std::ranges::find_first_of
定义在头文件 <algorithm> 中 |
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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) | (自 C++20) |
template< ranges::input_range R1, ranges::forward_range R2, class Pred = ranges::equal_to, |
(2) | (自 C++20) |
[
first1,
last1)
中的 任何 范围 [
first2,
last2)
中的元素。 使用二元谓词 pred 比较投影元素。此页面上描述的类函数实体是 niebloids,即
在实践中,它们可以实现为函数对象,或者使用特殊的编译器扩展。
内容 |
[编辑] 参数
first1, last1 | - | 要检查的元素范围(也称为haystack) |
first2, last2 | - | 要搜索的元素范围(也称为needles) |
r1 | - | 要检查的元素范围(也称为haystack) |
r2 | - | 要搜索的元素范围(也称为needles) |
pred | - | 用于比较元素的二元谓词 |
proj1 | - | 要应用于第一个范围中元素的投影 |
proj2 | - | 要应用于第二个范围中元素的投影 |
[编辑] 返回值
指向范围 [
first1,
last1)
中第一个元素的迭代器,该元素在投影后等于范围 [
first2,
last2)
中的元素。如果未找到此类元素,则返回与 last1 相等的迭代器。
[编辑] 复杂度
最多 S * N 次谓词和每个投影的应用,其中
(1) S = ranges::distance(first2, last2) 和 N = ranges::distance(first1, last1);
(2) S = ranges::distance(r2) 和 N = ranges::distance(r1)。
[编辑] 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 {}; |
[编辑] 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; }); // uses a binary comparator 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}}; // Compare only P::y data members by projecting them: 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 equivalent element {3, -3} was found at position 2.
[编辑] 另请参阅
搜索一组元素中的任何一个 (函数模板) | |
(C++20) |
查找第一个相邻的两个相等项(或满足给定谓词的项) (niebloid) |
(C++20)(C++20)(C++20) |
查找满足特定条件的第一个元素 (niebloid) |
(C++20) |
查找某个范围内最后一个元素序列 (niebloid) |
(C++20) |
搜索元素范围的第一个出现位置 (niebloid) |
(C++20) |
搜索某个范围内连续出现某个元素的第一个位置 (niebloid) |