模板类的std::is_base_of

您可以执行以下操作：

template <template <typename...> class C, typename...Ts>
std::true_type is_base_of_template_impl(const C<Ts...>*);
template <template <typename...> class C>
std::false_type is_base_of_template_impl(...);
template <typename T, template <typename...> class C>
using is_base_of_template = decltype(is_base_of_template_impl<C>(std::declval<T*>()));

实时演示

但从CCD_ 10继承多重或私有将失败。

使用Visual Studio 2017，当基类模板有多个模板参数，并且无法推导Ts... 时，这将失败

演示

VS错误报告

重构解决了VS的问题。

template < template <typename...> class base,typename derived>
struct is_base_of_template_impl
{
    template<typename... Ts>
    static constexpr std::true_type  test(const base<Ts...> *);
    static constexpr std::false_type test(...);
    using type = decltype(test(std::declval<derived*>()));
};
template < template <typename...> class base,typename derived>
using is_base_of_template = typename is_base_of_template_impl<base,derived>::type;

实时演示

这里的聚会有点晚，但我想给出上面的变体

template < template <typename...> class Base,typename Derived>
struct is_base_of_template
{
    // A function which can only be called by something convertible to a Base<Ts...>*
    // We return a std::variant here as a way of "returning" a parameter pack
    template<typename... Ts> static constexpr std::variant<Ts...> is_callable( Base<Ts...>* );
    // Detector, will return type of calling is_callable, or it won't compile if that can't be done
    template <typename T> using is_callable_t = decltype( is_callable( std::declval<T*>() ) );
    // Is it possible to call is_callable which the Derived type
    static inline constexpr bool value = std::experimental::is_detected_v<is_callable_t,Derived>;
    // If it is possible to call is_callable with the Derived type what would it return, if not type is a void
    using type = std::experimental::detected_or_t<void,is_callable_t,Derived>;
};
template < template <typename...> class Base,typename Derived> 
using is_base_of_template_t = typename is_base_of_template<Base,Derived>::type;
template < template <typename...> class Base,typename Derived>
inline constexpr bool is_base_of_template_v = is_base_of_template<Base,Derived>::value;

这使用了所提出的is_detected机制，我认为这使测试的意图更加清晰。然而，我现在可以获得基类实例化时使用的类型，我觉得这很有用。所以我可以写

template <typename T, typename U> struct Foo { };
struct Bar : Foo<int,std::string> { };
static_assert( is_base_of_template_v<Foo,Bar> );
// The variant type contains the types with which the Foo base is instantiated 
static_assert( std::is_same_v<std::variant<int,std::string>,is_base_of_template_t<Foo,Bar>> );

以下解决方案适用于受保护的继承。

template <template <typename...> class BaseTemplate, typename Derived, typename TCheck = void>
struct test_base_template;
template <template <typename...> class BaseTemplate, typename Derived>
using is_base_template_of = typename test_base_template<BaseTemplate, Derived>::is_base;
//Derive - is a class. Let inherit from Derive, so it can cast to its protected parents
template <template <typename...> class BaseTemplate, typename Derived>
struct test_base_template<BaseTemplate, Derived, std::enable_if_t<std::is_class_v<Derived>>> : Derived
{
    template<typename...T>
    static constexpr std::true_type test(BaseTemplate<T...> *);
    static constexpr std::false_type test(...);
    using is_base = decltype(test((test_base_template *) nullptr));
};
//Derive - is not a class, so it is always false_type
template <template <typename...> class BaseTemplate, typename Derived>
struct test_base_template<BaseTemplate, Derived, std::enable_if_t<!std::is_class_v<Derived>>>
{
    using is_base = std::false_type;
};

令人惊讶的是，在VS2017上，它可以从同一模板进行多重继承，比如C<int>和C<float>both。（不知道怎么做！）

检查链接到测试代码

根据叶夫根尼·马蒙托夫的回答，我认为正确的解决方案是

template <template <typename...> class BaseTemplate, typename Derived>
struct test_base_template<BaseTemplate, Derived, std::enable_if_t<std::is_class_v<Derived>>> : Derived
{
    template<typename...T>
    static constexpr std::true_type test(BaseTemplate<T...> *);
    static constexpr std::false_type test(...);
    using is_base = decltype(test((Derived *) nullptr));
};