std::ranges::reverse_copy, std::ranges::reverse_copy_result - cppreference.com

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Defined in header <algorithm>

Call signature

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 );

(1) (since C++20)

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 );

(2) (since C++20)

Helper types

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

(3) (since C++20)

1) Copies the elements from the source range [firstlast) to the destination range [resultresult + N), where N is ranges::distance(first, last), 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 [0N). The behavior is undefined if 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 algorithm function objects (informally known as niebloids), that is:

Parameters

first, last - the iterator-sentinel pair defining the source range of elements to copy
r - the source range of elements to copy
result - the beginning of the destination range.

Return value

{last, result + N}.

Complexity

Exactly N assignments.

Notes

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

Possible implementation

See also the implementations in MSVC STL and libstdc++.

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 {};

Example

#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:

See also