How can I initialise a constexpr array with values using std::generate

Question

For example, if I wanted a constexpr std::array<int,100> initialised with all the multiples of 3 from 1-300 at compile time how can I do this?

My first thought was to use std::generate, something like:

constexpr std::array<int,100> a { std::generate(a.begin(), a.end(), [n=0]()mutable{ return n+=3; });

I get an error such as <source>:9:52: error: void value not ignored as it ought to be

and I can't use std::generate after this because of course, it's read only at that point

Thanks for any help

Does this answer your question? [Is it possible to initialize constexpr std::array member in a programtic way](https://stackoverflow.com/questions/50263306/is-it-possible-to-initialize-constexpr-stdarray-member-in-a-programtic-way) — gkhaos, Mar 10 '21 at 21:14

score 2 · Answer 1 · answered Mar 10 '21 at 21:12

2

You can use index_sequence:

template <size_t ... Indices>
constexpr auto gen(std::index_sequence<Indices...>) {
    return std::array<int,100>{ (Indices * 3)... };
}

int main() {
    constexpr std::array<int,100> a = gen(std::make_index_sequence<100>());
}

answered Mar 10 '21 at 21:12

rafix07

20,001
3
20
33

Wow. The power of fold expressions! – Enlico Mar 10 '21 at 21:24
Syntax for fold expression is a bit different. In above, it is normal unpacking parameters pack with multiplying by 3 each value of pack. – rafix07 Mar 10 '21 at 21:32
Oh, I thought "fold" referred to that too. In this case, _the power of the `...`_! – Enlico Mar 10 '21 at 21:39

score 0 · Answer 2 · answered Mar 10 '21 at 21:18

The trick is to put the code into an immediately-invoked lambda. Then it doesn't matter if you use std::generate or a plain loop:

constexpr std::array<int,100> a = []{
    std::array<int,100> ret{};
    std::generate(ret.begin(), ret.end(),  [n=0]() mutable {return n += 3;});
    return ret;
}();

constexpr std::array<int,100> a = []{
    constexpr std::array<int,100> ret{};
    for (std::size_t i = 0; i < ret.size(); i++)
        ret[i] = 3 * (1+i);
    return ret;
}();

How can I initialise a constexpr array with values using std::generate

2 Answers2