Is it possible to use the using keyword for aliasing a template template parameter?
template <template<int> class T>
struct Foo {
using type = T;
};
Thanks
using (or typedef) always provide an alias for a type, never for an higher-kinded type (template template parameter). What you can do is templatize the alias itself on the int:
template <template<int> class T>
struct Foo {
template <int X>
using type = T<X>;
};
Related
In my class, I have a tuple object which I want to access the objects of in that as reference.
Actually, I want to know how can I write a get function to access the objects in my tuple?
I wrote a function which return the lvalue of my object in the tuple. I want to know how can I change it to return rvalue?
I have a template named GetIndex, which returns the index of type object in my tuple. In my tuple I definitely sure there are unique types object.
//return just lvalue
template <typename T>
T get_module()
{
return std::get<Private::GetIndex<T, Args...>::value>(types);
}
//compilation error
template <typename T>
T& get_module()
{
return &std::get<Private::GetIndex<T, Args...>::value>(types);
}
'''
You can just put & after T. It will be enough.
template <typename T>
T &get_module()
{
return std::get<Private::GetIndex<T, Args...>::value>(types);
}
(I had no idea how to name this question and I couldn't find anything similar. Sorry if this is duplicate)
If I want to inherit from some base template class, I can do this that way:
template<typename A=int, typename B=char> class C {};
template<typename... Args> class D : public C<Args...> {}; //it works!
This way I can change in project passed parameters to template class C and I don't have to change every usage of class D. Great. But what if I have template class using not only types as parameters but also values? For example:
template<int dim=3, typename float_t=double> class GeometricObject{};
template<typename... Args> class Point : public GeometricObject<Args...>{}; //it doesnt work
Of course I could define last template with integer type on the beginning. But this is not a way, if I would have 100 different classes all inheriting from GeometricObject and then I would change default dim value to 2, I would have to change every single class definition.
I also hope that there is the way without using any #define, #else and similar preprocessor commands. I know that templates are in fact also preprocessor commands, but... well, let's be modern here ;)
You can not mix type and non-type parameters in a template parameter pack. But it seems that your Point and other derived classes don't need to access the parameter pack arguments separately. In such cases it's easier, as well as more semantically correct, to pass the base class:
template<int dim=3, typename float_t=double> class GeometricObject{};
template<class GeometricObject=GeometricObject<>> class Point : public GeometricObject{};
Instantiating a Point could then look like:
Point<> a{}; // the same as Point<GeometricObject<>> a{};
Point<GeometricObject<4>> b{};
Point<GeometricObject<2, float>> c{};
Of course the GeometricObject<...> could be typedef'd to something shorter. Also, it can be made to look like a namespace instead of providing parameters to each geometric object separately:
template<int dim = 3, typename float_t = double>
struct GeometricObjects {
using Base = GeometricObject<dim, float_t>;
using Point = ::Point<Base>;
// ...
};
using TwoDim = GeometricObjects<2>;
TwoDim::Point a{};
I suppose you have multiple template classes and you want your Point object to be able to inherit from them all.
Instead of doing:
template <typename ... Args>
class Point : public GeometricObject<Args...>{};
I would instead do:
template <typename T>
class Point : public T {};
Now we just have to define proper traits to access the types template parameters in case they are needed. These types should be factored into a std::tuple (for instance).
The burden to fill this trait is on the GeometricObject class. For example, with your definition we would have:
template <typename T>
struct type_parameters;
template <int N, typename Float>
struct type_parameters<GeometricObject<N, Float> {
typedef std::tuple<Float> types;
};
The main scenario: a method of Point needs the type template parameters of GeometricObject (to forward them to a method of GeometricObject). To achieve this, you will have to pass in a tuple that will be unfold to call the inner method. To do so I make use of features added in the STL for C++14. You could still rewrite them yourself but I spared me the hassle for this question...
template <typename T>
class Point : public T {
template <typename Method, typename ... Args, std::size_t ... Is>
auto call_unfold(Method method, std::tuple<Args...> const& tuple, std::integer_sequence<std::size_t, Is...>) {
return (this->*method)(std::get<Is>(tuple)...);
}
template <typename Method, typename Tuple>
auto call_unfold(Method method, Tuple const& tuple) {
return call_unfold(method, tuple, std::make_index_sequence<std::tuple_size<Tuple>::value>());
}
public:
typedef typename type_parameters<T>::types types;
void some_method(types const& args) {
return call_unfold(&T::some_method, args);
}
};
This example is quite meaningless but the same technique could be useful with constructors of Point that need to call a base class constructor.
A live demo showing how it works is available on Coliru
Ok, so I figured it out how I should include variable-type template parameters into tuples. Basically I need to 'encapsulate' them into new parameter. This example works perfectly well AND solves my problem:
#include <type_traits>
template<int n = 2> struct Dim {
const int dim = n;
};
template<typename T> class SillyBaseClass {
public:
typedef typename T dim;
};
template<typename... Args> class SillyDerivedClass : public SillyBaseClass<Args...>{
public:
typedef typename SillyBaseClass::dim dim;
SillyDerivedClass() {
static_assert(std::is_same<dim,Dim<2>>::value,
"Number of dimensions must be equal to 2");
}
};
int main() {
SillyDerivedClass<Dim<2>> Class2d; //this works
SillyDerivedClass<Dim<3>> Class3d; //this returns expected error
}
In situations, when you have a base and many derived classes, all of which encode a variadic pack, something like the following
template<typename ... Args>
struct base
{};
template<typename ... Args>
struct derived_1: base <Args...>
{};
template<typename ... Args>
struct derived_2: base <Args...>
{};
How does one check if objects of different derived classes were instantiated using the same pack. For instance given
derived_1<int,float,double> d1_obj;
derived_2<int,float,double> d2_obj;
I want a mechanism to tell me that both objects are equal, in the sense that they contain the same types in the same order.
You can do this using template template parameters and partial specialization for when arguments match:
template <class T, class U>
struct same_args : std::false_type{};
template <template <typename...> class T, template <typename...> class U,
typename... Ts>
struct same_args<T<Ts...>, U<Ts...>> : std::true_type{};
Then same_args<derived_1<int,float>, derived_2<int,float>>::value will be true.
Live Demo
template <template <class ...args1> class T1, template <class ...args2> class T2>
class CollisionManager
{
public:
CollisionManager(void);
~CollisionManager(void);
bool checkCollision(T1<args1...> firstArray, T2<args2...> secondArray);
};
For some reasons compliler alerts args1 and args2 in checkCollision() as undeclared identifier
args1 and args2 are just placeholders. You can just omit them. Also notice that the parameters to your class template are also templates, not types. When CollisionManager gets instantiated args1 and args2 are not even bound to any types.
Update:
It seems to me, that you want to write something like this:
template <class T1, class T2> class CollisionManager;
template <template <class...> class T1,
template <class...> class T2,
class... args1,
class... args2>
class CollisionManager<T1<args1...>, T2<args2...>>
{
public:
CollisionManagerImpl(void);
~CollisionManagerImpl(void);
bool checkCollision(T1<args1...> firstArray, T2<args2...> secondArray);
};
Given a decorator class with standard decorator CTOR
explicit TheDecorator(std::unique_ptr<BaseClass> &&p_rrBase);
I want to create a member variable that is a unique ptr of such a decorator. I thus tried
unique_ptr<DerivedClass> spToDeco(make_unique<DerivedClass>() ); // class to decorate
m_spDecoration = make_unique<TheDecorator>( move(spToDeco) ); // unique ptr of decorator
Using VS 2010 and Scott Meyers implentation of make_unique (without variadic templates, implementing both a version with zero and one argument(s) instead), I get the error
error C2780: _Ty &&std::forward(...): expects 1 argument -- 0 provided
which I understand, as "TheDecorator" expects the argument to forward. But where the heck do I have to code it into, how do I form the syntax correctly? Am I supposed to include a move in the <> brackets? Thanks a lot for help!
Make_Unique:
namespace std
{
template<typename T>
std::unique_ptr<T> make_unique()
{
return std::unique_ptr<T>( new T() );
}
template<typename T, typename Ts>
std::unique_ptr<T> make_unique(Ts&& params)
{
return std::unique_ptr<T>( new T(std::forward<Ts>()) );
}
}
EDIT 2: SOLVED already, it must read
template<typename T, typename Ts>
std::unique_ptr<T> make_unique(Ts&& params)
{
return std::unique_ptr<T>( new T(std::forward<Ts>(params)) );
}