c++ - 在没有预处理器的情况下从 hana 元组创建函数签名

Question

有没有办法在没有预处理器的情况下做到这一点？

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

namespace ba = boost::hana;

template <typename Arguments, unsigned ArgCount>
struct FunctionSigCreatorImpl {};

template <typename Arguments>
struct FunctionSigCreator : FunctionSigCreatorImpl<Arguments, decltype(ba::length(Arguments{}))::value>
{
};

#define DEF_ARG(z, n, data) \
    typename decltype(+ba::at(Arguments{}, ba::int_c<n>))::type

#define DEF_FUN_CREATOR(z, argCount, data)                                                  \
    template <typename Arguments>                                                         \
    struct FunctionSigCreatorImpl<Arguments, argCount>                                    \
    {                                                                                       \
        using Type = void(BOOST_PP_ENUM(argCount, DEF_ARG,));      \
    };

BOOST_PP_REPEAT(19, DEF_FUN_CREATOR,)

int main(int argc, char **argv)
{
    using MyTuple = ba::tuple<ba::type<int>, ba::type<long>, ba::type<char>>;
    static_assert(std::is_same<typename FunctionSigCreator<MyTuple>::Type, void(int, long, char)>::value);

    return 0;
}

Answer 1

Boost.CallableTraits 可以用 apply_return 做到这一点.

#include <boost/callable_traits.hpp>
#include <boost/hana.hpp>
#include <tuple>
#include <type_traits>

namespace hana = boost::hana;
namespace ct   = boost::callable_traits;

template <typename Tuple>
using to_function_sig = ct::apply_return_t<
  typename decltype(hana::unpack(Tuple{}, hana::template_<std::tuple>))::type,
  void
>;

int main()
{
  using MyTuple = hana::tuple<hana::type<int>, hana::type<long>, hana::type<char>>;
  static_assert(std::is_same<to_function_sig<MyTuple>, void(int, long, char)>::value, "");
}