如何获取可变参数模板类中的函数指针的参数类型？ [英] How do I get the argument types of a function pointer in a variadic template class?

问题描述

这是此问题的后续：函数的泛型函子与任何参数列表

我有这个functor类（全代码见上面的链接）：

I have this functor class (full code see link above):

template<typename... ARGS>
class Foo
{
    std::function<void(ARGS...)> m_f;
  public:
    Foo( std::function<void(ARGS...)> f ) : m_f(f) {}
    void operator()(ARGS... args) const { m_f(args...); }
};

在operator（）中，我可以使用递归剥离函数轻松访问args ...请在此处 http://www2.research.att.com/~bs/C ++ 0xFAQ.html＃variadic-templates

In operator() I can access the args... easily with a recursive "peeling" function as described here http://www2.research.att.com/~bs/C++0xFAQ.html#variadic-templates

我的问题是：我想访问f的参数类型，即ARGS ...，在构造函数中。显然我不能访问值，因为没有到目前为止，但参数类型列表以某种方式在f中插入，不是吗？

My problem is: I want to access the types of the arguments of f, i.e. ARGS..., in the constructor. Obviously I can't access values because there are none so far, but the argument type list is somehow burried in f, isn't it?

推荐答案

您可以写如下所示的 function_traits 类，以发现参数类型，返回类型和参数数量：

You can write function_traits class as shown below, to discover the argument types, return type, and number of arguments:

template<typename T> 
struct function_traits;  

template<typename R, typename ...Args> 
struct function_traits<std::function<R(Args...)>>
{
    static const size_t nargs = sizeof...(Args);

    typedef R result_type;

    template <size_t i>
    struct arg
    {
        typedef typename std::tuple_element<i, std::tuple<Args...>>::type type;
    };
};

测试代码：

struct R{};
struct A{};
struct B{};

int main()
{
   typedef std::function<R(A,B)> fun;

   std::cout << std::is_same<R, function_traits<fun>::result_type>::value << std::endl;
   std::cout << std::is_same<A, function_traits<fun>::arg<0>::type>::value << std::endl;
   std::cout << std::is_same<B, function_traits<fun>::arg<1>::type>::value << std::endl;
} 

演示： http://ideone.com/YeN29

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