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 — gkhaos
– gkhaos, Commented Mar 10, 2021 at 21:14

rafix07 · Accepted Answer · 2021-03-10 21:12:40Z

4

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, 2021 at 21:12

rafix07

21.3k3 gold badges26 silver badges36 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Enlico Over a year ago

Wow. The power of fold expressions!

rafix07 Over a year ago

Syntax for fold expression is a bit different. In above, it is normal unpacking parameters pack with multiplying by 3 each value of pack.

Enlico Over a year ago

Oh, I thought "fold" referred to that too. In this case, the power of the ...!

HolyBlackCat · Accepted Answer · 2024-03-29 03:21:54Z

0

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

edited Mar 29, 2024 at 3:21

answered Mar 10, 2021 at 21:18

HolyBlackCat

102k13 gold badges182 silver badges306 bronze badges

1 Comment

TimeVariant Over a year ago

The plain loop version is nice, but I had to drop the second "constexpr", and make the second line simply as follows: std::array<int,100> ret{}; (in order to compile)

Collectives™ on Stack Overflow

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

2 Answers 2

3 Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related