std::unordered_multiset<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
.
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 testing macro: __cpp_lib_generic_unordered_lookup
(for overload (2)).
Example
#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) |
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