std::map::find
From cppreference.com
iterator find( const Key& key ); |
(1) | |
const_iterator find( const Key& key ) const; |
(2) | |
template< class K > iterator find( const K& x ); |
(3) | (since C++14) |
template< class K > const_iterator find( const K& x ) const; |
(4) | (since C++14) |
1,2) Finds an element with key equivalent to
key
. 3,4) Finds an element with key that compares equivalent to the value
x
. This overload only participates in overload resolution if the qualified-id Compare::is_transparent is valid and denotes a type. It allows calling this function without constructing an instance of Key
Parameters
key | - | key value of the element to search for |
x | - | a value of any type that can be transparently compared with a key |
Return value
Iterator to an element with key equivalent to key
. If no such element is found, past-the-end (see end()) iterator is returned.
Complexity
Logarithmic in the size of the container.
Example
Run this code
Output:
Found 2 b
See also
returns the number of elements matching specific key (public member function) | |
returns range of elements matching a specific key (public member function) |
Example
Demonstrates the risk of accessing non-existing elements via operator [].
Run this code
#include <string> #include <iostream> #include <map> int main() { std::map<std::string,int> my_map; my_map["x"] = 11; my_map["y"] = 23; auto it = my_map.find("x"); if (it != my_map.end()) std::cout << "x: " << it->second << "\n"; it = my_map.find("z"); if (it != my_map.end()) std::cout << "z1: " << it->second << "\n"; // Accessing a non-existing element creates it if (my_map["z"] == 42) std::cout << "Oha!\n"; it = my_map.find("z"); if (it != my_map.end()) std::cout << "z2: " << it->second << "\n"; }
Output:
x: 11 z2: 0