std::ranges::count() 算法
- 自 C++20 起
- 简化
- 详细
// (1)
constexpr std::iter_difference_t<I>
count( I first, S last, const T& value, Proj proj = {} );
// (2)
constexpr ranges::range_difference_t<R>
count( R&& r, const T& value, Proj proj = {} );
参数类型是泛型的,并具有以下约束
I
-std::input_iterator
S
-std::sentinel_for<I>
R
-std::ranges::input_range
T
- (无)P
- (无)
对于所有重载,Proj
模板参数的默认类型为 std::identity
。
此外,每个重载都有以下约束
- (1) -
indirect_binary_predicate<ranges::equal_to, projected<I, Proj>, const T*>
- (2) -
indirect_binary_predicate<ranges::equal_to, projected<ranges::iterator_t<R>, Proj>, const T*>
(为方便阅读,此处省略了 std::
命名空间)
// (1)
template<
std::input_iterator I,
std::sentinel_for<I> S,
class T,
class Proj = std::identity
>
requires std::indirect_binary_predicate<ranges::equal_to, std::projected<I, Proj>, const T*>
constexpr std::iter_difference_t<I>
count( I first, S last, const T& value, Proj proj = {} );
// (2)
template<
ranges::input_range R,
class T,
class Proj = std::identity
>
requires std::indirect_binary_predicate<ranges::equal_to, std::projected<ranges::iterator_t<R>, Proj>, const T*>
constexpr ranges::range_difference_t<R>
count( R&& r, const T& value, Proj proj = {} );
返回给定范围内等于 value
的元素数量。
- (1) 计算等于
value
的元素(使用operator==
)。 - (2) 与 (1) 相同,但使用
r
作为源范围,如同使用ranges::begin(r)
作为first
和ranges::end(r)
作为last
。
本页描述的函数类实体是niebloids。
参数
first last | 要检查的元素范围。 |
r | 要检查的元素范围。 |
值 | 要搜索的值。 |
proj | 应用于元素的投影。 |
返回值
- (1 - 2) 等于
value
的元素数量。
复杂度
给定 N
为 ranges::distance(first, last)
- (1 - 2) 恰好
N
次比较和投影。
异常
(无)
可能的实现
ranges::count
struct count_fn
{
template<std::input_iterator I, std::sentinel_for<I> S,
class T, class Proj = std::identity>
requires std::indirect_binary_predicate<ranges::equal_to, std::projected<I, Proj>,
const T*>
constexpr std::iter_difference_t<I>
operator()(I first, S last, const T& value, Proj proj = {}) const
{
std::iter_difference_t<I> counter = 0;
for (; first != last; ++first)
if (std::invoke(proj, *first) == value)
++counter;
return counter;
}
template<ranges::input_range R, class T, class Proj = std::identity>
requires std::indirect_binary_predicate<ranges::equal_to,
std::projected<ranges::iterator_t<R>, Proj>,
const T*>
constexpr ranges::range_difference_t<R>
operator()(R&& r, const T& value, Proj proj = {}) const
{
return (*this)(ranges::begin(r), ranges::end(r), value, std::ref(proj));
}
};
inline constexpr count_fn count;
备注
如果你想获取范围 [first
; last
) 或 r
中元素的数量,且没有任何附加条件,请使用 ranges::distance
。
示例
Main.cpp
#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(v.begin(), v.end(), target1);
int num_items2 = ranges::count(v, target2);
std::cout << "number: " << target1 << " count: " << num_items1 << '\n';
std::cout << "number: " << target2 << " count: " << num_items2 << '\n';
// use a lambda expression to count elements divisible by 3.
int num_items3 = ranges::count_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 elements divisible by 11.
int num_items11 = ranges::count_if(v, [](int i) {return i % 11 == 0;});
std::cout << "number divisible by eleven: " << num_items11 << '\n';
}
输出
number: 3 count: 2
number: 5 count: 0
number divisible by three: 3
number divisible by eleven: 0