std::reverse_copy - cppreference.com

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# std::reverse_copy

From cppreference.com
< cpp‎ | algorithm

Defined in header `<algorithm>` |  |
template< class BidirIt, class OutputIt > OutputIt reverse_copy( BidirIt first, BidirIt 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 elements in the new range are in reverse order.

Behaves as if by executing the assignment *(d_first + (last - first) - 1 - i) = *(first + i) once for each non-negative `i < (last - first)`

If the source and destination ranges (that is, `[first, last)` and `[d_first, d_first+(last-first))` respectively) overlap, the behavior is undefined.

### [edit] Parameters

first, last | - | the range of elements to copy
d_first | - | the beginning of the destination range
Type requirements
- `BidirIt` must meet the requirements of `BidirectionalIterator`.
- `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 BidirIt, class OutputIt>
    OutputIt reverse_copy(BidirIt first, BidirIt last, OutputIt d_first)
    {
        while (first != last) {
            *(d_first++) = *(--last);
        }
        return d_first;
    }

### [edit] Example

Run this code

    #include <vector>
    #include <iostream>
    #include <algorithm>

    int main()
    {
        std::vector<int> v({1,2,3});
        for (const auto& value : v) {
            std::cout << value << " ";
        }
        std::cout << '\n';

        std::vector<int> destiny(3);
        std::reverse_copy(std::begin(v), std::end(v), std::begin(destiny));
        for (const auto& value : destiny) {
            std::cout << value << " ";
        }
        std::cout << '\n';
    }

Output:

    1 2 3
    3 2 1

### [edit] Complexity

Linear in the distance between `first` and `last`

### [edit] See also

reverse | reverses the order elements in a range (function template) [edit]

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