std::ranges::find_first_of
来自cppreference.com
<tbody>
</tbody>
| 在标头 <algorithm> 定义
|
||
| 调用签名 |
||
template< std::input_iterator I1, std::sentinel_for<I1> S1, std::forward_iterator I2, std::sentinel_for<I2> S2, class Pred = ranges::equal_to, class Proj1 = std::identity, class Proj2 = std::identity > requires std::indirectly_comparable<I1, I2, Pred, Proj1, Proj2> constexpr I1 find_first_of( I1 first1, S1 last1, I2 first2, S2 last2, Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {} ); |
(1) | (C++20 起) |
template< ranges::input_range R1, ranges::forward_range R2, class Pred = ranges::equal_to, class Proj1 = std::identity, class Proj2 = std::identity > requires std::indirectly_comparable<ranges::iterator_t<R1>, ranges::iterator_t<R2>, Pred, Proj1, Proj2> constexpr ranges::borrowed_iterator_t<R1> find_first_of( R1&& r1, R2&& r2, Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {} ); |
(2) | (C++20 起) |
1) 在范围
[first1, last1) 中搜索范围 [first2, last2) 中的任何元素，分别以 proj1 与 proj2 投影范围。用二元谓词 pred 比较投影后的元素。2) 同 (1)，但以
r1 为第一源范围并以 r2 为第二源范围，如同以 ranges::begin(r1) 为 first1，以 ranges::end(r1) 为 last1，以 ranges::begin(r2) 为 first2，并以 ranges::end(r2) 为 last2。此页面上描述的函数式实体是算法函数对象（非正式地称为 niebloid），即：
参数
| first1, last1 | - | 要检验的（又称草堆）元素范围的迭代器-哨位对 |
| first2, last2 | - | 要搜索的（又称针）元素范围的迭代器-哨位对 |
| r1 | - | 要检验的元素范围（又称草堆） |
| r2 | - | 要搜索的元素范围（又称针） |
| pred | - | 比较元素的二元谓词 |
| proj1 | - | 应用到第一范围中元素的投影 |
| proj2 | - | 应用到第二范围中元素的投影 |
返回值
指向范围 [first1, last1) 中首个在投影后等于范围 [first2, last2) 中某个元素的迭代器。若找不到这种元素，则返回等于 last1 的迭代器。
复杂度
至多应用 (S * N) 次比较和各自的投影，其中
(1) S = ranges::distance(first2, last2) 而 N = ranges::distance(first1, last1)；
(2) S = ranges::size(r2) 而 N = ranges::size(r1)。
可能的实现
struct find_first_of_fn
{
template<std::input_iterator I1, std::sentinel_for<I1> S1,
std::forward_iterator I2, std::sentinel_for<I2> S2,
class Pred = ranges::equal_to,
class Proj1 = std::identity,
class Proj2 = std::identity>
requires std::indirectly_comparable<I1, I2, Pred, Proj1, Proj2>
constexpr I1 operator()(I1 first1, S1 last1, I2 first2, S2 last2, Pred pred = {},
Proj1 proj1 = {}, Proj2 proj2 = {}) const
{
for (; first1 != last1; ++first1)
for (auto i = first2; i != last2; ++i)
if (std::invoke(pred, std::invoke(proj1, *first1), std::invoke(proj2, *i)))
return first1;
return first1;
}
template<ranges::input_range R1, ranges::forward_range R2,
class Pred = ranges::equal_to,
class Proj1 = std::identity,
class Proj2 = std::identity>
requires std::indirectly_comparable<ranges::iterator_t<R1>,
ranges::iterator_t<R2>,
Pred, Proj1, Proj2>
constexpr ranges::borrowed_iterator_t<R1>
operator()(R1&& r1, R2&& r2, Pred pred = {},
Proj1 proj1 = {}, Proj2 proj2 = {}) const
{
return (*this)(ranges::begin(r1), ranges::end(r1),
ranges::begin(r2), ranges::end(r2),
std::move(pred), std::move(proj1), std::move(proj2));
}
};
inline constexpr find_first_of_fn find_first_of {};
|
示例
运行此代码
#include <algorithm>
#include <iostream>
#include <iterator>
int main()
{
namespace rng = std::ranges;
constexpr static auto haystack = {1, 2, 3, 4};
constexpr static auto needles = {0, 3, 4, 3};
constexpr auto found1 = rng::find_first_of(haystack.begin(), haystack.end(),
needles.begin(), needles.end());
static_assert(std::distance(haystack.begin(), found1) == 2);
constexpr auto found2 = rng::find_first_of(haystack, needles);
static_assert(std::distance(haystack.begin(), found2) == 2);
constexpr static auto negatives = {-6, -3, -4, -3};
constexpr auto not_found = rng::find_first_of(haystack, negatives);
static_assert(not_found == haystack.end());
constexpr auto found3 = rng::find_first_of(haystack, negatives,
[](int x, int y) { return x == -y; }); // 使用二元比较器
static_assert(std::distance(haystack.begin(), found3) == 2);
struct P { int x, y; };
constexpr static auto p1 = {P{1, -1}, P{2, -2}, P{3, -3}, P{4, -4}};
constexpr static auto p2 = {P{5, -5}, P{6, -3}, P{7, -5}, P{8, -3}};
// 仅比较 P::y 数据成员，通过投影它们：
const auto found4 = rng::find_first_of(p1, p2, {}, &P::y, &P::y);
std::cout << "在位置 " << std::distance(p1.begin(), found4)
<< " 找到首个等价元素 {" << found4->x << ", " << found4->y
<< "}。\n";
}
输出：
在位置 2 找到首个等价元素 {3, -3}。
参阅
| 搜索一组元素中任一元素 (函数模板) | |
(C++20) |
查找首对相同（或满足给定谓词）的相邻元素 (算法函数对象) |
(C++20)(C++20)(C++20) |
查找首个满足特定条件的元素 (算法函数对象) |
(C++20) |
查找元素序列在特定范围中最后一次出现 (算法函数对象) |
(C++20) |
搜索元素范围的首次出现 (算法函数对象) |
(C++20) |
搜索元素在范围中首次连续若干次出现 (算法函数对象) |