std::ranges::rotate_copy,std::ranges::rotate_copy_result(3)

NAME
std::ranges::rotate_copy,std::ranges::rotate_copy_result - std::ranges::rotate_copy,std::ranges::rotate_copy_result

Synopsis
   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> (1) (since C++20)
   constexpr rotate_copy_result<I, O>
       rotate_copy( I first, I middle, S last, O result );
   template< ranges::forward_range R, std::weakly_incrementable O >
   requires std::indirectly_copyable<ranges::iterator_t<R>, O> (2) (since C++20)
   constexpr rotate_copy_result<ranges::borrowed_iterator_t<R>, O>
       rotate_copy( R&& r, ranges::iterator_t<R> middle, O result );

Helper types
   template< class I, class O > (3) (since C++20)
   using rotate_copy_result = in_out_result<I, O>;
   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.
   The behavior is undefined if either [first, middle) or [middle, last) 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:
     * Explicit template argument lists cannot be specified when calling any of them.
     * None of them are visible to argument-dependent lookup.
     * When any of them are found by normal unqualified lookup as the name to the left
       of the function-call operator, argument-dependent lookup is inhibited.
   In practice, they may be implemented as function objects, or with special compiler
   extensions.

Parameters
   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).

Complexity
   Linear: exactly N assignments.

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.

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 {};

Example
// Run this code
 #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';
 }

Output:
 3 4 5 1 2
 1 2 3 4 5

See also
   ranges::rotate rotates the order of elements in a range
   (C++20) (niebloid)
   ranges::copy
   ranges::copy_if copies a range of elements to a new location
   (C++20) (niebloid)
   (C++20)
   rotate_copy copies and rotate a range of elements
                   (function template)