如何在编译时测试B类是否是从std::vector派生的?
How to test at compile time whether class B is derived from std::vector?
template<class A>
struct is_derived_from_vector {
static const bool value = ????;
};
如何在编译时测试B类是否派生自模板族?
How to test at compile time whether class B is derived from template family?
template<class A, template< class > class Family>
struct is_derived_from_template {
static const bool value = ????;
};
使用:
template<class T> struct X {};
struct A : X<int> {}
struct B : std::vector<char> {}
struct D : X<D> {}
int main() {
std::cout << is_derived_from_template<A, X>::value << std::endl; // true
std::cout << is_derived_from_template<D, X>::value << std::endl; // true
std::cout << is_derived_from_vector<A>::value << std::endl; // false
std::cout << is_derived_from_vector<B>::value << std::endl; // true
}
试试这个:
#include <type_traits>
template <typename T, template <typename> class Tmpl> // #1 see note
struct is_derived
{
typedef char yes[1];
typedef char no[2];
static no & test(...);
template <typename U>
static yes & test(Tmpl<U> const &);
static bool const value = sizeof(test(std::declval<T>())) == sizeof(yes);
};
用法:
#include <iostream>
template<class T> struct X {};
struct A : X<int> {};
int main()
{
std::cout << is_derived<A, X>::value << std::endl;
std::cout << is_derived<int, X>::value << std::endl;
}
注意:在标记为 #1
的行中,您还可以让您的 trait 接受任何 模板,该模板至少有一个,但可能writint 的更多类型参数:
Note: In the line marked #1
, you could also make your trait accept any template that has at least one, but possibly more type arguments by writint:
template <typename, typename...> class Tmpl
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