std::ranges::reverse_copy() algorithm

od C++20

Simplified

// (1)
constexpr reverse_copy_result<I, O>
    reverse_copy( I first, S last, O result );

// (2)
constexpr reverse_copy_result<ranges::borrowed_iterator_t<R>, O>
    reverse_copy( R&& r, O result );

The type of arguments are generic and have following constraints:

• I - std::bidirectional_iterator
• S - std::sentinel_for<I>
• R - ranges::bidirectional_range
• O - std::weakly_incrementable

Additionally, each overload has the following constraints:

• (1) - std::indirectly_copyable<I, O>
• (2) - std::indirectly_copyable<ranges::iterator_t<R>, O>

Detailed

// (1)
template<
  std::bidirectional_iterator I,
  std::sentinel_for<I> S,
  std::weakly_incrementable O
>
  requires std::indirectly_copyable<I, O>
constexpr reverse_copy_result<I, O>
    reverse_copy( I first, S last, O result );

// (2)
template<
  ranges::bidirectional_range R,
  std::weakly_incrementable O
>
  requires std::indirectly_copyable<ranges::iterator_t<R>, O>
constexpr reverse_copy_result<ranges::borrowed_iterator_t<R>, O>
    reverse_copy( R&& r, O result );

With the helper types defined as follows:

template< class I, class O >
using reverse_copy_result = ranges::in_out_result<I, O>;

• (1) Given N as ranges::distance(first, last):

  Copies the elements from the source range [first; last) to the destination range [result; result + N), in such a way that the elements in the new range are in reverse order.

  Behaves as if by executing the assignment *(result + N - 1 - i) = *(first + i) once for each integer i in [0; N).

• (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.

Undefined Behaviour

The behavior is undefined

if 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.
r	The range of elements to copy.
result	The beginning of the destination range.

Return value

A value of type ranges::reverse_copy_result initialized as follows:

{
  last,
  result + N
}

Where N is ranges::distance(first, last).

Complexity

Exactly ranges::distance(first, last) assignments.

Exceptions

(none)

Possible implementation

reverse_copy(1)

struct reverse_copy_fn
{
    template<std::bidirectional_iterator I, std::sentinel_for<I> S,
             std::weakly_incrementable O>
    requires std::indirectly_copyable<I, O>
    constexpr ranges::reverse_copy_result<I, O>
        operator()(I first, S last, O result) const
    {
        auto ret = ranges::next(first, last);
        for (; last != first; *result = *--last, ++result);
        return {std::move(ret), std::move(result)};
    }

    template<ranges::bidirectional_range R, std::weakly_incrementable O>
    requires std::indirectly_copyable<ranges::iterator_t<R>, O>
    constexpr ranges::reverse_copy_result<ranges::borrowed_iterator_t<R>, O>
        operator()(R&& r, O result) const
    {
        return (*this)(ranges::begin(r), ranges::end(r), std::move(result));
    }
};

inline constexpr reverse_copy_fn reverse_copy {};

Notes

Implementations (e.g. MSVC STL) may enable vectorization when the both iterator types satisfy LegacyContiguousIterator and have the same value type, and the value type is TriviallyCopyable.

Examples

Main.cpp

#include <algorithm>
#include <iostream>
#include <string>

int main()
{
    std::string x {"12345"}, y(x.size(), ' ');
    std::cout << x << " → ";
    std::ranges::reverse_copy(x.begin(), x.end(), y.begin());
    std::cout << y << " → ";
    std::ranges::reverse_copy(y, x.begin());
    std::cout << x << '\n';
}

Output

12345 → 54321 → 12345