std::move
From cppreference.com
Defined in header <algorithm>
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(1) | ||
template< class InputIt, class OutputIt > OutputIt move( InputIt first, InputIt last, OutputIt d_first ); |
(since C++11) (until C++20) |
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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 trueParameters
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.
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policy | - | the execution policy to use. See execution policy for details. |
Type requirements | ||
-InputIt must meet the requirements of InputIterator.
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-OutputIt must meet the requirements of OutputIterator.
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-ForwardIt1, ForwardIt2 must meet the requirements of ForwardIterator.
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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 otherExecutionPolicy
, 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.
Run this code
#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
See also
(C++11) |
moves a range of elements to a new location in backwards order (function template) |
(C++11) |
obtains an rvalue reference (function template) |