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Initialize a constexpr array as sum of other two constexpr arrays

Given two constexpr arrays (type[N] or std::array<type, N>)

constexpr int A[5] { 0, 1, 2, 3, 4 };
constexpr int B[5] { 5, 4, 3, 2, 1 };

is it possible to initialize a new constexpr array performing an element-wise operation (or constexpr function)?

For example, can this code

constexpr int sum(int i) { return A[i] + B[i]; }
constexpr int S[5] { sum(0), sum(1), sum(2), sum(3), sum(4) };

be rewritten more conveniently in a form that calls sum(i) for every element in S?

like image 857
AkiRoss Avatar asked Nov 12 '13 18:11

AkiRoss


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1 Answers

Similar things have been done many times already, but here's a solution for this particular compile-time operation on arrays ;)

template<int... Is>
struct seq {};
template<int I, int... Is>
struct gen_seq : gen_seq<I-1, I-1, Is...> {};
template<int... Is>
struct gen_seq<0, Is...> : seq<Is...> {};

#include <array>

template<class T, int N, int... Is>
constexpr std::array<T, N> sum(T const (&lhs)[N], T const (&rhs)[N], seq<Is...>)
{
    return {{lhs[Is]+rhs[Is]...}};
}

template<class T, int N>
constexpr auto sum(T const (&lhs)[N], T const (&rhs)[N])
-> decltype( sum(lhs, rhs, gen_seq<N>{}) )
{
    return sum(lhs, rhs, gen_seq<N>{});
}

#include <iostream>
int main()
{
    constexpr int a[] = {1,2,3,4,5};
    constexpr int b[] = {1,2,3,4,5};
    constexpr auto c = sum(a,b);

    for(auto e : c) std::cout << e << ", ";
}

N.B. std::array::operator[] is not constexpr in C++11, therefore I've used raw arrays as input.


For arbitrary binary functions:

template<class T, int N, class F, int... Is>
constexpr auto transform(T const (&lhs)[N], T const (&rhs)[N], F f,
                         seq<Is...>)
-> std::array<decltype( f(lhs[0], rhs[0]) ), N>
{
    return {{ f(lhs[Is], rhs[Is])... }};
}

template<class T, int N, class F>
constexpr auto transform(T const (&lhs)[N], T const (&rhs)[N], F f)
-> decltype( transform(lhs, rhs, f, gen_seq<N>{}) )
{
    return transform(lhs, rhs, f, gen_seq<N>{});
}

constexpr int sum(int l, int r) { return l+r; }
// ...
constexpr auto c = transform(a,b,sum);

For arbitrary n-ary functions and arbitrary array-like types:

template<class F, class... Args>
constexpr auto index_invoke(F f, int i, Args&&... args)
-> decltype( f(args[i]...) )
{
    return f(args[i]...);
}

template<class F, int... Is, class... Args>
constexpr auto transform_impl(F f, seq<Is...>, Args&&... args)
-> std::array<decltype( f(args[0]...) ), sizeof...(Is)>
{
    return {{ index_invoke(f, Is, std::forward<Args>(args)...)... }};
}

template <class T, class...>
struct get_extent_helper
: std::integral_constant<int,
                         std::extent<typename std::remove_reference<T>::type>{}>
{};

template<class F, class... Args>
constexpr auto transform(F f, Args&&... args)
-> decltype( transform_impl(f, gen_seq< get_extent_helper<Args...>{} >{},
                            std::forward<Args>(args)...) )
{
    using N = get_extent_helper<Args...>;
    return transform_impl(f, gen_seq<N{}>{}, std::forward<Args>(args)...);
}

More lightweight with an alias-template:

template <class T, class...>
using get_extent_helper =
  std::integral_constant<int,
                         std::extent<typename std::remove_reference<T>::type>{}>;

But this is buggy under g++4.8.1

like image 97
dyp Avatar answered Nov 15 '22 07:11

dyp