std::count() algorithm
- od C++20
- od C++17
- do C++17
// (1)
template< class InputIt, class T >
constexpr typename iterator_traits<InputIt>::difference_type
count( InputIt first, InputIt last, const T& value );
// (2)
template< class ExecutionPolicy, class ForwardIt, class T >
typename iterator_traits<ForwardIt>::difference_type
count( ExecutionPolicy&& policy,
ForwardIt first, ForwardIt last, const T& value );
// (1)
template< class InputIt, class T >
typename iterator_traits<InputIt>::difference_type
count( InputIt first, InputIt last, const T& value );
// (2)
template< class ExecutionPolicy, class ForwardIt, class T >
typename iterator_traits<ForwardIt>::difference_type
count( ExecutionPolicy&& policy,
ForwardIt first, ForwardIt last, const T& value );
// (1)
template< class InputIt, class T >
typename iterator_traits<InputIt>::difference_type
count( InputIt first, InputIt last, const T& value );
Returns the number of elements equal to value
in the range [first1
; last1
).
- (1) Counts the elements that are equal to value (using
operator==
). - (2) Same as (1), but executed according to policy.
Overload Resolution
These overloads participate in overload resolution only if
std::is_execution_policy_v<std::decay_t<ExecutionPolicy>>
(do C++20)std::is_execution_policy_v<std::remove_cvref_t<ExecutionPolicy>>
(od C++20) istrue
.
Parameters
first last | The range of elements to examine. |
value | The value to search for. |
policy | The execution policy to use. See execution policy for details. |
Type requirements
InputIt | LegacyInputIterator |
ForwardIt | LegacyForwardIterator |
Return value
- (1 - 2) The number of elements that are equal to
value
.
Complexity
Given N
as std::distance(first, last)
:
- (1 - 2) exactly N comparisons with
value
usingoperator==
Exceptions
The overloads with a template parameter named ExecutionPolicy
report errors as follows:
- If execution of a function invoked as part of the algorithm throws an exception and
ExecutionPolicy
is one of the standard policies,std::terminate
is called. For any otherExecutionPolicy
, the behavior is implementation-defined. - If the algorithm fails to allocate memory,
std::bad_alloc
is thrown.
Possible implementation
count (1)
template<class InputIt, class T>
typename iterator_traits<InputIt>::difference_type
count(InputIt first, InputIt last, const T& value)
{
typename iterator_traits<InputIt>::difference_type ret = 0;
for (; first != last; ++first)
if (*first == value)
++ret;
return ret;
}
Notes
If you want to obtain the number of elements in range [first
; last
) without any additional criteria, use std::distance
Examples
#include <algorithm>
#include <array>
#include <iostream>
#include <iterator>
int main()
{
constexpr std::array v {1, 2, 3, 4, 4, 3, 7, 8, 9, 10};
std::cout << "v: ";
std::copy(v.cbegin(), v.cend(), std::ostream_iterator<int>(std::cout, " "));
std::cout << '\n';
// determine how many integers match a target value.
for (const int target: {3, 4, 5})
{
const int num_items = std::count(v.cbegin(), v.cend(), target);
std::cout << "number: " << target << ", count: " << num_items << '\n';
}
// use a lambda expression to count elements divisible by 4.
int count_div4 = std::count_if(v.begin(), v.end(), [](int i) { return i % 4 == 0; });
std::cout << "numbers divisible by four: " << count_div4 << '\n';
// A simplified version of `distance` with O(N) complexity:
auto distance = [](auto first, auto last)
{
return std::count_if(first, last, [](auto) { return true; });
};
static_assert(distance(v.begin(), v.end()) == 10);
}
v: 1 2 3 4 4 3 7 8 9 10
number: 3, count: 2
number: 4, count: 2
number: 5, count: 0
numbers divisible by four: 3
Hover to see the original license.