Convert between C 20 NTTP and type-CodePudding

I'm playing around with C 20 NTTPs (non-type template parameters), and I was wondering, is there a way to convert between the elements of an NTTP std::array and types in the form of T<int>?

Consider this example:

template<int X>
struct Func;

template<auto ints>
struct Foo{

    // This doesn't work because within the std::apply the ints isn't a constant expression.
    constexpr auto funcs() const {
        return std::apply([](auto... xs){ return std::tuple{Func<xs>{}...};}, ints);
    }
};

constexpr Foo<std::array{1,2,3}> f;
constexpr auto funcs = f.funcs();

It easily works if the elements are values in a variadic NTTP pack, but having them in a std::array would allow to use the full power of the standard library algorithms rather than having to operate on parameter packs through TMP (template metaprogramming).

template<int... ints>
struct Foo2{

    // This here works, of course.
    constexpr auto funcs() const {
         return std::tuple{Func<ints>{}...};
    }
};

constexpr Foo2<1,2,3> f2;
constexpr auto funcs = f2.funcs();

Full code here.

CodePudding user response：

You can get a parameter pack with indices to the ints array as template parameter pack using the std::index_sequence method:

return []<std::size_t... I>(std::index_sequence<I...>){
    return std::tuple{Func<ints[I]>{}...};
}(std::make_index_sequence<std::size(ints)>{});

Then ints[I] can be used exactly as you intend xs to be used in your example.