shift_right（）打算如何在 C++20 中实现？-解网

问：

在 C++20 中，标头获得了两个新算法：shift_left（） 和 shift_right（）。它们都接受任何 LegacyForwardIterator。对于，指定“移动按从 ”开始的递增顺序执行“;对于，指定“如果满足 LegacyBidirectionalIterator 要求，则移动将按从开始的降序执行。<algorithm>shift_left()i0shift_right()ForwardItilast - first - n - 1

我能想到一个相当简单的方法来实现：shift_left()

template <typename ForwardIt>
constexpr inline ForwardIt shift_left(ForwardIt first, ForwardIt last, typename std::iterator_traits<ForwardIt>::difference_type n) {
    if (n <= 0) return last;
    ForwardIt it = first;
    for (; n > 0; --n, ++it) {
        if (it == last) return first;
    }
    return std::move(it, last, first);
}

如果满足 LegacyBidirectionalIterator 的要求，我可以看到它可以以与 .然而，目前还不清楚如何实现非双向前向迭代器。ForwardItshift_right()shift_left()shift_right()

我想出了一种算法，它使用空间作为暂存空间来交换元素，但它似乎比上面的算法更浪费：[first, first+n)shift_left()

template <typename ForwardIt>
constexpr inline ForwardIt shift_right(ForwardIt first, ForwardIt last, typename std::iterator_traits<ForwardIt>::difference_type n) {
    if (n <= 0) return first;
    ForwardIt it = first;
    for (; n > 0; --n, ++it) {
        if (it == last) return last;
    }
    ForwardIt ret = it;
    ForwardIt ret_it = first;
    for (; it != last; ++it) {
        std::iter_swap(ret_it, it);
        ret_it++;
        if (ret_it == ret) ret_it = first;
    }
    return ret;
}

会有更好或“预期”的实施方式吗？shift_right()

C++ 算法 +-标准库 C +20

#include <algorithm>
#include <iterator>
#include <type_traits>
#include <utility>

template<class I>
using difference_type_t = typename std::iterator_traits<I>::difference_type;

template<class I>
using iterator_category_t = typename std::iterator_traits<I>::iterator_category;

template<class I, class Tag, class = void>
constexpr bool is_category = false;
template<class I, class Tag>
constexpr bool is_category<I, Tag, std::enable_if_t<
    std::is_convertible_v<iterator_category_t<I>, Tag>>> = true;

/// Increment (decrement for negative n) i |n| times or until i == bound,
/// whichever comes first. Returns n - the difference between i's final position
/// and its initial position. (Note: "advance" has overloads with this behavior
/// in the Ranges TS.)
template<class I>
constexpr difference_type_t<I> bounded_advance(
    I& i, difference_type_t<I> n, I const bound)
{
    if constexpr (is_category<I, std::bidirectional_iterator_tag>) {
        for (; n < 0 && i != bound; ++n, void(--i)) {
            ;
        }
    }

    for(; n > 0 && i != bound; --n, void(++i)) {
        ;
    }

    return n;
}

template<class ForwardIt>
ForwardIt shift_left(ForwardIt first, ForwardIt last, difference_type_t<ForwardIt> n)
{
    if (n <= 0) {
        return last;
    }

    auto mid = first;
    if (::bounded_advance(mid, n, last)) {
        return first;
    }

    return std::move(std::move(mid), std::move(last), std::move(first));
}

template<class ForwardIt>
ForwardIt shift_right(ForwardIt first, ForwardIt last, difference_type_t<ForwardIt> n)
{
    if (n <= 0) {
        return first;
    }

    if constexpr (is_category<ForwardIt, std::bidirectional_iterator_tag>) {
        auto mid = last;
        if (::bounded_advance(mid, -n, first)) {
            return last;
        }
        return std::move_backward(std::move(first), std::move(mid), std::move(last));
    } else {
        auto result = first;
        if (::bounded_advance(result, n, last)) {
            return last;
        }

        // Invariant: next(first, n) == result
        // Invariant: next(trail, n) == lead

        auto lead = result;
        auto trail = first;

        for (; trail != result; ++lead, void(++trail)) {
            if (lead == last) {
                // The range looks like:
                //
                //   |-- (n - k) elements --|-- k elements --|-- (n - k) elements --|
                //   ^-first          trail-^                ^-result          last-^
                //
                // Note that distance(first, trail) == distance(result, last)
                std::move(std::move(first), std::move(trail), std::move(result));
                return result;
            }
        }

        for (;;) {
            for (auto mid = first; mid != result; ++lead, void(++trail), ++mid) {
                if (lead == last) {
                    // The range looks like:
                    //
                    //   |-- (n - k) elements --|-- k elements --|-- ... --|-- n elements --|
                    //   ^-first            mid-^         result-^         ^-trail     last-^
                    //
                    trail = std::move(mid, result, std::move(trail));
                    std::move(std::move(first), std::move(mid), std::move(trail));
                    return result;
                }
                std::iter_swap(mid, trail);
            }
        }
    }
}

shift_right（）打算如何在 C++20 中实现？

How is shift_right() intended to be implemented in C++20?

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shift_right（） 打算如何在 C++20 中实现？

How is shift_right() intended to be implemented in C++20?

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shift_right（）打算如何在 C++20 中实现？