Defined in header <algorithm> | ||||
---|---|---|---|---|
| (since C++11) |
Moves the elements from the range [first, last)
, to another range ending at d_last
. The elements are moved in reverse order (the last element is moved first), but their relative order is preserved.
The behavior is undefined if d_last
is within (first, last]
. std::move must be used instead of std::move_backward
in that case.
Parameters
first, last | - | the range of the elements to move |
d_last | - | end of the destination range |
Type requirements | ||
- BidirIt1, BidirIt2 must meet the requirements of BidirectionalIterator . |
Return value
Iterator in the destination range, pointing at the last element moved.
Complexity
Exactly last - first
move assignments.
Possible implementation
|
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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 | #include <algorithm> #include <vector> #include <string> #include <iostream> int main() { std::vector<std::string> src{ "foo" , "bar" , "baz" }; std::vector<std::string> dest(src.size()); std::cout << "src: " ; for ( const auto &s : src) { std::cout << s << ' ' ; } std::cout << "\ndest: " ; for ( const auto &s : dest) { std::cout << s << ' ' ; } std::cout << '\n' ; std::move_backward(src.begin(), src.end(), dest.end()); std::cout << "src: " ; for ( const auto &s : src) { std::cout << s << ' ' ; } std::cout << "\ndest: " ; for ( const auto &s : dest) { std::cout << s << ' ' ; } std::cout << '\n' ; } |
Output:
1 2 3 4 | src: foo bar baz dest: src: dest: foo bar baz |
See also
(C++11) | moves a range of elements to a new location (function template) |
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