How to invoke a templated static class method having tuple input in a constexpr way

Question

How can a static constexpr class::method (int i1, int i2, int i3) be invoked, having input data available as tuple<int, int, int> in a constexpr way.

The default approach is using std::apply to apply each tuple element as argument to a function.

A minimal example to visualize, what I try to achieve looks like:

struct a {
    template <typename T>
    static constexpr void test(int i1, int i2, int i3) {
        // ...
    }
};

struct b : a {};
struct c {};

template <typename T>
struct test_functor {
    constexpr test_functor_t() {} // just for testing to express constexpr desire
    constexpr void operator()(auto... args) {
        T::test<c>(args...);
    }
};

constexpr std::tuple<int, int, int> tupl{ 1,2,3 };
constexpr test_functor<b> f;
std::apply(f, tupl);

this works at runtime, but fails to compile constexpr. How can this be implemented?

std::apply is constexpr. If the function you want to call is constexpr, what's the issue? — Nicol Bolas
– Nicol Bolas, Commented Apr 24, 2022 at 17:54
@NicolBolas the missing const declaration of operator() was the issue as ildjarn pointed out in 2. — Chris G.
– Chris G., Commented Apr 24, 2022 at 18:25

ildjarn · Accepted Answer · 2022-04-24 17:54:06Z

5

Working test_functor:

template <typename T>
struct test_functor {
    constexpr void operator()(auto... args) const {
        T::template test<c>(args...);
    }
};

The problems:

Your constructor was misnamed, and ultimately unnecessary – without a constructor your type is an aggregate and can be constexpr-constructed just fine.
Your operator() was not const – the primary problem, as you can't invoke a non-const member on a constexpr object.
Missing template keyword when invoking T::test – see this FAQ answer for a good explanation of dependent names.

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edited Apr 24, 2022 at 17:54

answered Apr 24, 2022 at 17:49

ildjarn

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2 Comments

Chris G. Over a year ago

Thanks for pointing out the missing template keyword too. To improve 3. further, it should be T::template test<c>(std::forward<decltype(args)>(args)...);.

ildjarn Over a year ago

@Chris : Indeed, but only if you also change args into a pack of forwarding references; otherwise that std::forward is really a misleading/confusing std::move. ;-]

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