我需要能够从基本CRTP类中访问派生类的静态方法。有什么方法可以实现这一目标?
这里是示例代码:
#define REQUIRES(...) std::enable_if_t<(__VA_ARGS__), bool> = true
template<typename Derived>
struct ExpressionBase {
Derived& derived() { return static_cast<Derived&>(*this); }
const Derived& derived() const { return static_cast<const Derived&>(*this); }
constexpr static int size()
{
return Derived::size();
}
template<typename T, REQUIRES(size() == 1)>
operator T() const;
};
struct Derived : public ExpressionBase<Derived>
{
constexpr static int size()
{
return 1;
}
};
从ExpressionBase<Derived>
派生涉及ExpressionBase<Derived>
的实例化,因此涉及实体的声明
template<typename T, REQUIRES(size() == 1)>
operator T() const;
[这里,std::enable_if_t
的模板参数格式错误(因为Derived
尚未完成)。 SFINAE规则不适用于此处,因为格式错误的表达式不在模板参数类型的直接上下文中,因此将其视为硬错误。
为了使不正确的形式立即发生,请使用以下代码:
#include <type_traits>
template <bool B, class T>
struct lazy_enable_if_c {
typedef typename T::type type;
};
template <class T>
struct lazy_enable_if_c<false, T> {};
template <class Cond, class T>
struct lazy_enable_if : public lazy_enable_if_c<Cond::value, T> {};
template <class T>
struct type_wrapper {
using type = T;
};
#define REQUIRES(...) std::enable_if_t<(__VA_ARGS__), bool> = true
template<typename Derived>
struct ExpressionBase {
Derived& derived() { return static_cast<Derived&>(*this); }
const Derived& derived() const { return static_cast<const Derived&>(*this); }
struct MyCond {
static constexpr bool value = Derived::size() == 1;
};
template<typename T, typename = typename lazy_enable_if<MyCond, type_wrapper<T>>::type>
operator T () const {
return T{};
}
};
struct Derived : public ExpressionBase<Derived>
{
constexpr static int size() {
return 1;
}
};
int main() {
Derived d;
int i = d;
return 0;
}
实际上是根据boost
改编而成的,您可以在here中找到更多详细信息。