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

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< cpp‎ | algorithm‎ | ranges

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

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