std::ranges::rotate_copy, std::ranges::rotate_copy_result

Defined in header <algorithm>
Call signature
template< std::forward_iterator I, std::sentinel_for<I> S,
          std::weakly_incrementable O >
requires std::indirectly_copyable<I, O>
constexpr rotate_copy_result<I, O>
    rotate_copy( I first, I middle, S last, O result );
(1) (since C++20)
template< ranges::forward_range R, std::weakly_incrementable O >
requires std::indirectly_copyable<ranges::iterator_t<R>, O>
constexpr rotate_copy_result<ranges::borrowed_iterator_t<R>, O>
    rotate_copy( R&& r, ranges::iterator_t<R> middle, O result );
(2) (since C++20)
Helper types
template< class I, class O >
using rotate_copy_result = in_out_result<I, O>;
(3) (since C++20)
1) Copies the elements from the source range [firstlast), to the destination range beginning at result in such a way, that the element *middle becomes the first element of the destination range and *(middle - 1) becomes the last element. The result is that the destination range contains a left rotated copy of the source range.
The behavior is undefined if either [firstmiddle) or [middlelast) is not a valid range, or the source and destination ranges overlap.
2) Same as (1), but uses r as the source range, as if using ranges::begin(r) as first and ranges::end(r) as last.

The function-like entities described on this page are niebloids, that is:

In practice, they may be implemented as function objects, or with special compiler extensions.


first, last - the source range of elements to copy from
r - the source range of elements to copy from
middle - the iterator to the element that should appear at the beginning of the destination range
result - beginning of the destination range

Return value

{last, result + N}, where N = ranges::distance(first, last).


Linear: exactly N assignments.


If the value type is TriviallyCopyable and the iterator types satisfy contiguous_iterator, implementations of ranges::rotate_copy usually avoid multiple assignments by using a "bulk copy" function such as std::memmove.

Possible implementation

See also the implementations in libstdc++ and MSVC STL.

struct rotate_copy_fn
    template<std::forward_iterator I, std::sentinel_for<I> S, std::weakly_incrementable O>
    requires std::indirectly_copyable<I, O>
    constexpr ranges::rotate_copy_result<I, O>
        operator()(I first, I middle, S last, O result) const
        auto c1 {ranges::copy(middle, std::move(last), std::move(result))};
        auto c2 {ranges::copy(std::move(first), std::move(middle), std::move(c1.out))};
        return {std::move(c1.in), std::move(c2.out)};
    template<ranges::forward_range R, std::weakly_incrementable O>
    requires std::indirectly_copyable<ranges::iterator_t<R>, O>
    constexpr ranges::rotate_copy_result<ranges::borrowed_iterator_t<R>, O>
        operator()(R&& r, ranges::iterator_t<R> middle, O result) const
        return (*this)(ranges::begin(r), std::move(middle),
                       ranges::end(r), std::move(result));
inline constexpr rotate_copy_fn rotate_copy {};


#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>
int main()
    std::vector<int> src {1, 2, 3, 4, 5};
    std::vector<int> dest(src.size());
    auto pivot = std::ranges::find(src, 3);
    std::ranges::rotate_copy(src, pivot, dest.begin());
    for (int i : dest)
        std::cout << i << ' ';
    std::cout << '\n';
    // copy the rotation result directly to the std::cout
    pivot = std::ranges::find(dest, 1);
    std::ranges::rotate_copy(dest, pivot, std::ostream_iterator<int>(std::cout, " "));
    std::cout << '\n';


3 4 5 1 2
1 2 3 4 5

See also

rotates the order of elements in a range
copies a range of elements to a new location
copies and rotate a range of elements
(function template)

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