我有以下代码,其中我使用模板化的 static
方法 random
定义了 struct quick
并进行了一些专门化:
(我使用了其他 SO 答案中的 function_traits
。附在底部以供引用。)
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;
}
我想在 check
中获取所有类型的 F
参数,这样我就可以生成一个带有随机条目的 tuple
(基于在我的随机
方法特化上)。
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;
但惨败:
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;
我使用了来自 this 的 function traits
实用程序所以回答。
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.
};
};
最佳答案
请注意,在 function_traits
中,您已经拥有所有参数类型。您所要做的就是公开它们:
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>;
};
现在,获取所有函数参数只是function_traits<F>::all_args
.
如果你不想改变function_traits
,我们只需要添加一个外部元函数:
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;
关于c++ - 如何从参数包中获取所有参数的类型?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38688041/