Public 继承的静态断言

Question

我构建了一个助手class，它可以通过模板构建自定义class，这个自定义class必须继承自某个class，我可以检查一下std::is_base_of.

但是我还需要检查继承是public，如何实现？

作为参考，这是 class 的精简版，我在里面有 std::is_base_of。

template<class CustomSink>
class Sink
{
    static_assert(std::is_base_of<BaseSink, CustomSink>::value, "CustomSink must derive from BaseSink");
    //Some static assert here to check if custom sink has publicly inherited BaseSink 
    //static_assert(is_public.....
public:
    template<class... Args>
    Sink(Args&&... args)
    {
    }
    ~Sink()
    {
    }       
};

Answer 1

据我所知，public继承是唯一可以执行隐式指针转换的情况（可以通过重载运算符实现引用转换）。

template <class T>
std::true_type is_public_base_of_impl(T*);

template <class T>
std::false_type is_public_base_of_impl(...);

template <class B, class D>
using is_public_base_of = decltype(is_public_base_of_impl<B>(std::declval<D*>()));

See it live on Coliru

Answer 2

感谢两个 Quentin and cpplearner for pointing me in the right direction. I found Quentins 如果断言应该通过，答案工作正常，但在失败的情况下 static_assert 不会捕获错误，而是在模板内生成，删除清晰 static_assert 消息的好处。

然后cpplearner提到了std::is_convertible，我以前尝试使用但忘记了需要*，而且B和D似乎是错误的。

所有这些让我创造了：

static_assert(std::is_convertible<Derived*, Base*>::value, "Derived must inherit Base as public");

这似乎可以完成工作，下面是作为完整示例的完整代码。

#include <type_traits>

class Base { };
class Derived : Base { };
class DerivedWithPublic : public Base { };

int main() {
    static_assert(std::is_convertible<DerivedWithPublic*, Base*>::value, "Class must inherit Base as public");
    static_assert(std::is_convertible<Derived*, Base*>::value, "Derived must inherit Base as public");
}

Public 继承的静态断言

Static Assert for Public Inheritance

c++

static-assert

typetraits

c++11