std::{{{1}}}::count
From cppreference.com
size_type count( const Key& key ) const; |
(1) | (since {std}) |
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
.Contents |
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.
Example
Run this code
#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
See also
finds element with specific key (public member function of std::{{{1}}} )
| |
(C++20) |
checks if the container contains element with specific key (public member function of std::{{{1}}} )
|
returns range of elements matching a specific key (public member function of std::{{{1}}} )
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