boost::hana 用于可变模板实例化的元组解包

Question

与相关我想知道是否可以使用 boost::hana:

以直接的方式实现这样的事情

#include <boost/hana.hpp>
#include <boost/hana/unpack.hpp>

namespace hana = boost::hana;

template<typename ... T>
struct A {};

int main() {

  auto my_tuple = hana::tuple_t<int, double, float>;

  // Is there any way to unpack my_tuple as template arguments for A?
  // Something like
  using MyStruct = A<hana::unpack_types(my_tuple)...>;

  static_assert(std::is_same<MyStruct, A<int, double, float>>::value, "Ooops!");
}

Answer 1

你可以自己做：

 template <template <typename...> class C, typename Tuple> struct RebindImpl;

 template <template <typename...> class C, typename ... Ts>
 struct RebindImpl<C, hana::tuple_t<Ts...>>{
     using type = C<Ts...>;
};

 template <template <typename...> class C, typename Tuple> 
 using Rebind = typename RebindImpl<C, Tuple>::type;

Answer 2

如果您只关心获得正确的类型，那么一个简单的方法可以做到这一点：

template <typename... Args>
constexpr A<Args...> make_A(hana::tuple_t<Args...>) {
    return {};
}

Answer 3

使用template_ to lift A into a metafunction, then call unpack:

using MyStruct = decltype(hana::unpack(my_tuple, hana::template_<A>))::type;

Example.

boost::hana 用于可变模板实例化的元组解包

boost::hana tuple unpacking for variadic template instantiation

c++

boost

c++14

boost-hana