Reputation: 1386
I am trying to write a container class that changes behavior depending on the based on a given dimension of the class as follow:
template <typename T, size_t DIM>
class Container { ... };
I tried to achieve that by using type_triats as follow:
template<typename T, size_t DIM, typename ENABLE = typename std::enable_if<
std::integral_constant<bool, DIM == 1>::value>::type>
{ // implementation for 1D };
and for 2 d container as follow:
template<typename T, size_t DIM, typename ENABLE = typename std::enable_if<
std::integral_constant<bool, DIM == 2>::value>::type>
{ // implementation for 2D };
by I get redefenition error ?!
Any idea how I can implement this customization.
Also, is there anyway that I can mask the implementaion of method using enable_if ??!!!
Upvotes: 2
Views: 70
Reputation: 10939
You can do this with SFINAE, but I don't see any reason to not just use explicit specialisation.
#include <iostream>
#include <type_traits>
// enable_if_t only available since C++14
#if ! ( __cplusplus > 201103L )
namespace std {
template< bool B, class T = void >
using enable_if_t = typename enable_if<B,T>::type;
}
#endif
template < typename T, size_t DIM, typename = void >
class Container
{
public:
Container() { std::cout << "Hello from generic\n"; }
};
template < typename T, size_t DIM >
class Container < T, DIM, std::enable_if_t<DIM == 1> >
{
public:
Container() { std::cout << "Hello from DIM = 1\n"; }
};
template < typename T, size_t DIM >
class Container < T, DIM, std::enable_if_t<DIM == 2> >
{
public:
Container() { std::cout << "Hello from DIM = 2\n"; }
};
int main()
{
Container<int, 1> c1;
Container<int, 2> c2;
Container<int, 3> c3;
}
Upvotes: 1
Reputation: 476940
That's not what enable_if
is for. What you most likely want is a straight-up specialization:
template <typename, size_t> class Container;
template <typename T> class Container<T, 1> {
// 1-D impl
};
template <typename T> class Container<T, 2> {
// 2-D impl
};
// etc.
Common behaviour can be factored into base classes if needed.
Upvotes: 4