是否有可能有一个静态断言,提供为模板参数的类型是否实现了参数包中列出的所有类型,即。一个参数包知道std :: is_base_of()?
template <typename Type, typename... Requirements>
class CommonBase
{
static_assert(is_base_of<Requirements..., Type>::value, "Invalid.");
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
parameter pack aware version of std::is_base_of()
public:
template <typename T> T* as()
{
static_assert(std::is_base_of<Requirements..., T>::value, "Invalid.");
return reinterpret_cast<T*>(this);
}
};
答案 0 :(得分:16)
更新C ++ 17: 使用C ++ 17的折叠表达式,这几乎变得微不足道了:
template <typename Type, typename... Requirements>
class CommonBase
{
static_assert((std::is_base_of_v<Type, Requirements> && ...), "Invalid.");
};
原始答案(C ++ 11/14):
您可以使用包扩展和std::all_of
的某些静态版本:
template <bool... b> struct static_all_of;
//implementation: recurse, if the first argument is true
template <bool... tail>
struct static_all_of<true, tail...> : static_all_of<tail...> {};
//end recursion if first argument is false -
template <bool... tail>
struct static_all_of<false, tail...> : std::false_type {};
// - or if no more arguments
template <> struct static_all_of<> : std::true_type {};
template <typename Type, typename... Requirements>
class CommonBase
{
static_assert(static_all_of<std::is_base_of<Type, Requirements>::value...>::value, "Invalid.");
// pack expansion: ^^^
};
struct Base {};
struct Derived1 : Base {};
struct Derived2 : Base {};
struct NotDerived {};
int main()
{
CommonBase <Base, Derived1, Derived2> ok;
CommonBase <Base, Derived1, NotDerived, Derived2> error;
}
通过在Requirements...
中为问号添加std::is_base_of<Type, ?>::value
中的每个类型,包扩展将扩展为您获得的值列表,即对于main中的第一行,它将扩展为{{1对于第二行,它将是static_all_of<true, true>
答案 1 :(得分:3)
仅供将来参考,因为我刚遇到这个问题,使用C ++ 17,您现在可以使用这样的折叠表达式:
template<typename Base, typename... Args>
constexpr auto all_base_of()
{
return (std::is_base_of<Base, Args>::value && ...);
}
static_assert(all_base_of<Base, A, B, C>());