Ifd_first is within the range[first, last), the behavior is undefined. In this case,std::move_backward may be used instead.

[edit]Parameters

[edit]Return value

The iterator to the element past the last element moved.

[edit]Complexity

Exactlystd::distance(first, last) move assignments.

[edit]Exceptions

The overload with a template parameter namedExecutionPolicy reports errors as follows:

• If execution of a function invoked as part of the algorithm throws an exception andExecutionPolicy is one of thestandard 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) whilestd::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));}

thread 1 endedthread 2 endedthread 3 ended

[edit]See also

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