std::unordered_map count() method

since C++20

// (1) Non const version
iterator find( const Key& key );

// (2) Const version
const_iterator find( const Key& key ) const;

// (3) Non const version
template< class K > iterator find( const K& x );

// (4) Const version
template< class K > const_iterator find( const K& x ) const;

since C++11

// (1) Non const version
iterator find( const Key& key );

// (2) Const version
const_iterator find( const Key& key ) const;

• (1) Returns the number of elements with key that compares equal to the specified argument key, which is either 1 or 0 since this container does not allow duplicates.
• (2) Returns the number of elements with key that compares equivalent to the specified argument x. This overload participates in overload resolution only if Hash::is_transparent and KeyEqual::is_transparent are valid and each denotes a type. This assumes that such Hash is callable with both K and Key type, and that the KeyEqual is transparent, which, together, allows calling this function without constructing an instance of Key.

Parameters

• key - key value of the elements to count
• x - a value of any type that can be transparently compared with a key

Return value

• (1) Number of elements with key key, that is either 1 or 0.
• (2) Number of elements with key that compares equivalent to x.

Complexity

Average case, constant - O(1).
Worst case, linear in size of the container - O(size()).

Exceptions

(none)

Notes

Feature testing macro: __cpp_lib_generic_unordered_lookup (for overload (2))

Example

Main.cpp

#include <string>
#include <iostream>
#include <unordered_map>
 
int main()
{
    std::unordered_map<int, std::string> dict = {
        {1, "one"}, {6, "six"}, {3, "three"}
    };
    dict.insert({4, "four"});
    dict.insert({5, "five"});
    dict.insert({6, "six"});
 
    std::cout << "dict: { ";
    for (auto const& [key, value] : dict)
    {
        std::cout << "[" << key << "]=" << value << " ";
    }
    std::cout << "}\n\n";
 
    for (int i{1}; i != 8; ++i)
    {
        std::cout << "dict.count(" << i << ") = " << dict.count(i) << '\n';
    }
}

Possible output

dict: { [5]=five [4]=four [1]=one [6]=six [3]=three }
 
dict.count(1) = 1
dict.count(2) = 0
dict.count(3) = 1
dict.count(4) = 1
dict.count(5) = 1
dict.count(6) = 1
dict.count(7) = 0