Filling an std::array using math formula in compile time

Question

I want to fill a constexpr std::array in compile time using a math function. Is it possible in an easy way?

I found this solution: C++11: Compile Time Calculation of Array. However, is there any other modern solution using only std? This one seems too confusing for me.

int main()
{
   // This is the array I need to fill
   constexpr std::array<double, 100000> elements;
   for (int i=0; i!=100000; ++i)
   {
      // Each element is calculated using its position with long exponential maths.
      elements[i] = complexFormula(i); // complexFormula is constexpr
   }


   double anyVal = elements[43621];
   // ...
}

Yes, it is a constexpr – Juan JuezSarmiento Aug 30 '19 at 06:18 — Juan JuezSarmiento, Aug 30 '19 at 06:18
Also, FWIW, `int` may not be enough to hold `100000`. – L. F. Aug 30 '19 at 06:19 — L. F., Aug 30 '19 at 06:19
@L.F. a compiler with a 16-bit `int` is pretty rare – Alan Birtles Aug 30 '19 at 06:41 — Alan Birtles, Aug 30 '19 at 06:41
@AlanBirtles Rare != nonexistent – L. F. Aug 30 '19 at 06:43 — L. F., Aug 30 '19 at 06:43

score 6 · Accepted Answer · answered Aug 30 '19 at 06:22

Here's a non-confusing approach: wrap the calculation in a function:

template <int N>
constexpr std::array<double, N> generate()
{
    std::array<double, N> arr{};
    for (int i = 0; i < N; ++i)
        arr[i] = complexFormula(i);
    return arr;
}

Usage example:

constexpr std::array<double, 10000> arr = generate<10000>();

(live demo)

This works because, since C++14, loops are allowed in a constexpr function, and variables can be modified as long as their lifetime starts within the evaluation of the constant expression.

Filling an std::array using math formula in compile time

1 Answers1