c++ - 如何实现is_polymorphic_functor？

Question

我正在尝试实现 is_polymorphic_functor 元函数以获得以下结果:

//non-polymorphic functor
template<typename T> struct X { void operator()(T); };

//polymorphic functor 
struct Y { template<typename T> void operator()(T); };

std::cout << is_polymorphic_functor<X<int>>::value << std::endl; //false
std::cout << is_polymorphic_functor<Y>::value << std::endl; //true

这只是一个例子。理想情况下，它应该适用于任何 数量的参数，即operator()(T...)。 Here are few more test cases我用它来测试@Andrei Tita 的解决方案，该解决方案在两个测试用例中都失败了。

我试过这个:

template<typename F>
struct is_polymorphic_functor
{
  private:
     typedef struct { char x[1]; }  yes;
     typedef struct { char x[10]; } no;

     static yes check(...);

     template<typename T >
     static no check(T*, char (*) [sizeof(functor_traits<T>)] = 0 );            
  public:
     static const bool value = sizeof(check(static_cast<F*>(0))) == sizeof(yes);
};

它试图利用以下 functor_traits 的实现:

//functor traits
template <typename T>
struct functor_traits : functor_traits<decltype(&T::operator())>{};

template <typename C, typename R, typename... A>
struct functor_traits<R(C::*)(A...) const> : functor_traits<R(C::*)(A...)>{};

template <typename C, typename R, typename... A>
struct functor_traits<R(C::*)(A...)>
{
   static const size_t arity = sizeof...(A) };

   typedef R result_type;

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

对于多态仿函数给出了以下错误:

error: decltype cannot resolve address of overloaded function

如何解决这个问题并使 is_polymorphic_functor 按预期工作？

Answer 1

这对我有用:

template<typename T>
struct is_polymorphic_functor
{
private:
    //test if type U has operator()(V)
    template<typename U, typename V>
    static auto ftest(U *u, V* v) -> decltype((*u)(*v), char(0));
    static std::array<char, 2> ftest(...);

    struct private_type { };

public:
    static const bool value = sizeof(ftest((T*)nullptr, (private_type*)nullptr)) == 1;
};