std::ranges::rotate_copy() algorithm
- od C++20
- Simplified
- Detailed
// (1)
constexpr rotate_copy_result<I, O>
rotate_copy( I first, I middle, S last, O result );
// (2)
constexpr rotate_copy_result<ranges::borrowed_iterator_t<R>, O>
rotate_copy( R&& r, ranges::iterator_t<R> middle, O result );
The type of arguments are generic and have the following constraints:
I
-std::forward_iterator
S
-std::sentinel_for<I>
O
-std::weakly_incrementable
R
-std::forward_range
Additionally, each overload has the following constraints:
- (1) -
std::indirectly_copyable<I, O>
- (2) -
std::indirectly_copyable<ranges::iterator_t<R>, O>
// (1)
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 );
// (2)
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 );
With the helper types defined as follows:
template< class I, class O >
using rotate_copy_result = in_out_result<I, O>;
-
(1) Copies the elements from the source range [
first
;last
), to the destination range beginning atresult
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.
-
(2) Same as (1), but uses
r
as the source range, as if usingranges::begin(r)
asfirst
andranges::end(r)
aslast
.
The behavior is undefined
if either [first
; middle
) or [middle
; last
) is not a valid range, or the source and destination ranges overlap.The function-like entities described on this page are niebloids.
Parameters
first last | The range of elements to copy from. |
r | The range of elements to copy from. |
result | The beginning of the destination range. |
middle | The iterator to the element that should appear at the beginning of the destination range. |
Return value
{
last,
result + N
}
Where N is ranges::distance(first, last)
.
Complexity
Exactly ranges::distance(first, last)
assignments.
Exceptions
(none)
Possible implementation
rotate_copy(1) and rotate_copy(2)
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 {};
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
.
Examples
#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';
}
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