std:: move

If d_first is within the range [ first , last ) , the behavior is undefined. In this case, std::move_backward may be used instead.

Parameters

Return value

The iterator to the element past the last element moved.

Complexity

Exactly std:: distance ( first, last ) 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 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)
{
    for (; first != last; ++d_first, ++first)
        *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 <algorithm>
#include <chrono>
#include <iostream>
#include <iterator>
#include <list>
#include <thread>
#include <vector>
 
void f(int n)
{
    std::this_thread::sleep_for(std::chrono::seconds(n));
    std::cout << "thread " << n << " ended" << std::endl;
}
 
int main()
{
    std::vector<std::jthread> v;
    v.emplace_back(f, 1);
    v.emplace_back(f, 2);
    v.emplace_back(f, 3);
    std::list<std::jthread> l;
 
    // copy() would not compile, because std::jthread is noncopyable
    std::move(v.begin(), v.end(), std::back_inserter(l));
}

thread 1 ended
thread 2 ended
thread 3 ended

See also