std::ranges::find_first_of
| 在标头 <algorithm> 定义
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| 调用签名 |
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| template< std::input_iterator I1, std::sentinel_for<I1> S1, std::forward_iterator I2, std::sentinel_for<I2> S2, |
(1) | (C++20 起) |
| template< ranges::input_range R1, ranges::forward_range R2, class Pred = ranges::equal_to, |
(2) | (C++20 起) |
[first1, last1) 中搜索任何范围 [first2, last2) 中的元素，分别以 proj1 与 proj2 投影范围。用二元谓词 pred 比较投影后的元素。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 。此页面上描述的仿函数实体是 niebloid，即：
实际上，它们能以函数对象，或者某些特殊编译器扩展实现。
参数
| first1, last1 | - | 要检验的元素范围（又称草堆） |
| first2, last2 | - | 要搜索的元素范围（又称针） |
| 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来自范围 [first2, last2) 中元素的迭代器。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
#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 << "First equivalent element {" << found4->x << ", " << found4->y << "} was found at position " << std::distance(p1.begin(), found4) << ".\n"; }
输出：
First equavalent element {3, -3} was found at position 2.参阅
| 搜索元素集合中的任意元素 (函数模板) | |
| (C++20) |
查找首对相邻的相同（或满足给定谓词的）元素 (niebloid) |
| (C++20)(C++20)(C++20) |
查找满足特定条件的的第一个元素 (niebloid) |
| (C++20) |
查找特定范围中最后出现的元素序列 (niebloid) |
| (C++20) |
搜索一个元素范围 (niebloid) |
| (C++20) |
在范围中搜索一定量的某个元素的连续副本 (niebloid) |