std::ranges::move_backward, std::ranges::move_backward_result

template< std::bidirectional_iterator I1, std::sentinel_for<I1> S1,
          std::bidirectional_iterator I2 >
    requires std::indirectly_movable<I1, I2>
constexpr move_backward_result<I1, I2>
    move_backward( I1 first, S1 last, I2 d_last );

template< ranges::bidirectional_range R, std::bidirectional_iterator I >
    requires std::indirectly_movable<ranges::iterator_t<R>, I>
constexpr move_backward_result<ranges::borrowed_iterator_t<R>, I>
    move_backward( R&& r, I d_last );

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

Moves all elements from the source range to the destination range ending at d_last, starting from the end and proceeding to the beginning (reverse order).

If d_last is in (first, last], the behavior is undefined.

The function-like entities described on this page are algorithm function objects (informally known as niebloids), that is:

• Explicit template argument lists cannot be specified when calling any of them.
• None of them are visible to argument-dependent lookup.
• When any of them are found by normal unqualified lookup as the name to the left of the function-call operator, argument-dependent lookup is inhibited.

Parameters

Return value

A ranges::move_backward_result object where:

• The data member in holds the past-the-end iterator of the source range.
• The data member out holds and an iterator to the last move-assigned element in the destination range, or d_last if no element is moved.

Complexity

Notes

When moving overlapping ranges, ranges::move is appropriate when moving to the left (beginning of the destination range is outside the source range) while ranges::move_backward is appropriate when moving to the right (end of the destination range is outside the source range).

Possible implementation

struct move_backward_fn
{
    template<std::bidirectional_iterator I1, std::sentinel_for<I1> S1,
             std::bidirectional_iterator I2>
        requires std::indirectly_movable<I1, I2>
    constexpr ranges::move_backward_result<I1, I2>
        operator()(I1 first, S1 last, I2 d_last) const
    {
        auto i {last};
        for (; i != first; *--d_last = ranges::iter_move(--i))
        {}
        return {std::move(last), std::move(d_last)};
    }
    
    template<ranges::bidirectional_range R, std::bidirectional_iterator I>
        requires std::indirectly_movable<ranges::iterator_t<R>, I>
    constexpr ranges::move_backward_result<ranges::borrowed_iterator_t<R>, I>
        operator()(R&& r, I d_last) const
    {
        return (*this)(ranges::begin(r),
                       ranges::next(ranges::begin(r), ranges::end(r)),
                       std::move(d_last));
    }
};

inline constexpr move_backward_fn move_backward{};

Example

#include <algorithm>
#include <iostream>
#include <string>
#include <string_view>
#include <vector>

using Vec = std::vector<std::string>;

void print(std::string_view rem, const Vec& vec)
{
    std::cout << rem << "[" << vec.size() << "]: ";
    for (const std::string& s : vec)
        std::cout << (s.size() ? s : std::string{"·"}) << ' ';
    std::cout << '\n';
}

int main()
{
    Vec a{"▁", "▂", "▃", "▄", "▅", "▆", "▇", "█"};
    Vec b(a.size());
    
    print("Before move:\n" "a", a);
    print("b", b);
    
    std::ranges::move_backward(a, b.end());
    
    print("\n" "Move a >> b:\n" "a", a);
    print("b", b);
    
    std::ranges::move_backward(b.begin(), b.end(), a.end());
    print("\n" "Move b >> a:\n" "a", a);
    print("b", b);
    
    std::ranges::move_backward(a.begin(), a.begin()+3, a.end());
    print("\n" "Overlapping move a[0, 3) >> a[5, 8):\n" "a", a);
}

Before move:
a[8]: ▁ ▂ ▃ ▄ ▅ ▆ ▇ █
b[8]: · · · · · · · ·

Move a >> b:
a[8]: · · · · · · · ·
b[8]: ▁ ▂ ▃ ▄ ▅ ▆ ▇ █

Move b >> a:
a[8]: ▁ ▂ ▃ ▄ ▅ ▆ ▇ █
b[8]: · · · · · · · ·

Overlapping move a[0, 3) >> a[5, 8):
a[8]: · · · ▄ ▅ ▁ ▂ ▃

See also