Say you have a tuple type and you want to extract its template parameter pack in order to instantiate another template. If that is a type template, then I can have a utility like this:
template < typename Tuple, template <typename...> typename What >
struct PutTupleInT;
template < typename... Types, template <typename...> typename What >
struct PutTupleInT<std::tuple<Types...>, What>
{
using Result = What<Types...>;
};
But what if the desired template is a variable template? While template <typename...> typename What is the "placeholder" for a type template, then what is the "placeholder" for a variable template?
I've tried the following for clang-4.0.0 (the only compiler by now that supports non-type template parameters with auto type), but it failed. Actually I am not sure if this is a correct syntax for C++17.
template < typename Tuple, template <typename...> auto What >
struct PutTupleInV;
template < typename... Types, template <typename...> auto What >
struct PutTupleInV<std::tuple<Types...>, What>
{
static constexpr auto value = What<Types...>;
};
I don't think you can do that. Quoting N4606:
§14.3.3 [temp.arg.template]/1
A template-argument for a template template-parameter shall be the name of a class template or an alias template, expressed as id-expression.
A variable template doesn't fit this requirement.
You could cheat a little and use a proxy type to select the template:
template < typename Tuple, class Proxy>
struct PutTupleInTV;
template < typename... Types, class Proxy>
struct PutTupleInTV<std::tuple<Types...>, Proxy>
{
static constexpr auto value = Proxy::template value<Types...>;
};
and then for
template<typename...> struct foo{};
template<typename... Ts> constexpr foo<Ts...> foo_v{};
struct use_foo
{
template<typename... Ts>
static constexpr auto value = foo_v<Ts...>;
};
you could say
PutTupleInTV<tup, use_foo>::value