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std::unordered_set count() method

// (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;
  • (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
This article originates from this CppReference page. It was likely altered for improvements or editors' preference. Click "Edit this page" to see all changes made to this document.
Hover to see the original license.

std::unordered_set count() method

// (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;
  • (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
This article originates from this CppReference page. It was likely altered for improvements or editors' preference. Click "Edit this page" to see all changes made to this document.
Hover to see the original license.