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

Since C++20

Header: <algorithm>

  1. Copies the elements from the source range [first,last), 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.

# Declarations

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 );

(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 );

(since C++20)

Helper types

template< class I, class O >
using rotate_copy_result = in_out_result<I, O>;

(since C++20)

# Parameters

# Return value

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

# Notes

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.

# Example

#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';
}

# See also