Sum of array elements as constexpr

Question

I am trying to get the sum of an array const intas constexprso that I can use the 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 since it std::accumulate()does not return constexpr. In the end, I have a workaround

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 a simpler solution?

Answer 1

Since C ++ 14 relaxes 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);
}

Answer 2

Using C ++ 14:

template<typename T, std::size_t N>
constexpr T array_sum(T (&array)[N]) {
    T sum = 0;
    for (std::size_t i = 0; i < N; i++) {
        sum += array[i];
    }
    return sum;
};

And use it like:

int arr[] = {1, 2, 3};
int brr[array_sum(arr)];

Answer 3

+1 ++ 14, ++ 11 constexpr

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{}; }

,

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

sum1 6

Answer 4

//With C++14
#include <iostream>
using namespace std;

template <typename T, std::size_t N>
constexpr T arraysum(const T (&array)[N])  {
    T sum = 0;
    for (size_t i = 0; i < N; ++i) {
        sum += array[i];
    }
    return sum;
}

template <typename T, std::size_t N>
constexpr T arraysize(T (&)[N])  {
    return N;
}

int main() {
    // your code goes here
    constexpr int arr[] = {1, 2, 3};
    constexpr int size = arraysize(arr);
    cout<<size<<endl;
    constexpr int sum1 = arraysum(arr);
    cout <<sum1;
    int arr2[sum1];
    return 0;
}

Check the code here

Answer 5

there is a little compiled code, but it worked with C ++ 11 and only deep recursion O (log (N)) is required.

template<typename Iterator >
constexpr size_t c_dist(Iterator first, Iterator last){ return (last - first); }

template< typename Iterator, typename U>
constexpr U c_accumulate(Iterator first, Iterator last, U u)
{
    return (first == last) 
            ? u 
            : c_dist(first , last) == 1 
                 ? ( u + *first ) 
                 :  
                    c_accumulate(first, first + c_dist(first, last ) / 2, u ) +  
                    c_accumulate(first + c_dist(first, last)/2, last, u);

}

constexpr int a[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21};
constexpr int sum = c_accumulate(a, a + sizeof(a)/sizeof(a[0]), 0);

static_assert(sum == 21*22/2, "!");