Getting the number of elements in std::array at compile time

Question

Is the following a valid C++ code, and why not?

std::array a1;
std::array a2;

It doesn\'t compile

Answer 1

array<T>::size() is constexpr, but you can't use it in this way because a1 isn't a constexpr value. Additionally, it can't be constexpr because string isn't a literal type.

However, you can work around this if you want, by deducing the size_t template parameter. Example:

#include <string>
#include <array>
#include <iostream>
using namespace std;

template<typename>
struct array_size;
template<typename T, size_t N>
struct array_size<array<T,N> > {
    static size_t const size = N;
};

array<string, 42> a1;
array<string, array_size<decltype(a1)>::size> a2;

int main() {
    cout << a2.size() << endl;
}

Answer 2

It is not static but it is a constexpr http://www.cplusplus.com/reference/array/array/size/
EDIT: this may not be a bug, take a look at this Error using a constexpr as a template parameter within the same class

Answer 3

std::array::size is actually required to be constexpr per § 23.3.2.1 of the C++11 standard:

23.3.2.4 array::size [array.size]  
template <class T, size_t N> constexpr size_type array<T,N>::size() noexcept;  
Returns: N

I'm guessing this just slipped past whoever implemented it in GCC.

---

After testing, this works:

std::array<int, 42> a1;
std::array<int, a1.size()> a2;

This may actually have something to do with std::string not being a valid constexpr type to make compile-time instances of, whereas int is.

Answer 4

You can use the same template-inference method as has always been used for C++98 array bound detection.

template<size_t N, typename T>
constant_integer<N> array_size( const std::array<T, N>& );

• Demo: http://ideone.com/R4k1vG

Make a nice macro wrapper and enjoy!

Many variations are also possible, such as:

• http://ideone.com/Hn46ei