std::move
Defined in header <algorithm> | ||
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. | ||
-ForwardIt1, ForwardIt2 must meet the requirements of LegacyForwardIterator. |
[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
ExecutionPolicy
is 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) |