NAME

std::rank - std::rank

Synopsis

Defined in header <type_traits>
template< class T > (since C++11)
struct rank;

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.

Helper variable template

template< class T > (since C++17)
inline constexpr std::size_t rank_v = rank<T>::value;

Inherited from std::integral_constant

Member constants

value the number of dimensions of T or zero
[static] (public static member constant)

Member functions

operator std::size_t converts the object to std::size_t, returns value
(public member function)
operator() returns value
(C++14) (public member function)

Member types

Type Definition
value_type std::size_t
type std::integral_constant<std::size_t, value>

Possible implementation

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> {};

Example

// Run this code

#include <iostream>
#include <type_traits>

int main()
{
std::cout << std::rank<int>{} << "\n\n";

std::cout << std::rank<int[5]>{} << '\n';
std::cout << std::rank<int[5][5]>{} << '\n';
std::cout << std::rank<int[][5][5]>{} << '\n';

[[maybe_unused]] int ary[][3]={ {1,2,3} };
// The reason of rank of "ary[0]" is calculated as 0
std::cout << std::rank<decltype(ary[0])>::value << '\n';
// 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
std::cout << std::rank<std::remove_cvref_t<decltype(ary[0])>>::value << '\n';
}

Output:

0

1
2
3
0
1