带有继承的C ++检测习惯用法失败

Question

The below code fails to compile. For some reason inheriting from HasFoo causes IsWrapper to fail. It has something to do with the friend function foo() because inheriting from other classes seems to work fine. I don't understand why inheriting from HasFoo causes the detection idiom to fail.

What is the proper way to detect WithFoo as a Wrapper ?

https://godbolt.org/z/VPyarN

#include <type_traits>
#include <iostream>

template<typename TagType, typename ValueType>
struct Wrapper {
  ValueType V;
};

// Define some useful metafunctions.

template<typename Tag, typename T>
T BaseTypeImpl(const Wrapper<Tag, T> &);

template<typename T>
using BaseType = decltype(BaseTypeImpl(std::declval<T>()));

template<typename Tag, typename T>
Tag TagTypeImpl(const Wrapper<Tag, T> &);

template<typename T>
using TagType = decltype(TagTypeImpl(std::declval<T>()));

// Define VoidT.  Not needed with C++17.
template<typename... Args>
using VoidT = void;

// Define IsDetected.
template<template <typename...> class Trait, class Enabler, typename... Args>
struct IsDetectedImpl
  : std::false_type {};

template<template <typename...> class Trait, typename... Args>
struct IsDetectedImpl<Trait, VoidT<Trait<Args...>>, Args...>
  : std::true_type {};

template<template<typename...> class Trait, typename... Args>
using IsDetected = typename IsDetectedImpl<Trait, void, Args...>::type;

// Define IsWrapper true if the type derives from Wrapper.

template<typename T>
using IsWrapperImpl =
  std::is_base_of<Wrapper<TagType<T>, BaseType<T>>, T>;

template<typename T>
using IsWrapper = IsDetected<IsWrapperImpl, T>;

// A mixin.

template<typename T>
struct HasFoo {
  template<typename V,
           typename std::enable_if<IsWrapper<T>::value &&
                                   IsWrapper<V>::value>::type * = nullptr>
  friend void foo(const T &This, const V &Other) {
    std::cout << typeid(This).name() << " and " << typeid(Other).name()
              << " are wrappers\n";
  }
};

template<typename Tag>
struct WithFoo : public Wrapper<WithFoo<Tag>, int>,
                 public HasFoo<WithFoo<Tag>> {};

int main(void) {
  struct Tag {};

  WithFoo<Tag> WrapperFooV;

  // Fails.  Why?
  static_assert(IsWrapper<decltype(WrapperFooV)>::value,
                "Not a wrapper");

  return 0;
}

Answer 1

I don't understand why inheriting from HasFoo causes the detection idiom to fail.

Isn't completely clear to me also but surely a problem is that you use IsWrapper<T> inside the body of HasFoo<T> and, when you inherit HasFoo<WithFoo<Tag>> from WithFoo<Tag> you have that WithFoo<Tag> is incomplete when you check it with IsWrapper .

A possible solution (I don't know if acceptable for you) is define (and SFINAE enable/disable) foo() outside HasFoo .

I mean... try rewriting HasFoo as follows

template <typename T>
struct HasFoo {
  template <typename V>
  friend void foo(const T &This, const V &Other);
};

and defining foo() outside

template <typename T, typename V>
std::enable_if_t<IsWrapper<T>::value && IsWrapper<V>::value>
      foo(const T &This, const V &Other) {
  std::cout << typeid(This).name() << " and " << typeid(Other).name()
            << " are wrappers\n";
}

What is the proper way to detect WithFoo as a Wrapper?

Sorry but your code is too complicated for me.

I propose the following (simpler, I hope) alternative

#include <type_traits>
#include <iostream>

template<typename TagType, typename ValueType>
struct Wrapper {
  ValueType V;
};

template <typename T1, typename T2>
constexpr std::true_type IW_helper1 (Wrapper<T1, T2> const &);

template <typename T>
constexpr auto IW_helper2 (T t, int) -> decltype( IW_helper1(t) );

template <typename T>
constexpr std::false_type IW_helper2 (T, long);

template <typename T>
using IsWrapper = decltype(IW_helper2(std::declval<T>(), 0));

template <typename T>
struct HasFoo {
  template <typename V>
  friend void foo(const T &This, const V &Other);
};

template <typename T, typename V>
std::enable_if_t<IsWrapper<T>::value && IsWrapper<V>::value>
      foo(const T &This, const V &Other) {
  std::cout << typeid(This).name() << " and " << typeid(Other).name()
            << " are wrappers\n";
}

template<typename Tag>
struct WithFoo : public Wrapper<WithFoo<Tag>, int>,
                 public HasFoo<WithFoo<Tag>> {};

int main () {
  struct Tag {};

  WithFoo<Tag> WrapperFooV;

  static_assert(IsWrapper<decltype(WrapperFooV)>::value,
                "Not a wrapper");
}