如果索引超出范围,如何从std :: tuple_element返回void？

Question

我有一个函数traits结构,它使用std::tuple_element以下方法提供函数参数的类型:

#include <iostream>
#include <tuple>
#include <typeinfo>

template <typename T>
struct function_traits;

template <typename T_Ret, typename ...T_Args>
struct function_traits<T_Ret(T_Args...)> {
    // Number of arguments.
    enum { arity = sizeof...(T_Args) };
    // Argument types.
    template <size_t i>
    struct args {
        using type
            = typename std::tuple_element<i, std::tuple<T_Args...>>::type;
    };
};

int main() {
    using Arg0 = function_traits<int(float)>::args<0>::type;
    //using Arg1 = function_traits<int(float)>::args<1>::type; // Error, should be void.

    std::cout << typeid(Arg0).name() << std::endl;
    //std::cout << typeid(Arg1).name() << std::endl;
}

工作示例:Ideone

如果索引i超出范围(>= arity),则会产生编译时错误.相反,我想args<i>::type是void任何i超出范围.

虽然我可以专注args于具体的i,例如i == arity,我怎么能专注args于所有人i >= arity？

Answer 1

使用std::conditional和额外的间接：

struct void_type { using type = void; };


template <typename T_Ret, typename ...T_Args>
struct function_traits<T_Ret(T_Args...)> {
    // Number of arguments.
    enum { arity = sizeof...(T_Args) };

    // Argument types.
    template <size_t i>
    struct args {
        using type
            = typename std::conditional<(i < sizeof...(T_Args)),
                                        std::tuple_element<i, std::tuple<T_Args...>>,
                                        void_type>::type::type;
    };
};

演示版