Passing a constexpr function to use at compile time

Question

I'm trying to make a simple way to generate std::array's at compile time. It's been a struggle so far to figure out a good way to pass a constexpr function in at compile time. The workaround I have found so far is like this.

#include <iostream> #include <array> namespace a { constexpr int f(const int & i) { return i * i * i;} #include "generate_with_function.hpp" } int main() { auto arr = a::generator<false,10,0>::array; for (auto i : arr) { std::cout << i << " "; } return 0; }

this basically assumes you will define a function called f and I wrapped it in a namespace incase I wanted to do a different one. I wanted to know if there is a more clever way to pass in the function and use it at compile time. Also here is the code that makes the list.

template <bool B, size_t Count,int ... Nums> struct generator; template <size_t Count> struct generator<false,Count,0> { constexpr static std::array<int,Count> array = generator<false,Count,1,f(0)>::array; }; template <size_t Count, int Current, int ... Results> struct generator<false,Count,Current, Results...> { constexpr static std::array<int,Count> array = generator<Current+1==Count,Count,Current+1,f(Current), Results...>::array; }; template <size_t Count, int Current, int ... Results> struct generator<true,Count,Current,Results...> { constexpr static std::array<int,Count> array{{Results...}}; };

and before you ask no I don't actually have a real need for this.

As noted by @us2012 I should be specific about what I would rather have.

nowrapping in namespace
having to write the function but not actually passing it anywhere
and not requiring the function to be named f

When you say "a more clever way", which particular property of your current solution are you trying to avoid? — us2012
– us2012, Commented Oct 10, 2013 at 1:05
@us2012 couple things, wrapping in namespace, having to write the function but not actually passing it anywhere and requiring the function to be named f — aaronman
– aaronman, Commented Oct 10, 2013 at 1:06
@sth if you want to reuse it with a different function it is — aaronman
– aaronman, Commented Oct 10, 2013 at 1:15
Btw, here's my solution to that problem. It does pass the function (as a pointer), though. — dyp
– dyp, Commented Oct 10, 2013 at 6:16

sth · Accepted Answer · 2013-10-10 01:40:55Z

You can actually use the function as a template parameter, here called Gen:

template <bool B, size_t Count, int Current, int Gen(size_t), int ... Nums> struct generator; template <size_t Count, int Current, int Gen(size_t), int ... Results> struct generator<false,Count,Current, Gen, Results...> { constexpr static std::array<int,Count> array generator<Current+1==Count,Count,Current+1,Gen, Gen(Current), Results...>::array; }; template <size_t Count, int Current, int Gen(size_t), int ... Results> struct generator<true,Count,Current,Gen,Results...> { constexpr static std::array<int,Count> array{{Results...}}; };

With this it actually works to pass a constexpr function to the template (if the type matches exactly):

// helper to hide the "internal" template parameters template <size_t Count, int Gen(size_t)> struct gen { constexpr static std::array<int, Count> array = generator<false, Count, 0, Gen>::array; }; constexpr int f(size_t i) { return i * i * i; } int main() { auto arr = gen<10,f>::array; for (auto i : arr) { std::cout << i << " "; } return 0; }

You can also switch around the Gen(Current) and Results... parameters to get the values at the "correct" array indexes.

This looks like it might work, the function takes an int though
@aaronman: Like written here it only works if the function takes size_t, but it could probably be generalized by adding even another template parameter for the parameter type of the function, or so.
Thanks, my problem from before is that I thought a function was not allowed to be a non-type template param
@aaronman A function type is not allowed as type of a non-type template parameter, but a function type decays to a pointer-to-function type (even) in template parameters. And pointer-to-function (as well as reference to function) is allowed as type for a non-type template parameter. Additionally, integral, enumeration, lvalue reference, pointer-to-object, pointer-to-member and nullptr_t are allowed as types [temp.param]/4.

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