std::move
|   Defined in header  <algorithm>
  | 
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|   template< class InputIt, class OutputIt > OutputIt move( InputIt first, InputIt last,  | 
(1) |  (since C++11)  (constexpr since C++20)  | 
|   template< class ExecutionPolicy, class ForwardIt1, class ForwardIt2 > ForwardIt2 move( ExecutionPolicy&& policy,  | 
(2) | (since C++17) | 
[first, last), to another range beginning at d_first, starting from first and proceeding to last. 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.| 
 std::is_execution_policy_v<std::decay_t<ExecutionPolicy>> is true.  | 
(until C++20) | 
| 
 std::is_execution_policy_v<std::remove_cvref_t<ExecutionPolicy>> is true.  | 
(since C++20) | 
If d_first is within the range [first, last), the behavior is undefined. In this case, std::move_backward may be used instead.
Contents | 
[edit] Parameters
| first, last | - | the pair of iterators defining the source range of elements to move | 
| d_first | - | the beginning of the destination range | 
| policy | - | the execution policy to use | 
| Type requirements | ||
 -InputIt must meet the requirements of LegacyInputIterator.
 | ||
 -OutputIt must meet the requirements of LegacyOutputIterator.
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 -ForwardIt1, ForwardIt2 must meet the requirements of LegacyForwardIterator.
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[edit] Return value
The iterator to the element past the last element moved.
[edit] Complexity
Exactly std::distance(first, last) move assignments.
[edit] 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 
ExecutionPolicyis one of the standard policies, std::terminate is called. For any otherExecutionPolicy, the behavior is implementation-defined. - If the algorithm fails to allocate memory, std::bad_alloc is thrown.
 
[edit] 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; }  | 
[edit] 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).
[edit] 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)); }
Output:
thread 1 ended thread 2 ended thread 3 ended
[edit] See also
|    (C++11)  | 
  moves a range of elements to a new location in backwards order  (function template)  | 
|    (C++11)  | 
  converts the argument to an xvalue  (function template)  | 
|   (C++20)  | 
 moves a range of elements to a new location (algorithm function object)  |