std::unordered_multiset count() method

od 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;

od 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.
• (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.
• (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_multiset set{2, 7, 1, 8, 2, 8, 1, 8, 2, 8};
 
    std::cout << "The set is:\n";
    for (int e: set) { std::cout << e << ' '; }
 
    const auto [min, max] = std::ranges::minmax(set);
 
    std::cout << "\nNumbers [" << min << ".." << max << "] frequency:\n";
 
    for (int i{min}; i <= max; ++i) {
        std::cout << i << ':' << set.count(i) << "; ";
    }
}

Possible output

The set is:
8 8 8 8 1 1 7 2 2 2
Numbers [1..8] frequency:
1:2; 2:3; 3:0; 4:0; 5:0; 6:0; 7:1; 8:4;