std::rotate_copy

From cppreference.com
< cpp‎ | algorithm
 
 
 
Defined in header <algorithm>
template< class ForwardIt, class OutputIt >

OutputIt rotate_copy( ForwardIt first, ForwardIt n_first,

                      ForwardIt last, OutputIt d_first );

Copies the elements from the range [first, last), to another range beginning at d_first in such a way, that the element n_first becomes the first element of the new range and n_first - 1 becomes the last element.

Contents

[edit] Parameters

first, last - the range of elements to copy
n_first - an iterator to an element in [first, last) that should appear at the beginning of the new range
d_first - beginning of the destination range
Type requirements
-
ForwardIt must meet the requirements of ForwardIterator.
-
OutputIt must meet the requirements of OutputIterator.

[edit] Return value

Output iterator to the element past the last element copied.

[edit] Possible implementation

template<class ForwardIt, class OutputIt>
OutputIt rotate_copy(ForwardIt first, ForwardIt n_first,
                           ForwardIt last, OutputIt d_first)
{
    d_first = std::copy(n_first, last, d_first);
    return std::copy(first, n_first, d_first);
}

[edit] Example

#include <algorithm>
#include <vector>
#include <iostream>
 
int main()
{
    std::vector<int> src = {1, 2, 3, 4, 5}; 
    auto pivot = std::find(src.begin(), src.end(), 3); 
    std::vector<int> dest(src.size());                                          
 
    std::rotate_copy(src.begin(), pivot, src.end(), dest.begin());
 
    for (const auto &i : dest) {
        std::cout << i << ' ';
    }   
    std::cout << '\n';
}

Output:

3 4 5 1 2

[edit] Complexity

linear in the distance between first and last

[edit] See also

rotates the order of elements in a range
(function template)