std::ranges::rotate_copy, std::ranges::rotate_copy_result
From cppreference.com
| Defined in header <algorithm>
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| Call signature |
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| template< std::forward_iterator I, std::sentinel_for<I> S, std::weakly_incrementable O > |
(1) | (since C++20) |
| template< ranges::forward_range R, std::weakly_incrementable O > requires std::indirectly_copyable<ranges::iterator_t<R>, O> |
(2) | (since C++20) |
| Helper types |
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| 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
| (C++20) |
rotates the order of elements in a range (niebloid) |
| (C++20)(C++20) |
copies a range of elements to a new location (niebloid) |
| copies and rotate a range of elements (function template) |