Create N-element constexpr array in C++11

Question

Hello i'm learning C++11, I'm wondering how to make a constexpr 0 to n array, for example:

n = 5;

int array[] = {0 ... n};

so array may be {0, 1, 2, 3, 4, 5}

Answer 1

#include <array>
#include <iostream>

template<int... N>
struct expand;

template<int... N>
struct expand<0, N...>
{
    constexpr static std::array<int, sizeof...(N) + 1> values = {{ 0, N... }};
};

template<int L, int... N> struct expand<L, N...> : expand<L-1, L, N...> {};

template<int... N>
constexpr std::array<int, sizeof...(N) + 1> expand<0, N...>::values;

int main()
{
    std::cout << expand<100>::values[9];
}

Answer 2

Based on @Xeo's excellent idea, here is an approach that lets you fill an array of

• constexpr std::array<T, N> a = { fun(0), fun(1), ..., fun(N-1) };
• where T is any literal type (not just int or other valid non-type template parameter types), but also double, or std::complex (from C++14 onward)
• where fun() is any constexpr function
• which is supported by std::make_integer_sequence from C++14 onward, but easily implemented today with both g++ and Clang (see Live Example at the end of the answer)
• I use @JonathanWakely 's implementation at GitHub (Boost License)

Here is the code

template<class Function, std::size_t... Indices>
constexpr auto make_array_helper(Function f, std::index_sequence<Indices...>) 
-> std::array<typename std::result_of<Function(std::size_t)>::type, sizeof...(Indices)> 
{
    return {{ f(Indices)... }};
}

template<int N, class Function>
constexpr auto make_array(Function f)
-> std::array<typename std::result_of<Function(std::size_t)>::type, N> 
{
    return make_array_helper(f, std::make_index_sequence<N>{});    
}

constexpr double fun(double x) { return x * x; }

int main() 
{
    constexpr auto N = 10;
    constexpr auto a = make_array<N>(fun);

    std::copy(std::begin(a), std::end(a), std::ostream_iterator<double>(std::cout, ", ")); 
}

Live Example

Answer 3

Try boost::mpl::range_c<int, 0, N> docs.

Answer 4

Use C++14 integral_sequence, or its invariant index_sequence

#include <iostream>

template< int ... I > struct index_sequence{ 

    using type = index_sequence;
    using value_type = int;

    static constexpr std::size_t size()noexcept{ return sizeof...(I); }
};

// making index_sequence
template< class I1, class I2> struct concat;

template< int ...I, int ...J> 
struct concat< index_sequence<I...>, index_sequence<J...> > 
        :  index_sequence< I ... , ( J + sizeof...(I) )... > {};

template< int N > struct make_index_sequence_impl;

template< int N > 
using make_index_sequence = typename make_index_sequence_impl<N>::type;

template< > struct make_index_sequence_impl<0> : index_sequence<>{};
template< > struct make_index_sequence_impl<1> : index_sequence<0>{};

template< int N > struct make_index_sequence_impl 
     : concat< make_index_sequence<N/2>, make_index_sequence<N - N/2> > {};



// now, we can build our structure.   
template < class IS > struct mystruct_base;

template< int ... I >
struct mystruct_base< index_sequence< I ... > >
{

   static constexpr int array[]{I ... };
};

template< int ... I >
constexpr int mystruct_base< index_sequence<I...> >::array[] ;

template< int N > struct mystruct 
   : mystruct_base< make_index_sequence<N > > 
{};

int main()
{
    mystruct<20> ms;

    //print
    for(auto e : ms.array)
    {
        std::cout << e << ' ';
    }
    std::cout << std::endl;

    return 0;
}

output: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

---

UPDATE: You may use std::array:

template< int ... I >
static constexpr std::array< int, sizeof...(I) >  build_array( index_sequence<I...> ) noexcept 
{ 
   return std::array<int, sizeof...(I) > { I... };
}

int main()
{
    std::array<int, 20> ma = build_array( make_index_sequence<20>{} );

    for(auto e : ma) std::cout << e << ' ';
    std::cout << std::endl;
}

Answer 5

In C++14 it can be easily done with a constexpr constructor and a loop:

#include <iostream>

template<int N>
struct A {
    constexpr A() : arr() {
        for (auto i = 0; i != N; ++i)
            arr[i] = i; 
    }
    int arr[N];
};

int main() {
    constexpr auto a = A<4>();
    for (auto x : a.arr)
        std::cout << x << '\n';
}

Answer 6

using boost preprocessor, it's very simple.

 #include <cstdio>
 #include <cstddef>

 #include <boost/preprocessor/repeat.hpp>
 #include <boost/preprocessor/comma_if.hpp>

 #define IDENTITY(z,n,dummy)   BOOST_PP_COMMA_IF(n) n

 #define INITIALIZER_n(n)   { BOOST_PP_REPEAT(n,IDENTITY,~)  }

 int main(int argc, char* argv[])
 {
     int array[] = INITIALIZER_n(25);

     for(std::size_t i = 0; i < sizeof(array)/sizeof(array[0]); ++i)
        printf("%d ",array[i]);

     return 0;
 }

OUTPUT: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24