std::move

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

Parameters

Return value

The iterator to the element past the last element moved.

Complexity

Exactly std::distance(first, last)

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
• 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

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