std::unordered_set::erase
From cppreference.com
< cpp | container | unordered set
| iterator erase( const_iterator pos );
|
(1) | (since C++11) |
| iterator erase( const_iterator first, const_iterator last );
|
(2) | (since C++11) |
| size_type erase( const key_type& key );
|
(3) | (since C++11) |
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 iterators and references are not invalidated.
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 order of the elements that are not erased is preserved (this makes it possible to erase individual elements while iterating through the container) |
(since C++14) |
Contents |
[edit] Parameters
| pos | - | iterator to the element to remove |
| first, last | - | range of elements to remove |
| key | - | key value of the elements to remove |
[edit] Return value
1-2) Iterator following the last removed element.
3) Number of elements removed.
[edit] Exceptions
1,2) (none)
3) Any exceptions thrown by the
Compare object.[edit] Complexity
Given an instance c of unordered_set:
1) Average case: constant, worst case: c.size()
3) Average case: c.count(key), worst case: c.size()
[edit] Example
Run this code
#include <unordered_set> #include <iostream> int main() { std::unordered_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
[edit] See also
| clears the contents (public member function) |