Defined in header <type_traits> | ||
---|---|---|
template< class T > struct rank; | (since C++11) |
If T
is an array type, provides the member constant value
equal to the number of dimensions of the array. For any other type, value
is 0
.
The behavior of a program that adds specializations for rank
or rank_v
(since C++17) is undefined.
template< class T > inline constexpr std::size_t rank_v = rank<T>::value; | (since C++17) |
value
[static] | the number of dimensions of T or zero (public static member constant) |
operator std::size_t | converts the object to std::size_t, returns value (public member function) |
operator()
(C++14) | returns value (public member function) |
Type | Definition |
---|---|
value_type | std::size_t |
type | std::integral_constant<std::size_t, value> |
template<class T> struct rank : public std::integral_constant<std::size_t, 0> {}; template<class T> struct rank<T[]> : public std::integral_constant<std::size_t, rank<T>::value + 1> {}; template<class T, std::size_t N> struct rank<T[N]> : public std::integral_constant<std::size_t, rank<T>::value + 1> {}; |
#include <type_traits> int main() { static_assert( std::rank<int>{} == 0 && std::rank<int[5]>{} == 1 && std::rank<int[5][5]>{} == 2 && std::rank<int[][5][5]>{} == 3 ); [[maybe_unused]] int ary[][3] = {{1, 2, 3}}; // The reason of rank of "ary[0]" is calculated as 0 static_assert(std::rank_v<decltype(ary[0])> == 0); // is that rank cannot deal with reference type. i.e. int(&)[3] static_assert(std::is_same_v<decltype(ary[0]), int(&)[3]>); // The solution is to remove reference type static_assert(std::rank_v<std::remove_cvref_t<decltype(ary[0])>> == 1); }
(C++11) | checks if a type is an array type (class template) |
(C++11) | obtains the size of an array type along a specified dimension (class template) |
(C++11) | removes one extent from the given array type (class template) |
(C++11) | removes all extents from the given array type (class template) |
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