std::unordered_map count() method
- od C++20
- od C++11
// (1) Non const version
iterator find( const Key& key );
// (2) Const version
const_iterator find( const Key& key ) const;
// (3) Non const version
template< class K > iterator find( const K& x );
// (4) Const version
template< class K > const_iterator find( const K& x ) const;
// (1) Non const version
iterator find( const Key& key );
// (2) Const version
const_iterator find( const Key& key ) 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 ifHash::is_transparent
andKeyEqual::is_transparent
are valid and each denotes a type. This assumes that suchHash
is callable with bothK
andKey
type, and that theKeyEqual
is transparent, which, together, allows calling this function without constructing an instance ofKey
.
Parameters
key
- key value of the elements to countx
- 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 <string>
#include <iostream>
#include <unordered_map>
int main()
{
std::unordered_map<int, std::string> dict = {
{1, "one"}, {6, "six"}, {3, "three"}
};
dict.insert({4, "four"});
dict.insert({5, "five"});
dict.insert({6, "six"});
std::cout << "dict: { ";
for (auto const& [key, value] : dict)
{
std::cout << "[" << key << "]=" << value << " ";
}
std::cout << "}\n\n";
for (int i{1}; i != 8; ++i)
{
std::cout << "dict.count(" << i << ") = " << dict.count(i) << '\n';
}
}
Possible output
dict: { [5]=five [4]=four [1]=one [6]=six [3]=three }
dict.count(1) = 1
dict.count(2) = 0
dict.count(3) = 1
dict.count(4) = 1
dict.count(5) = 1
dict.count(6) = 1
dict.count(7) = 0
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