std::move
Defined in header <algorithm>
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template< class InputIt, class OutputIt > OutputIt move( InputIt first, InputIt last, OutputIt d_first ); |
(since C++11) | |
Moves the elements in the range [first, last)
, to another range beginning at d_first
. 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.
Contents |
[edit] Parameters
first, last | - | the range of elements to move |
d_first | - | the beginning of the destination range. If d_first is within [first, last) , std::move_backward must be used instead of std::move.
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Type requirements | ||
-InputIt must meet the requirements of InputIterator .
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-OutputIt must meet the requirements of OutputIterator .
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[edit] Return value
Output iterator to the element past the last element moved (d_first + (last - first))
[edit] Complexity
Exactly last - first
move assignments.
[edit] 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; } |
[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 <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
[edit] 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) |