Sum of array elements as constexpr

Question

I'm trying to get the sum of a const int array as a constexpr so that I can use sum as the size of another array

constexpr int arr[] = {1, 2, 3};
constexpr int sum1 = std::accumulate(arr, arr + 3, 0); // not OK
int arr1[sum1];

The above does not compile as std::accumulate() does not return a constexpr. I end up have a workaround like this

template <size_t N>
constexpr int sum(int const a[])
{
    return a[N-1] + sum<N - 1>(a);
}

template <>
constexpr int sum<0>(int const a[])
{
    return 0;
}

constexpr int arr[] = {1, 2, 3};
constexpr int sum1 = sum<3>(arr);
int arr1[sum1];

Is there any simpler solution?

Answer 1

+1 for the C++14 solutions, but I propose a C++11 solution based on a simple recursive constexpr function

template <typename T, std::size_t N>
constexpr T aSum (T const (&a)[N], std::size_t i = 0U)
 { return i < N ? (a[i] + aSum(a, i+1U)) : T{}; }

So it's possible to write

constexpr int arr[] {1, 2, 3};
constexpr int sum1 { aSum(arr) };

and sum1 become 6

Answer 2

Since C++14's relaxed constexpr you can do something like this:

#include <iostream>

constexpr int arr[] = {1, 2, 3};

template <size_t Size>
constexpr int sum(const int (&arr)[Size])
{
    int ret = 0;
    for (int i = 0; i < Size; ++i)
        ret += arr[i];
    return ret;
}

int main()
{
    int arr1[sum(arr)];
    std::cout << sizeof(arr1) / sizeof(int);
}