std::unordered_set count() method

since C++20

// (1) Const version only
size_type count( const Key& key ) const;

// (2) Const version only
template< class K >
size_type count( const K& x ) const;

since C++11

// (1) Const version only
size_type count( 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 <algorithm>
#include <iostream>
#include <unordered_set>
 
int main() {
    std::unordered_set set{2, 7, 1, 8, 2, 8, 1, 8, 2, 8};
 
    std::cout << "The set is: ";
    for (int e: set) { std::cout << e << ' '; }
 
    const auto [min, max] = std::ranges::minmax(set);
 
    std::cout << "\nNumbers from " << min << " to " << max << " that are in the set: ";
    for (int i{min}; i <= max; ++i) {
        if (set.count(i) == 1) {
            std::cout << i << ' ';
        }
    }
}

Possible output

The set is: 8 1 7 2
Numbers from 1 to 8 that are in the set: 1 2 7 8