std::unordered_multiset<Key,Hash,KeyEqual,Allocator>::count
From cppreference.com
< cpp | container | unordered multiset
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
.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
linear in the number of elements with key key
on average, worst case linear in the size of the container.
Notes
Feature-test macro: | __cpp_lib_generic_unordered_lookup (for overload (2)) |
Example
Run this code
#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;
See also
(C++11) |
finds element with specific key (public member function) |
(C++20) |
checks if the container contains element with specific key (public member function) |
(C++11) |
returns range of elements matching a specific key (public member function) |