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std::move

Defined in header `<algorithm>`
	(1)
template< class InputIt, class OutputIt > OutputIt move( InputIt first, InputIt last, OutputIt d_first );		(since C++11) (until C++20)
template< class InputIt, class OutputIt > constexpr OutputIt move( InputIt first, InputIt last, OutputIt d_first );		(since C++20)
template< class ExecutionPolicy, class ForwardIt1, class ForwardIt2 > ForwardIt2 move( ExecutionPolicy&& policy, ForwardIt1 first, ForwardIt1 last, ForwardIt2 d_first );	(2)	(since C++17)

1) Moves the elements in the range [first, last), to another range beginning at d_first, starting from first and proceeding to last - 1. After this operation 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.

2) Same as (1), but executed according to policy. This overload only participates in overload resolution if std::is_execution_policy_v<std::decay_t<ExecutionPolicy>> is true

Parameters

first, last	-	the range of elements to move
d_first	-	the beginning of the destination range. The behavior is undefined if `d_first` is within the range `[first, last)`. In this case, `std::move_backward` may be used instead of std::move.
policy	-	the execution policy to use. See execution policy for details.
Type requirements
-`InputIt` must meet the requirements of InputIterator.
-`OutputIt` must meet the requirements of OutputIterator.
-`ForwardIt1, ForwardIt2` must meet the requirements of ForwardIterator.

Return value

Output iterator to the element past the last element moved (d_first + (last - first)).

Complexity

Exactly last - first move assignments.

Exceptions

The overload with a template parameter named ExecutionPolicy reports errors as follows:

If execution of a function invoked as part of the algorithm throws an exception and ExecutionPolicy is one of the three standard policies, std::terminate is called. For any other ExecutionPolicy, the behavior is implementation-defined.
If the algorithm fails to allocate memory, std::bad_alloc is thrown.

Possible implementation

template<class InputIt, class OutputIt>
OutputIt move(InputIt first, InputIt last, OutputIt d_first)
{
    while (first != last) {
        *d_first++ = std::move(*first++);
    }
    return d_first;
}

Notes

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

Example

The following code moves thread objects (which themselves are not copyable) from one container to another.

#include <iostream>
#include <vector>
#include <list>
#include <iterator>
#include <thread>
#include <chrono>
 
void f(int n)
{
    std::this_thread::sleep_for(std::chrono::seconds(n));
    std::cout << "thread " << n << " ended" << '\n';
}
 
int main() 
{
    std::vector<std::thread> v;
    v.emplace_back(f, 1);
    v.emplace_back(f, 2);
    v.emplace_back(f, 3);
    std::list<std::thread> l;
    // copy() would not compile, because std::thread is noncopyable
 
    std::move(v.begin(), v.end(), std::back_inserter(l)); 
    for (auto& t : l) t.join();
}

Output:

thread 1 ended
thread 2 ended
thread 3 ended

move_backward (C++11)	moves a range of elements to a new location in backwards order (function template)
move (C++11)	obtains an rvalue reference (function template)