std:: 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 ​ .

If the program adds specializations for std::rank or std::rank_v (since C++17) , the behavior is undefined.

Helper variable template

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

#include <type_traits>
 
static_assert(std::rank<int>{} == 0);
static_assert(std::rank<int[5]>{} == 1);
static_assert(std::rank<int[5][5]>{} == 2);
static_assert(std::rank<int[][5][5]>{} == 3);
 
int main()
{
    [[maybe_unused]] int ary[][3] = {{1, 2, 3}};
 
    // The rank of reference type, e.g., ary[0], that is int(&)[3], is 0:
    static_assert(std::rank_v<decltype(ary[0])> == 0);
    static_assert(std::is_same_v<decltype(ary[0]), int(&)[3]>);
 
    // The solution is to remove the reference type.
    static_assert(std::rank_v<std::remove_cvref_t<decltype(ary[0])>> == 1);
}

See also