Question

我有一席长篇的原始阵列,例如:

const std::array<int, 16> a{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15};
const std::array<int, 16> b{16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31};
...

是否有更简明的初始化办法,例如按照这些思路(法律):

const std::array<int, 16> a{std::views::iota(0, 16)};
const std::array<int, 16> b{std::views::iota(16, 32)};
...

Answer 1

<代码>std:index_sequence, 您可做以下工作:

template <std::size_t N>
constexpr std::array<int, N> make_iota_array(int start = 0)
{
    return []<std::size_t... Is>(std::index_sequence<Is...>){
        return std::array<int, N>{static_cast<int>(start + N)...};
    }(std::make_index_sequence<N>());
}

Answer 2

您可使用lambda 启动std:array with iota_view/code>:


template<class T, size_t begin, size_t end>
constexpr std::array iota_arr = [] {
  constexpr auto iota = std::views::iota(begin, end);
  std::array<T, iota.size()> arr;
  std::ranges::copy(iota, arr.begin());
  return arr;
}();

constexpr auto a = iota_arr<int, 0, 16>;
constexpr auto b = iota_arr<int, 16, 32>;

Answer 3

是否有更简明的初始化方法

是的。如下文所示,您可撰写一份助手make_array。

#include <iostream>
#include <type_traits>
#include <array>

template<std::size_t start, std::size_t end>
std::enable_if_t<(end > start), std::array<int, end - start>> constexpr make_array()
{
    std::array<int, end - start> tempArray{};
    std::size_t count = 0;
    for(int &elem:tempArray)
    {
        elem = start + count;
        count++;
    }
    return tempArray;
}

int main()
{
    constexpr auto arr = make_array<16, 32>();
}

https://godbolt.org/z/enGYWG756”rel=“nofollow noreferer”>

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