Question

当std::is_base_of<A, B>是模板类时，有没有办法测试A？

template <typename X, typename Y> class A {};

template <typename X> class B : public A<X, char> {};

我想静态地测试类似std::is_base_of<A, B<int>>的含义，B来自A的任何特化。（为了使其更通用，假设我们不知道B专门化A的方式，即B＆lt; X＆gt;派生自A＆lt; X， char ＆gt;）

解决的一种方法是从（非模板）类派生A说C，然后检查std::is_base_of<C, B<int>>。但还有另一种方法吗？

Answer 1

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*>()));

Live Demo

但会因多次继承或来自A的私有继承而失败。

使用Visual Studio 2017，当基类模板具有多个模板参数时，这将失败，并且无法推断Ts...

Demo

VS Bug Report

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;

Live Demo

Answer 2

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，它适用于来自同一模板的多重继承，例如C＆lt; int＆gt;和C＆lt;浮动＆gt;都。（不知道怎么做！）

检查 Link to test code

Answer 3

在这里参加聚会有点晚，但我想提供上述内容的一种变体

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;

这使用了提议的is_detected机制，我认为这会使测试的意图更加清晰。但是，我现在可以获得在我认为有用的同时实例化基类的类型。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>> );

Answer 4

根据Evgeny Mamontov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));
};

std :: is_base_of用于模板类

4 个答案: