在编译时将两个或多个不同大小的数组合并到一个数组中

Question

I was not able to find an answer on how to combine two or more arrays at compiletime in modern c++.

#include <array>
#include <cstdint>

const std::array<std::uint8_t, 1> one_elem = {1};
const std::array<std::uint8_t, 2> two_elem = {2, 3};
const std::array<std::uint8_t, 3> all = {one_elem, two_elem}; 
// expected: all == {1, 2, 3}

I would be glad with anything that is somewhat easy to read, eg

std::uint8_t one_elem[] = {1};
std::uint8_t two_elem[] = {2, 3};
std::uint8_t all[] = {one_elem, two_elem}; // cannot be that hard

Is there a way? What can I do to get this solved?

Answer 1

If you are using C++17, you can do this:

template <typename T, std::size_t N1, std::size_t N2>
constexpr std::array<T, N1 + N2> concat(std::array<T, N1> lhs, std::array<T, N2> rhs)
{
    std::array<T, N1 + N2> result{};
    std::size_t index = 0;

    for (auto& el : lhs) {
        result[index] = std::move(el);
        ++index;
    }
    for (auto& el : rhs) {
        result[index] = std::move(el);
        ++index;
    }

    return result;
}

constexpr std::array<std::uint8_t, 1> one_elem = {1};
constexpr std::array<std::uint8_t, 2> two_elem = {2, 3};
constexpr std::array<std::uint8_t, 3> all = concat(one_elem, two_elem);

It doesn't work in C++14, because std::array isn't constexpr friendly until C++17. However, if you don't care that the final result is constexpr , you can simply mark each variable as const , and this will work:

const std::array<std::uint8_t, 1> one_elem = {1};
const std::array<std::uint8_t, 2> two_elem = {2, 3};
const std::array<std::uint8_t, 3> all = concat(one_elem, two_elem);

The compiler will almost certainly optimize the concat away.

If you need a C++14 solution, we have to create it via std::array 's constructor, so it's not nearly as nice:

#include <array>
#include <cstdint>
#include <cstddef>
#include <type_traits>

// We need to have two parameter packs in order to
// unpack both arrays. The easiest way I could think of for
// doing so is by using a parameter pack on a template class
template <std::size_t... I1s>
struct ConcatHelper
{
    template <typename T, std::size_t... I2s>
    static constexpr std::array<T, sizeof...(I1s) + sizeof...(I2s)>
    concat(std::array<T, sizeof...(I1s)> const& lhs,
           std::array<T, sizeof...(I2s)> const& rhs,
           std::index_sequence<I2s...>)
    {
        return { lhs[I1s]... , rhs[I2s]... };
    }
};

// Makes it easier to get the correct ConcatHelper if we know a
// std::index_sequence. There is no implementation for this function,
// since we are only getting its type via decltype()
template <std::size_t... I1s>
ConcatHelper<I1s...> get_helper_type(std::index_sequence<I1s...>);

template <typename T, std::size_t N1, std::size_t N2>
constexpr std::array<T, N1 + N2> concat(std::array<T, N1> const& lhs, std::array<T, N2> const& rhs)
{
    return decltype(get_helper_type(std::make_index_sequence<N1>{}))::concat(lhs, rhs, std::make_index_sequence<N2>{});
}

constexpr std::array<std::uint8_t, 1> one_elem = {1};
constexpr std::array<std::uint8_t, 2> two_elem = {2, 3};
constexpr std::array<std::uint8_t, 3> all = concat(one_elem, two_elem);

Answer 2

There's already a way to concatenate arrays in C++: std::tuple_cat . The only problem is that it gives you a tuple<uint8_t, uint8_t, uint8_t> instead of a std::array<uint8_t, 3> . But that problem is solvable with a different standard library function: std::apply . That one is technically C++17, but is implementable in C++14. You just need a funject:

struct to_array_t {
    template <class T, class... Ts> 
    std::array<std::decay_t<T>, sizeof...(Ts)+1> operator()(T&& t, Ts&&... ts) const {
        return {{std::forward<T>(t), std::forward<Ts>(ts)...}};
    }   
} to_array{};

and then you can use it:

auto all = std::apply(to_array, std::tuple_cat(one_elem, two_elem));

Which might be easier to just hide behind a function:

template <class Target=void, class... TupleLike>
auto array_concat(TupleLike&&... tuples) {
    return std::apply([](auto&& first, auto&&... rest){
        using T = std::conditional_t<
            !std::is_void<Target>::value, Target, std::decay_t<decltype(first)>>;
        return std::array<T, sizeof...(rest)+1>{{
            decltype(first)(first), decltype(rest)(rest)...
        }};
    }, std::tuple_cat(std::forward<TupleLike>(tuples)...));
}

Perfect forwarding with lambdas is a bit ugly. The Target type is allow the user to specify a type for the resulting array - otherwise it will be selected as the decayed type of the first element.