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

std::deque<T,Allocator>::erase: std::deque<T,Allocator>::erase

NAME

std::deque<T,Allocator>::erase - std::deque<T,Allocator>::erase

Synopsis

iterator erase( iterator pos ); (until C++11)
iterator erase( const_iterator pos ); (1) (since C++11)
iterator erase( iterator first, iterator last ); (2) (until C++11)
iterator erase( const_iterator first, const_iterator last ); (since C++11)

Erases the specified elements from the container.
1) Removes the element at pos.
2) Removes the elements in the range [first, last).
All iterators and references are invalidated, unless the erased elements are at the end or the beginning of the container, in which case only the iterators and references to the erased elements are invalidated.

It is unspecified when the past-the-end iterator is invalidated. (until C++11)
The past-the-end iterator is also invalidated unless the erased elements are at the beginning of the container and the last element is not erased. (since C++11)

The iterator pos must be valid and dereferenceable. Thus the end() iterator (which is valid, but is not dereferencable) cannot be used as a value for pos.
The iterator first does not need to be dereferenceable if first==last: erasing an empty range is a no-op.

Parameters

pos - iterator to the element to remove
first, last - range of elements to remove

Type requirements

Return value

Iterator following the last removed element. If the iterator pos refers to the last element, the end() iterator is returned.

Exceptions

Does not throw unless an exception is thrown by the assignment operator of T.

Complexity

Linear: the number of calls to the destructor of T is the same as the number of elements erased, the number of calls to the assignment operator of T is no more than the lesser of the number of elements before the erased elements and the number of elements after the erased elements

Example

// Run this code

  #include <deque>
  #include <iostream>

  int main( )
  {
      std::deque<int> c{0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
      for (auto &i : c) {
          std::cout << i << " ";
      }
      std::cout << '\n';

      c.erase(c.begin());

      for (auto &i : c) {
          std::cout << i << " ";
      }
      std::cout << '\n';

      c.erase(c.begin()+2, c.begin()+5);

      for (auto &i : c) {
          std::cout << i << " ";
      }
      std::cout << '\n';

      // Erase all even numbers (C++11 and later)
      for (auto it = c.begin(); it != c.end(); ) {
          if (*it % 2 == 0) {
              it = c.erase(it);
          } else {
              ++it;
          }
      }

      for (auto &i : c) {
          std::cout << i << " ";
      }
      std::cout << '\n';
  }

Output:

  0 1 2 3 4 5 6 7 8 9
  1 2 3 4 5 6 7 8 9
  1 2 6 7 8 9
  1 7 9