std::move
From cppreference.com
| Defined in header <algorithm>
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template< class InputIt, class OutputIt >
OutputIt move( InputIt first, InputIt last, OutputIt d_first );
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(1) | (since C++11) (constexpr since C++20) |
template< class ExecutionPolicy, class ForwardIt1, class ForwardIt2 >
ForwardIt2 move( ExecutionPolicy&& policy,
ForwardIt1 first, ForwardIt1 last, ForwardIt2 d_first );
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(2) | (since C++17) |
Moves all elements in the source range [first, last) to the destination range [d_first, std::next(d_first, std::distance(first, last))).
1) Move starts from
first and proceeding to last. If
d_first is in the source range, the behavior is undefined.2) Same as (1), but the move order is determined by
policy. This overload participates in overload resolution only if the value of the following expression is
true:
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(until C++20) |
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(since C++20) |
If the source and destination ranges overlap, the behavior is undefined.
Parameters
| first, last | - | the pair of iterators defining the source range |
| d_first | - | the beginning of the destination range |
| policy | - | the execution policy to use |
| Type requirements | ||
-InputIt must meet the requirements of LegacyInputIterator.
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-OutputIt must meet the requirements of LegacyOutputIterator.
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-ForwardIt1, ForwardIt2 must meet the requirements of LegacyForwardIterator.
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Return value
The past-the-end iterator of the destination range.
Complexity
Exactly std::distance(first, last) assignments.
Exceptions
2) During the execution process:
- If the temporary memory resources required for parallelization are not available, std::bad_alloc is thrown.
- If an uncaught exception is thrown while accessing objects via an algorithm argument, the behavior is determined by the execution policy (for standard policies, std::terminate is invoked).
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;
}
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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 <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
See also
(C++20) |
moves a range of elements to a new location (algorithm function object) |
(C++11) |
moves a range of elements to a new location in backwards order (function template & algorithm function object) |
(C++20) |
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(C++11) |
converts the argument to an xvalue (function template) |