std::ranges::count_if() algorithm
- od C++20
- Simplified
- Detailed
// (1)
constexpr std::iter_difference_t<I>
count_if( I first, S last, Pred pred, Proj proj = {} );
// (2)
constexpr ranges::range_difference_t<R>
count_if( R&& r, Pred pred, Proj proj = {} );
The type of arguments are generic and have the following constraints:
I
-std::input_iterator
S
-std::sentinel_for<I>
R
-std::ranges::input_range
Pred
:- (1) -
std::indirect_unary_predicate<std::projected<I, Proj>>
- (2) -
std::projected<ranges::iterator_t<R>, Proj>>
- (1) -
P
- (none)
The Proj
template argument has a default type of std::identity
for all overloads.
// (1)
template<
std::input_iterator I,
std::sentinel_for<I> S,
std::indirect_unary_predicate<std::projected<I, Proj>> Pred
class Proj = std::identity
>
constexpr std::iter_difference_t<I>
count_if( I first, S last, Pred pred, Proj proj = {} );
// (2)
template<
ranges::input_range R,
std::indirect_unary_predicate<std::projected<I, Proj>> Pred
class Proj = std::identity
>
constexpr ranges::range_difference_t<R>
count_if( R&& r, Pred pred, Proj proj = {} );
Returns the number of elements in the given range that satisfy specific criteria.
- (1) Counts elements for which predicate
p
returnstrue
. - (2) Same as (1), but uses
r
as the source range, as if usingranges::begin(r)
asfirst
andranges::end(r)
aslast
.
The function-like entities described on this page are niebloids.
Parameters
first last | The range of elements to examine. |
r | The range of elements to examine. |
pred | Predicate to apply to the projected elements. |
proj | Projection to apply to the elements. |
Return value
- (1 - 2) The number of elements for which the predicate
p
holdstrue
.
Complexity
Given N
as ranges::distance(first, last)
:
- (1 - 2) exactly
N
calls top
and projections.
Exceptions
(none)
Possible implementation
ranges::count_if
struct count_if_fn
{
template<std::input_iterator I, std::sentinel_for<I> S,
class Proj = std::identity,
std::indirect_unary_predicate<std::projected<I, Proj>> Pred>
constexpr std::iter_difference_t<I>
operator()(I first, S last, Pred pred, Proj proj = {}) const
{
std::iter_difference_t<I> counter = 0;
for (; first != last; ++first)
if (std::invoke(pred, std::invoke(proj, *first)))
++counter;
return counter;
}
template<ranges::input_range R, class Proj = std::identity,
std::indirect_unary_predicate<
std::projected<ranges::iterator_t<R>, Proj>> Pred>
constexpr ranges::range_difference_t<R>
operator()(R&& r, Pred pred, Proj proj = {}) const
{
return (*this)(ranges::begin(r), ranges::end(r),
std::ref(pred), std::ref(proj));
}
};
inline constexpr count_if_fn count_if;
Notes
If you want to obtain the number of elements in range [first
; last
) or r
, without any additional criteria, use ranges::distance
.
Examples
#include <algorithm>
#include <iostream>
#include <vector>
int main()
{
std::vector<int> v {1, 2, 3, 4, 4, 3, 7, 8, 9, 10};
namespace ranges = std::ranges;
// determine how many integers in a std::vector match a target value.
int target1 = 3;
int target2 = 5;
int num_items1 = ranges::count_if(v.begin(), v.end(), target1);
int num_items2 = ranges::count_if(v, target2);
std::cout << "number: " << target1 << " count_if: " << num_items1 << '\n';
std::cout << "number: " << target2 << " count_if: " << num_items2 << '\n';
// use a lambda expression to count_if elements divisible by 3.
int num_items3 = ranges::count_if_if(v.begin(), v.end(), [](int i) {return i % 3 == 0;});
std::cout << "number divisible by three: " << num_items3 << '\n';
// use a lambda expression to count_if elements divisible by 11.
int num_items11 = ranges::count_if_if(v, [](int i) {return i % 11 == 0;});
std::cout << "number divisible by eleven: " << num_items11 << '\n';
}
number: 3 count_if: 2
number: 5 count_if: 0
number divisible by three: 3
number divisible by eleven: 0
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