std::unordered_map::equal_range
From cppreference.com
< cpp | container | unordered map
| std::pair<iterator,iterator> equal_range( const Key& key ); |
(1) | (since C++11) |
| std::pair<const_iterator,const_iterator> equal_range( const Key& key ) const; |
(2) | (since C++11) |
| template< class K > std::pair<iterator,iterator> equal_range( const K& x ); |
(3) | (since C++20) |
| template< class K > std::pair<const_iterator,const_iterator> equal_range( const K& x ) const; |
(4) | (since C++20) |
1,2) Returns a range containing all elements with key
key in the container. The range is defined by two iterators, the first pointing to the first element of the wanted range and the second pointing past the last element of the range.3,4) Returns a range containing all elements in the container with key equivalent to
x. This overload only participates in overload resolution if the qualified-id Hash::transparent_key_equal is valid and denotes a type. This assumes that such Hash is callable with both K and Key type, and that its key_equal is transparent, which, together, allows calling this function without constructing an instance of Key.Parameters
| key | - | key value to compare the elements to |
Return value
std::pair containing a pair of iterators defining the wanted range. If there are no such elements, past-the-end (see end()) iterators are returned as both elements of the pair.
Complexity
Average case linear in the number of elements with the key key, worst case linear in the size of the container.
Example
Run this code
#include <iostream> #include <unordered_map> int main() { std::unordered_map<int,char> map = {{1,'a'},{1,'b'},{1,'d'},{2,'b'}}; auto range = map.equal_range(1); for (auto it = range.first; it != range.second; ++it) { std::cout << it->first << ' ' << it->second << '\n'; } }
Output:
1 a
See also
| finds element with specific key (public member function) |