std::set<Key,Compare,Allocator>::erase (3) - Linux Manuals

std::set<Key,Compare,Allocator>::erase: std::set<Key,Compare,Allocator>::erase

NAME

std::set<Key,Compare,Allocator>::erase - std::set<Key,Compare,Allocator>::erase

Synopsis

void erase( iterator pos ); (until C++11)
iterator erase( const_iterator pos ); (since C++11)
iterator erase( iterator pos ); (1) (since C++17)
void erase( iterator first, iterator last ); (until C++11)
iterator erase( const_iterator first, const_iterator last ); (2) (since C++11)
size_type erase( const key_type& key ); (3)

Removes specified elements from the container.
1) Removes the element at pos.
2) Removes the elements in the range [first; last), which must be a valid range in *this.
3) Removes the element (if one exists) with the key equivalent to key.
References and iterators to the erased elements are invalidated. Other references and iterators are not affected.
The iterator pos must be valid and dereferenceable. Thus the end() iterator (which is valid, but is not dereferenceable) cannot be used as a value for pos.

Parameters

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

Return value

1-2) Iterator following the last removed element.
3) Number of elements removed.

Exceptions

1,2) (none)
3) Any exceptions thrown by the Compare object.

Complexity

Given an instance c of set:
1) Amortized constant
2) log(c.size()) + std::distance(first, last)
3) log(c.size()) + c.count(k)

Example

// Run this code

  #include <set>
  #include <iostream>
  int main()
  {
      std::set<int> c = {1, 2, 3, 4, 5, 6, 7, 8, 9};
      // erase all odd numbers from c
      for(auto it = c.begin(); it != c.end(); )
          if(*it % 2 == 1)
              it = c.erase(it);
          else
              ++it;
      for(int n : c)
          std::cout << n << ' ';
  }

Output:

  2 4 6 8