std::deque<T,Allocator>::operator= (3) - Linux Manuals

std::deque<T,Allocator>::operator=: std::deque<T,Allocator>::operator=

NAME

std::deque<T,Allocator>::operator= - std::deque<T,Allocator>::operator=

Synopsis

deque& operator=( const deque& other ); (1)
deque& operator=( deque&& other ); (since C++11)
                                                             (2) (until C++17)
deque& operator=( deque&& other ) noexcept(/* see below */); (since C++17)
deque& operator=( std::initializer_list<T> ilist ); (3) (since C++11)

Replaces the contents of the container.
1) Copy assignment operator. Replaces the contents with a copy of the contents of other.
If std::allocator_traits<allocator_type>::propagate_on_container_copy_assignment::value is true, the target allocator is replaced by a copy of the source allocator. If the target and the source allocators do not compare equal, the target (*this) allocator is used to deallocate the memory, then other's allocator is used to allocate it before copying the elements.
(since C++11).
2) Move assignment operator. Replaces the contents with those of other using move semantics (i.e. the data in other is moved from other into this container). other is in a valid but unspecified state afterwards. If std::allocator_traits<allocator_type>::propagate_on_container_move_assignment::value is true, the target allocator is replaced by a copy of the source allocator. If it is false and the source and the target allocators do not compare equal, the target cannot take ownership of the source memory and must move-assign each element individually, allocating additional memory using its own allocator as needed. In any case, all elements originally present in *this are either destroyed or replaced by elementwise move-assignment.
3) Replaces the contents with those identified by initializer list ilist.

Parameters

other - another container to use as data source
ilist - initializer list to use as data source

Return value

*this

Complexity

1) Linear in the size of *this and other.
2) Linear in the size of *this unless the allocators do not compare equal and do not propagate, in which case linear in the size of *this and other
3) Linear in the size of *this and ilist.

Exceptions

                                                                   (since C++17)
2)
noexcept specification:
noexcept(std::allocator_traits<Allocator>::is_always_equal::value)

Notes

After container move assignment (overload (2)), unless elementwise move assignment is forced by incompatible allocators, references, pointers, and iterators (other than the end iterator) to other remain valid, but refer to elements that are now in *this. The current standard makes this guarantee via the blanket statement in §23.2.1[container.requirements.general]/12, and a more direct guarantee is under consideration via LWG_2321.

Example

The following code uses to assign one std::deque to another:
// Run this code

  #include <deque>
  #include <iostream>

  void display_sizes(const std::deque<int>& nums1,
                     const std::deque<int>& nums2,
                     const std::deque<int>& nums3)
  {
      std::cout << "nums1: " << nums1.size()
                << " nums2: " << nums2.size()
                << " nums3: " << nums3.size() << '\n';
  }

  int main()
  {
      std::deque<int> nums1 {3, 1, 4, 6, 5, 9};
      std::deque<int> nums2;
      std::deque<int> nums3;

      std::cout << "Initially:\n";
      display_sizes(nums1, nums2, nums3);

      // copy assignment copies data from nums1 to nums2
      nums2 = nums1;

      std::cout << "After assigment:\n";
      display_sizes(nums1, nums2, nums3);

      // move assignment moves data from nums1 to nums3,
      // modifying both nums1 and nums3
      nums3 = std::move(nums1);

      std::cout << "After move assigment:\n";
      display_sizes(nums1, nums2, nums3);
  }

Output:

  Initially:
  nums1: 6 nums2: 0 nums3: 0
  After assigment:
  nums1: 6 nums2: 6 nums3: 0
  After move assigment:
  nums1: 0 nums2: 6 nums3: 6