How to detect if a method is virtual?

问题:

I tried to make a traits to find if a method is virtual: (https://ideone.com/9pfaCZ)

// Seveval structs which should fail depending if T::f is virtual or not.
template <typename T> struct Dvf : T { void f() final; };
template <typename T> struct Dvo : T { void f() override; };
template <typename T> struct Dnv : T { void f() = delete; };

template <typename U>
class has_virtual_f
{
private:
    template <std::size_t N> struct helper {};
    template <typename T>
    static std::uint8_t check(helper<sizeof(Dvf<T>)>*);
    template<typename T> static std::uint16_t check(...);
public:
    static
    constexpr bool value = sizeof(check<U>(0)) == sizeof(std::uint8_t);
};

Test cases:

struct V  { virtual void f(); };
struct NV {         void f(); };
struct E  {                   };
struct F  { virtual void f() final; }; // Bonus (unspecified expected output)

static_assert( has_virtual_f< V>::value, "");
static_assert(!has_virtual_f<NV>::value, "");
static_assert(!has_virtual_f< E>::value, "");

But I got error: 'void Dvf<T>::f() [with T = NV]' marked final, but is not virtual.
If I don't use sizeof and directly Dvf<T>* in check, I don't have compilation error, but check is not discarded for "bad" type in SFINAE :( .

What is the proper way to detect if a method is virtual ?

回答1:

The code isn't perfect but it basically passes the tests (at least in all clangs available on wandbox and gcc since 7.):

#include <type_traits>

template <class T>
using void_t = void;

template <class T, T v1, T v2, class = std::integral_constant<bool, true>>
struct can_be_compaired: std::false_type { };

template <class T, T v1, T v2>
struct can_be_compaired<T, v1, v2, std::integral_constant<bool, v1 == v2>>: std::true_type { };

template <class T, class = void>
struct has_virtual_f: std::false_type { };

template <class T>
struct has_virtual_f<T, void_t<decltype(&T::f)>>{
    constexpr static auto value = !can_be_compaired<decltype(&T::f), &T::f, &T::f>::value;
};

struct V  { virtual void f() { }      };
struct NV {         void f() { }      };
struct E  {                           };
struct F  { virtual void f() final{ } }; // Bonus (unspecified expected output)

int main() {
   static_assert( has_virtual_f< V>::value, "");
   static_assert(!has_virtual_f<NV>::value, "");
   static_assert(!has_virtual_f< E>::value, "");
   static_assert( has_virtual_f< F>::value, "");
}

[live demo]

---

The relevant standard parts that theoretically let the trait fly: [expr.eq]/4.3, [expr.const]/2.23

回答2:

There is probably no way to determine if a specific method is virtual. I say this because the Boost project researched traits for years and never produced such a traits test.

However, in C++11, or using the Boost library, you can use the is_polymorphic<> template to test a type to see if the type has virtual functions. See std::is_polymorphic<> or boost::is_polymorphic<> for reference.