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std::unordered_set<Key,Hash,KeyEqual,Allocator>::count

size_type count( const Key& key ) const;	(1)	(since C++11)
template< class K > size_type count( const K& x ) const;	(2)	(since C++20)

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

Constant on average, worst case linear in the size of the container.

Notes

Feature-test macro	Comment
`__cpp_lib_generic_unordered_lookup`	for overload (2)

Example

#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