标准库是否有办法检查两个模板化类型的基本模板类型的相等性？

Question

Having trouble coming up with the correct terminology to search for this adequately, but does the standard library have something to test for the same base template type?

template <typename T>
struct foo {};

template <typename T>
struct bar {};

static_assert(std::is_same_base_type<foo<int>, foo<float>>::value == 1);
static_assert(std::is_same_base_type<foo<int>, bar<int>>::value == 0);

Answer 1

In the standard library?

No, as far I know.

But is trivial to write it.

template <typename, typename>
struct is_same_template : public std::false_type
 { };

template <template <typename> class C, typename T, typename U>
struct is_same_template<C<T>, C<U>> : public std::true_type
 { };

So you can write

static_assert( true == is_same_template<foo<int>, foo<float>>::value, "!" ) ;
static_assert( false == is_same_template<foo<int>, bar<int>>::value, "!" );

The problem of this solution is that the specialization works only for template-template bases receiving only one template type parameter.

You can improve it, for bases (template-template arguments) receiving a variadic list of arguments

template <template <typename...> class C,
          typename ... Ts, typename ... Us>
struct is_same_template<C<Ts...>, C<Us...>> : public std::true_type
 { };

but this doesn't works to check, by examples, std::array

static_assert( true == is_same_template<std::array<int, 3u>,
                                        std::array<float, 5u>>::value, "!" ) ;

For std::array you have to add another specialization

template <template <typename, std::size_t> class C,
          typename T1, std::size_t S1, typename T2, std::size_t S2>
struct is_same_template<C<T1, S1>, C<T2, S2>> : public std::true_type
 { };

Unfortunately there are innumerable possible template-template signatures so you have to add innumerable is_same_template specializations.

This is the reason (I suppose) there isn't a standard library type-traits.