std::ranges::move, std::ranges::move_result - cppreference.com

template< std::input_iterator I, std::sentinel_for<I> S, std::weakly_incrementable O >
requires std::indirectly_movable<I, O>
constexpr move_result<I, O>
    move( I first, S last, O result );

template< ranges::input_range R, std::weakly_incrementable O >
requires std::indirectly_movable<ranges::iterator_t<R>, O>
constexpr move_result<ranges::borrowed_iterator_t<R>, O>
    move( R&& r, O result );

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

1) Moves the elements in the range, defined by [first, last), to another range beginning at result. The behavior is undefined if result is within the range [first, last). In such a case, ranges::move_backward may be used instead.

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 elements in the moved-from range will still contain valid values of the appropriate type, but not necessarily the same values as before the move.

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

{last, result + N}, where

1) N = ranges::distance(first, last).

2) N = ranges::distance(r).

Complexity

Exactly N move assignments.

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_fn
{
    template<std::input_iterator I, std::sentinel_for<I> S, std::weakly_incrementable O>
    requires std::indirectly_movable<I, O>
    constexpr ranges::move_result<I, O>
        operator()(I first, S last, O result) const
    {
        for (; first != last; ++first, ++result)
            *result = ranges::iter_move(first);
        return {std::move(first), std::move(result)};
    }
    template<ranges::input_range R, std::weakly_incrementable O>
    requires std::indirectly_movable<ranges::iterator_t<R>, O>
    constexpr ranges::move_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 move_fn move {};

Example

#include <algorithm>
#include <chrono>
#include <iostream>
#include <iterator>
#include <list>
#include <thread>
#include <vector>
using namespace std::literals::chrono_literals;

void f(std::chrono::milliseconds n)
{
    std::this_thread::sleep_for(n);
    std::cout << "thread with n=" << n.count() << "ms ended" << std::endl;
}

int main()
{
    std::vector<std::jthread> v;
    v.emplace_back(f, 400ms);
    v.emplace_back(f, 600ms);
    v.emplace_back(f, 800ms);

    std::list<std::jthread> l;

    // std::ranges::copy() would not compile, because std::jthread is non-copyable
    std::ranges::move(v, std::back_inserter(l));
}

thread with n=400ms ended
thread with n=600ms ended
thread with n=800ms ended

See also