std::list<T,Allocator>::swap
From cppreference.com
void swap( list& other ); |
(until C++17) | |
void swap( list& other ) noexcept(/* see below */); |
(since C++17) | |
Exchanges the contents of the container with those of other. Does not invoke any move, copy, or swap operations on individual elements.
All iterators and references remain valid. It is unspecified whether an iterator holding the end() value in this container will refer to this or the other container after the operation.
If std::allocator_traits<allocator_type>::propagate_on_container_swap::value is true, then the allocators are exchanged using an unqualified call to non-member |
(since C++11) |
Parameters
other | - | container to exchange the contents with |
Return value
(none)
Exceptions
(none) |
(until C++17) |
noexcept specification:
noexcept(std::allocator_traits<Allocator>::is_always_equal::value) |
(since C++17) |
Complexity
Constant.
Example
Run this code
#include <iostream> #include <list> template<class Os, class Co> Os& operator<<(Os& os, const Co& co) { os << '{'; for (auto const& i : co) os << ' ' << i; return os << " } "; } int main() { std::list<int> a1{1, 2, 3}, a2{4, 5}; auto it1 = std::next(a1.begin()); auto it2 = std::next(a2.begin()); int& ref1 = a1.front(); int& ref2 = a2.front(); std::cout << a1 << a2 << *it1 << ' ' << *it2 << ' ' << ref1 << ' ' << ref2 << '\n'; a1.swap(a2); std::cout << a1 << a2 << *it1 << ' ' << *it2 << ' ' << ref1 << ' ' << ref2 << '\n'; // Note that after swap the iterators and references stay associated with their // original elements, e.g. it1 that pointed to an element in 'a1' with value 2 // still points to the same element, though this element was moved into 'a2'. }
Output:
{ 1 2 3 } { 4 5 } 2 5 1 4 { 4 5 } { 1 2 3 } 2 5 1 4
See also
specializes the std::swap algorithm (function template) |