std::unordered_multiset<Key,Hash,KeyEqual,Allocator>::count
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Tables
std::unordered_multiset (C++23)
(C++17)
(C++17)
unordered_multiset::count
(C++20)
(until C++20)
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)
size_type count( const K& x ) 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 and KeyEqual are both transparent. 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.[edit] Parameters
key
-
key value of the elements to count
x
-
a value of any type that can be transparently compared with a key
[edit] Return value
1) Number of elements with key key.
2) Number of elements with key that compares equivalent to x.
[edit] Complexity
linear in the number of elements with key key on average, worst case linear in the size of the container.
[edit] Notes
| Feature-test macro | Value | Std | Feature |
|---|---|---|---|
__cpp_lib_generic_unordered_lookup |
201811L |
(C++20) | Heterogeneous comparison lookup in unordered associative containers, overload (2) |
[edit] 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) << "; "; std::cout << '\n'; }
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;