如何从参数包中获取所有参数的类型？

Question

我有以下代码片段,我struct quick用模板化static方法定义random了一些特殊化:

(我使用function_traits其他SO答案.附在底部供参考.)

struct quick
{
  template <typename T>
  static T random();

  template <typename F>
  static void check(F f)
  {

    constexpr auto arity = function_traits<F>::arity; // easy :)
    std::cout << arity << std::endl;
    typedef typename function_traits<F>::template arg<0>::type type0; // easy:)
    // how to get all types of all F's parameters?
  }
};

template <>
std::string quick::random<std::string>()
{
  return std::string("test");
}

template <>
int quick::random<int>()
{
  return 1;
}

我想获取所有类型F的参数,check以便我可以生成tuple随机条目(基于我的random方法特化).

像这样:

auto t0 = std::make_tuple(quick::random<AllTypes>()...); //pseudo code
auto t =
    std::make_tuple(quick::random <
                                  function_traits<F>::template arg<std::make_index_sequence<arity>>::type...
                                  >
                                  ()...
                     );

我尝试过类似的东西:

template<typename F, typename ...TIdxs>
using ArgTypes = typename function_traits<F>::template arg<TIdxs>::type...;

// ...
// inside check

typedef ArgTypes<F, std::make_index_sequence<arity>> types;

但悲惨地失败了:

main.cpp:80:72: error: expected ‘;’ before ‘...’ token
 using ArgTypes = typename function_traits<F>::template arg<TIdxs>::type...;
                                                                        ^
main.cpp: In static member function ‘static void quick::check(F, D)’:
main.cpp:98:15: error: ‘ArgTypes’ does not name a type
       typedef ArgTypes<F, std::make_index_sequence<arity>> types;

---

我已经使用function traits了这个 SO答案的实用工具.

template <typename T>
struct function_traits : function_traits<decltype(&T::operator())>
{};
// For generic types, directly use the result of the signature of its 'operator()'

template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) const>
// we specialize for pointers to member function
{
    enum { arity = sizeof...(Args) };
    // arity is the number of arguments.

    typedef ReturnType result_type;

    template <size_t i>
    struct arg
    {
        typedef typename std::tuple_element<i, std::tuple<Args...>>::type type;
        // the i-th argument is equivalent to the i-th tuple element of a tuple
        // composed of those arguments.
    };
};

Answer 1

请注意function_traits,您已经拥有所有参数类型.你所要做的就是揭露它们:

template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) const>
// we specialize for pointers to member function
{
    enum { arity = sizeof...(Args) };

    using result_type = ReturnType;

    using all_args = std::tuple<Args...>; // <-- add this

    template <size_t i> // <-- consider making this an alias template
    using arg = std::tuple_element_t<i, all_args>;
};

而现在,获取所有函数参数只是function_traits<F>::all_args.

如果您不想更改function_traits,我们只需添加外部元函数:

template <class F, class = std::make_index_sequence<function_traits<F>::arity>>
struct all_args;

template <class F, size_t... Is>
struct all_args<F, std::index_sequence<Is...>> {
    using type = std::tuple<typename function_traits<F>::template arg<Is>::type...>;
};

template <class F>
using all_args_t = typename all_args<F>::type;

Answer 2

不确定这就是您想要的但是...quick按以下方式修改怎么样？

struct quick
{
  template <typename T>
  static T random();

  template<typename F, std::size_t I>
     using ArgTypes = typename function_traits<F>::template arg<I>::type;

  template<typename F, std::size_t ... Is>
     using ArgTuple = std::tuple< ArgTypes<F, Is>... >;

  template <typename F, std::size_t ... Is>
  static ArgTuple<F, Is...> makeArgTuple ()
   { return make_tuple(quick::random<Is>()...); }

  template <typename F>
  static void check(F f)
  {

    constexpr auto arity = function_traits<F>::arity; // easy :)
    std::cout << arity << std::endl;

    typedef typename function_traits<F>::template arg<0>::type type0; // easy:)

    auto t = ArgTuple<F, std::make_index_sequence<arity>::type> ();

    auto t2 = makeArgTuple<F, std::make_index_sequence<arity>::type>();
  }

};

你错误地将typenames传递TIdxs给arg; arg需要一个std::size_t.

考虑到std::make_index_sequence它是 C++14 功能（但在 C++11 中创建它也很容易）。

ps：抱歉我的英语不好。