Question

This is a follow up of this problem: Generic functor for functions with any argument list

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...); }
};

In operator() I can access the args... easily with a recursive "peeling" function as described in Stroustrup’s C++11 FAQ

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?

Answer 1

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;
    };
};

Test code:

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;
} 

Demo : http://ideone.com/YeN29