std::swap_ranges
From cppreference.com
Defined in header
<algorithm>
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template< class ForwardIt1, class ForwardIt2 >
ForwardIt2 swap_ranges( ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2 ) |
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Exchanges elements between range [first1, last1)
and another range starting at first2
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Contents |
[edit] Parameters
first1, last1 | - | the first range of elements to swap |
first2 | - | beginning of the second range of elements to swap |
Type requirements | ||
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ForwardIt1, ForwardIt2 must meet the requirements of ForwardIterator .
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-The types of dereferenced ForwardIt1 and ForwardIt2 must meet the requirements of Swappable
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[edit] Return value
Iterator to the element past the last element exchanged in the range beginning with first2
.
[edit] Possible implementation
template<class ForwardIt1, class ForwardIt2> ForwardIt2 swap_ranges(ForwardIt1 first1, ForwardIt1 last1, ForwardIt2 first2) { while (first1 != last1) { std::iter_swap(first1++, first2++); } return first2; } |
[edit] Example
Demonstrates swapping of subranges from different containers
Run this code
#include <algorithm> #include <list> #include <vector> #include <iostream> int main() { std::vector<int> v = {1, 2, 3, 4, 5}; std::list<int> l = {-1, -2, -3, -4, -5}; std::swap_ranges(v.begin(), v.begin()+3, l.begin()); for(int n : v) std::cout << n << ' '; std::cout << '\n'; for(int n : l) std::cout << n << ' '; std::cout << '\n'; }
Output:
-1 -2 -3 4 5 1 2 3 -4 -5
[edit] Complexity
linear in the distance between first
and last
[edit] See also
swaps the elements pointed to by two iterators (function template) |
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swaps the values of two objects (function template) |