std::map<Key,T,Compare,Allocator>::equal_range (3) - Linux Manuals

std::map<Key,T,Compare,Allocator>::equal_range: std::map<Key,T,Compare,Allocator>::equal_range

NAME

std::map<Key,T,Compare,Allocator>::equal_range - std::map<Key,T,Compare,Allocator>::equal_range

Synopsis

std::pair<iterator,iterator> equal_range( const Key& key ); (1)
std::pair<const_iterator,const_iterator> equal_range( const Key& key ) const; (2)
template< class K > (3) (since C++14)
std::pair<iterator,iterator> equal_range( const K& x );
template< class K > (4) (since C++14)
std::pair<const_iterator,const_iterator> equal_range( const K& x ) const;

Returns a range containing all elements with the given key in the container. The range is defined by two iterators, one pointing to the first element that is not less than key and another pointing to the first element greater than key. Alternatively, the first iterator may be obtained with lower_bound(), and the second with upper_bound().
1,2) Compares the keys to key.
3,4) Compares the keys to the value x. This overload only participates in overload resolution if the qualified-id Compare::is_transparent is valid and denotes a type. They allow calling this function without constructing an instance of Key.

 This section is incomplete
 Reason: explain better

Parameters

key - key value to compare the elements to
x - alternative value that can be compared to Key

Return value

std::pair containing a pair of iterators defining the wanted range: the first pointing to the first element that is not less than key and the second pointing to the first element greater than key.
If there are no elements not less than key, past-the-end (see end()) iterator is returned as the first element. Similarly if there are no elements greater than key, past-the-end iterator is returned as the second element.

Complexity

Logarithmic in the size of the container.

Example

// Run this code

  #include <map>
  #include <iostream>

  int main()
  {
      const std::map<int, const char*> m{
          { 0, "zero" },
          { 1, "one" },
          { 2, "two" },
      };

      {
          auto p = m.equal_range(1);
          for (auto& q = p.first; q != p.second; ++q) {
              std::cout << "m[" << q->first << "] = " << q->second << '\n';
          }

          if (p.second == m.find(2)) {
              std::cout << "end of equal_range (p.second) is one-past p.first\n";
          } else {
              std::cout << "unexpected; p.second expected to be one-past p.first\n";
          }
      }

      {
          auto pp = m.equal_range(-1);
          if (pp.first == m.begin()) {
              std::cout << "pp.first is iterator to first not-less than -1\n";
          } else {
              std::cout << "unexpected pp.first\n";
          }

          if (pp.second == m.begin()) {
              std::cout << "pp.second is iterator to first element greater-than -1\n";
          } else {
              std::cout << "unexpected pp.second\n";
          }
      }

      {
          auto ppp = m.equal_range(3);
          if (ppp.first == m.end()) {
              std::cout << "ppp.first is iterator to first not-less than 3\n";
          } else {
              std::cout << "unexpected ppp.first\n";
          }

          if (ppp.second == m.end()) {
              std::cout << "ppp.second is iterator to first element greater-than 3\n";
          } else {
              std::cout << "unexpected ppp.second\n";
          }
      }
  }

Output:

  m[1] = one
  end of equal_range (p.second) is one-past p.first
  pp.first is iterator to first not-less than -1
  pp.second is iterator to first element greater-than -1
  ppp.first is iterator to first not-less than 3
  ppp.second is iterator to first element greater-than 3