides something like that.
Hope it helps.
Consider using a wrapper for your ObjectManager.
NOTE marks the lines which will fail to compile if T and BaseT are not compatible.
Assuming:
template<class T>
ObjectManager{
T* objectAtIndex(size_t i); // As an example.
void insertAtIndex(T* object, size_t i); // As an example.
};
Wrapper:
template<class T, class BaseT>
class MyObjectManager
{
public:
ObjectManager<BaseT>* manager; // public for simplicity only!
T* objectAtIndex(size_t i){
return static_cast<T*>(manager->objectAtIndex(i)); // <- NOTE
}
void insertAtIndex(T* object, size_t i){
manager->insertAtIndex(object, i); // <- NOTE
}
};
Usage:
class MyEngine : public Engine
{
MyObjectManager<MyBox, Box> my_manager;
MyEngine(){
// Setup the manager(s).
manager = new ObjectManager<Box>();
my_manager.manager = manager;
// Example usage.
my_manager.insertAtIndex(new MyBox(), 0);
MyBox* p = my_manager.objectAtIndex(0);
}
};
This isn't possible without altering the Design in some way I think.
ObjectManager<Box> just isn't ObjectManager<MyBox> and it is not a base class of it but only the template arguments are base and derived of each other.
If your ObjectManager uses a pointer or smart pointer for the "managed" object...
template <class T>
class ObjectManager
{
T * object;
public:
ObjectManager(T * ptr) : object(ptr) { }
};
... you can construct the ObjectManager using a pointer to a derived object:
class MyEngine : public Engine
{
MyEngine()
{
manager = new ObjectManager<Box>(new MyBox);
}
};
If Engine is an abstract interface you could also have an own manager in MyEngine and use it to implement MyEngine
class MyEngine : public Engine
{
ObjectManager<MyBox> * mymanager;
MyEngine() : Engine(), mymanager(new ObjectManager<MyBox>)
{ }
};
You could provide an implementation to support the conversion. This should be similar to the way a std::unique_ptr can be implicitly converted from a derived pointer type to a base pointer type.
Example Code
#include <iostream>
#include <memory>
#include <vector>
class Box
{
public:
virtual ~Box() {}
virtual void foo() = 0;
};
class MyBox : public Box
{
public:
virtual ~MyBox() {}
virtual void foo() override
{
std::cout << "MyBox::foo()\n";
}
};
template<typename T>
class ObjectManager
{
public:
ObjectManager() {}
void add(T *object)
{
objects.emplace_back(object);
}
template<typename U>
ObjectManager<T> &operator=(ObjectManager<U> &other)
{
return *this;
}
std::size_t size() const
{
return objects.size();
}
T& operator[](std::size_t i)
{
return *objects[i];
}
private:
std::vector<std::unique_ptr<T>> objects;
};
class Engine
{
public:
ObjectManager<Box> manager;
};
class MyEngine : public Engine
{
public:
MyEngine()
{
manager = ObjectManager<MyBox>();
manager.add(new MyBox());
}
};
int main()
{
MyEngine engine;
for (std::size_t i = 0; i < engine.manager.size(); ++i)
{
engine.manager[i].foo();
}
return 0;
}
Example Output
MyBox::foo()
Another Example
This might help illustrate how to implement the conversion.
Example Code
template<typename T>
class MyPointer
{
public:
MyPointer() :
mPointer(nullptr)
{
// Do nothing
}
template<typename U>
MyPointer(MyPointer<U> &other) :
mPointer(other.mPointer)
{
other.mPointer = nullptr;
}
MyPointer(T *pointer) :
mPointer(pointer)
{
// Do nothing
}
template<typename U>
MyPointer<T> &operator=(MyPointer<U> &other)
{
mPointer = other.mPointer;
other.mPointer = nullptr;
return *this;
}
~MyPointer()
{
delete mPointer;
}
T* operator->()
{
return mPointer;
}
private:
template<typename U> friend class MyPointer;
T* mPointer;
};
int main()
{
{
MyPointer<MyBox> myBox(new MyBox());
MyPointer<Box> box;
box = myBox;
box->foo();
}
{
MyPointer<MyBox> myBox(new MyBox());
MyPointer<Box> box(myBox);
box->foo();
}
return 0;
}
Example Output
MyBox::foo()
MyBox::foo()
Is it possible to declare some type of base class with template methods which i can override in derived classes? Following example:
#include <iostream>
#include <stdexcept>
#include <string>
class Base
{
public:
template<typename T>
std::string method() { return "Base"; }
};
class Derived : public Base
{
public:
template<typename T>
std::string method() override { return "Derived"; }
};
int main()
{
Base *b = new Derived();
std::cout << b->method<bool>() << std::endl;
return 0;
}
I would expect Derived as the output but it is Base. I assume it is necessary to make a templated wrapper class which receives the implementing class as the template parameter. But i want to make sure.
1) Your functions, in order to be polymorphic, should be marked with virtual
2) Templated functions are instantiated at the POI and can't be virtual (what is the signature??How many vtable entries do you reserve?). Templated functions are a compile-time mechanism, virtual functions a runtime one.
Some possible solutions involve:
Change design (recommended)
Follow another approach e.g. multimethod by Andrei Alexandrescu (http://www.icodeguru.com/CPP/ModernCppDesign/0201704315_ch11.html)
Template methods cannot be virtual. One solution is to use static polymorphism to simulate the behavior of "template virtual" methods:
#include <iostream>
#include <stdexcept>
#include <string>
template<typename D>
class Base
{
template<typename T>
std::string _method() { return "Base"; }
public:
template<typename T>
std::string method()
{
return static_cast<D&>(*this).template _method<T>();
}
};
class Derived : public Base<Derived>
{
friend class Base<Derived>;
template<typename T>
std::string _method() { return "Derived"; }
public:
//...
};
int main()
{
Base<Derived> *b = new Derived();
std::cout << b->method<bool>() << std::endl;
return 0;
}
where method is the interface and _method is the implementation. To simulate a pure virtual method, _method would absent from Base.
Unfortunately, this way Base changes to Base<Derived> so you can no longer e.g. have a container of Base*.
Also note that for a const method, static_cast<D&> changes to static_cast<const D&>. Similarly, for an rvalue-reference (&&) method, it changes to static_cast<D&&>.
Another possible aproach to make your example work as you expect is to use std::function:
class Base {
public:
Base() {
virtualFunction = [] () -> string { return {"Base"}; };
}
template <class T> string do_smth() { return virtualFunction(); }
function<string()> virtualFunction;
};
class Derived : public Base {
public:
Derived() {
virtualFunction = [] () -> string { return {"Derived"}; };
}
};
int main() {
auto ptr = unique_ptr<Base>(new Derived);
cout << ptr->do_smth<bool>() << endl;
}
This outputs "Derived". I'm not sure that this is what you realy want, but I hope it will help you..
I had the same problem, but I actually came up with a working solution. The best way to show the solution is by an example:
What we want(doesn't work, since you can't have virtual templates):
class Base
{
template <class T>
virtual T func(T a, T b) {};
}
class Derived
{
template <class T>
T func(T a, T b) { return a + b; };
}
int main()
{
Base* obj = new Derived();
std::cout << obj->func(1, 2) << obj->func(std::string("Hello"), std::string("World")) << obj->func(0.2, 0.1);
return 0;
}
The solution(prints 3HelloWorld0.3):
class BaseType
{
public:
virtual BaseType* add(BaseType* b) { return {}; };
};
template <class T>
class Type : public BaseType
{
public:
Type(T t) : value(t) {};
BaseType* add(BaseType* b)
{
Type<T>* a = new Type<T>(value + ((Type<T>*)b)->value);
return a;
};
T getValue() { return value; };
private:
T value;
};
class Base
{
public:
virtual BaseType* function(BaseType* a, BaseType* b) { return {}; };
template <class T>
T func(T a, T b)
{
BaseType* argA = new Type<T>(a);
BaseType* argB = new Type<T>(b);
BaseType* value = this->function(argA, argB);
T result = ((Type<T>*)value)->getValue();
delete argA;
delete argB;
delete value;
return result;
};
};
class Derived : public Base
{
public:
BaseType* function(BaseType* a, BaseType* b)
{
return a->add(b);
};
};
int main()
{
Base* obj = new Derived();
std::cout << obj->func(1, 2) << obj->func(std::string("Hello"), std::string("World")) << obj->func(0.2, 0.1);
return 0;
}
We use the BaseType class to represent any datatype or class you would usually use in a template. The members(and possibly operators) you would use in a template are described here with the virtual tag. Note that the pointers are necessary in order to get the polymorphism to work.
Type is a template class that extends Derived. This actually represents a specific type, for example Type<int>. This class is very important, since it allows us to convert any type into the BaseType. The definition of the members we described described in BaseType are implemented here.
function is the function we want to override. Instead of using a real template we use pointers to BaseType to represent a typename. The actual template function is in the Base class defined as func. It basically just calls function and converts T to Type<T>. If we now extend from Base and override function, the new overridden function gets called for the derived class.
I want to have a container (let's say an std::vector) that would hold various inherited types, and would instantiate them,.i.e. vector of classes --> vector of objects.
For instance:
class A{};
class B: public class A
{};
class C: public class A
{};
void main()
{
std::vector<of inherited A types> typesVec;
std::vector<A*> objectsVec;
typesVec.push_back(class B);
typesVec.push_back(class C);
for (int i = 0; i < typesVec.size(); i++)
{
A* pA = new typesVec.at(i);
objectsVec.push_back(pA);
}
}
Thanks in advance..
This isn't possible in C++ (at least not directly). I can see this happening in a language that has reflection, but C++ doesn't.
What you can do instead is create a factory or simply methods that create objects of the specified type.
Instead of having a vector of types, you'd have a vector of object generators (close enough, right?):
class A{};
class B: public class A
{};
class C: public class A
{};
struct AFactory
{
virtual A* create() { return new A; }
};
struct BFactory : AFactory
{
virtual A* create() { return new B; }
};
struct CFactory : AFactory
{
virtual A* create() { return new C; }
};
//...
typesVec.push_back(new BFactory);
typesVec.push_back(new CFactory);
for (int i = 0; i < typesVec.size(); i++)
{
A* pA = typesVec.at(i)->create();
objectsVec.push_back(pA);
}
There is a reusable approach with templates. This is a generic factory for derived types that comes with an install and a create method which lets you write code like this:
int main() {
TypeVector<Base> t;
t.install<Foo>("Foo");
t.install<Bar>("Bar");
t.create("Foo")->hello();
}
Note it's a sketch implementation. In the real world, you may provide another template parameter to specify the underlying container type (for few types, vector is probably more efficient than set).
The type-vector is this:
template <typename Base>
class Creator;
template <typename Base>
class TypeVector {
public:
template <typename Derived>
void install (std::string const &name) ;
std::shared_ptr<Base> create (std::string const &name) const;
private:
struct Meta {
Meta(std::shared_ptr<Creator<Base>> creator, std::string const &name)
: creator(creator), name(name) {}
std::shared_ptr<Creator<Base>> creator;
std::string name;
};
std::vector<Meta> creators_;
};
We somehow need a way to store the type in an allocatable manner. We do it like boost::shared_ptr, which combines an abstract base class and a template derived class:
template <typename Base>
class Creator {
public:
virtual ~Creator() {}
virtual std::shared_ptr<Base> create() const = 0;
};
template <typename Base, typename Derived>
class ConcreteCreator : public Creator<Base> {
public:
virtual std::shared_ptr<Base> create() const {
return std::shared_ptr<Base>{new Derived()};
}
};
The "concrete creator" is able to allocate an actual object, and return a pointer-to-base of it.
Finally, here are the implementations of TypeVector::install and TypeVector::create:
template <typename Base>
template <typename Derived>
void
TypeVector<Base>::install (std::string const &name)
{
creators_.emplace_back(
std::shared_ptr<Creator<Base>>(new ConcreteCreator<Base, Derived>()),
name);
}
template <typename Base>
std::shared_ptr<Base>
TypeVector<Base>::create (std::string const &name) const
{
for (auto m : creators_) {
if (name == m.name) return m.creator->create();
}
throw std::runtime_error("...");
}
and finally, here's a test:
#include <iostream>
struct Base {
virtual ~Base() {}
virtual void hello() const = 0;
};
struct Foo : Base {
virtual void hello() const { std::cout << "I am a Foo\n"; }
};
struct Bar : Base {
virtual void hello() const { std::cout << "I am a Bar\n"; }
};
int main() {
TypeVector<Base> t;
t.install<Foo>("Foo");
t.install<Bar>("Bar");
t.create("Foo")->hello();
}
You can go further and make any constructor callable for code like ...
...
Bar(Color, Age, int)
...
t.create("Foo", Color::Red, Age::TooOld, 42)
... but this requires an awesome grasp of variadic template argument lists, and how to fold them into a constructor call (can be done and has been done, but it would explode this answer).
Just a quick solution sketch:
The C++ standard does not provide direct calls to constructors. As such you can't have function pointers to constructors; you can, however, have a wrapper function "create", something like:
template<typename T>
T* create () {
return (new T();
}
Provide overloaded create definitions for one argument, two arguments, ... or try to use variadic templates; or, if you already know what types you need, you can create the create functions specifically. Then you can have a function pointer to the create function:
&create<TheType>
Mind that the signature of this function however depends on the type used. You can however create a struct that contains typdefs for the templated type, a typedef for the type pointer, and the create function as a functor operator().
Thus you can have two vectors, one for the function pointers to the create function, or, alternatively to the structs mentioned before, and one with the actual objects. In your case where you only have inherited types, you might be able to define functions A* createB() { return new B(); }, A* createC() { return new C(); }, ... for each inherited type B, C, ... and have a vector for pointers to these create functions and the second vector for the A pointers.
I might point you Andrei Alesandrescu´s book Modern C++ Design (or the Loki library he describes in the book) and the chapter about type lists. This would require you to do the typeVec.insert( type ) at compile time.