std::ranges::search
| 定义于头文件 <algorithm>
|
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| 调用签名 |
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template< std::forward_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 ranges::subrange<I1>
search( I1 first1, S1 last1, I2 first2, S2 last2, Pred pred = {},
Proj1 proj1 = {}, Proj2 proj2 = {} );
|
(1) | (自 C++20 起) |
template< ranges::forward_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_subrange_t<R1>
search( R1&& r1, R2&& r2, Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {} );
|
(2) | (自 C++20 起) |
[first2, last2)在范围[first1, last1)中。元素通过二元谓词pred在分别用proj2和proj1投影后进行比较。r1作为第一源范围,r2作为第二源范围,如同使用ranges::begin(r1)作为first1,ranges::end(r1)作为last1,ranges::begin(r2)作为first2,ranges::end(r2)作为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 niebloids),即:
参数
| first1, last1 | - | 定义待检查元素范围的迭代器-哨位对(称为 haystack ) |
| first2, last2 | - | 定义待查找元素范围的迭代器-哨位对(称为 needle ) |
| r1 | - | 待检查的元素范围(称为 haystack ) |
| r2 | - | 待查找的元素范围(称为 needle ) |
| pred | - | 应用于投影元素的二元谓词 |
| proj1 | - | 应用于第一个范围元素的投影 |
| proj2 | - | 应用于第二个范围元素的投影 |
返回值
[
first2
,
last2
)
(称为
needle
)在范围
[
first1
,
last1
)
(称为
haystack
)中的首次出现位置,其中已分别对两个序列的元素应用投影
proj1
和
proj2
,并随后应用二元谓词
pred
来比较投影后的元素。
如果未找到该序列,则返回 ranges:: subrange { last1, last1 } 。
如果要搜索的范围(称为 needle )为空,即 first2 == last2 ,则返回 ranges:: subrange { first1, first1 } 。复杂度
最多进行
S * N
次对应谓词和每个投影的应用,其中
(1)
S
=
ranges::
distance
(
first2, last2
)
且
N
=
ranges::
distance
(
first1, last1
)
;
(2)
S
=
ranges::
distance
(
r2
)
且
N
=
ranges::
distance
(
r1
)
。
可能的实现
struct search_fn { template<std::forward_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 ranges::subrange<I1> operator()(I1 first1, S1 last1, I2 first2, S2 last2, Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {}) const { for (;; ++first1) { I1 it1 = first1; for (I2 it2 = first2;; ++it1, ++it2) { if (it2 == last2) return {first1, it1}; if (it1 == last1) return {it1, it1}; if (!std::invoke(pred, std::invoke(proj1, *it1), std::invoke(proj2, *it2))) break; } } } template<ranges::forward_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_subrange_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 search_fn search {}; |
示例
#include <algorithm> #include <cctype> #include <iostream> #include <iterator> #include <string_view> using namespace std::literals; void print(int id, const auto& haystack, const auto& needle, const auto& found) { std::cout << id << ") search(\"" << haystack << "\", \"" << needle << "\"); "; const auto first = std::distance(haystack.begin(), found.begin()); const auto last = std::distance(haystack.begin(), found.end()); if (found.empty()) std::cout << "not found;"; else { std::cout << "found: \""; for (const auto x : found) std::cout << x; std::cout << "\";"; } std::cout << " subrange: {" << first << ", " << last << "}\n"; } int main() { constexpr auto haystack {"abcd abcd"sv}; constexpr auto needle {"bcd"sv}; // 使用迭代器对 begin()/end() 进行搜索: constexpr auto found1 = std::ranges::search( haystack.begin(), haystack.end(), needle.begin(), needle.end()); print(1, haystack, needle, found1); // 使用范围 r1, r2 进行搜索: constexpr auto found2 = std::ranges::search(haystack, needle); print(2, haystack, needle, found2); // 'needle' 范围为空: constexpr auto none {""sv}; constexpr auto found3 = std::ranges::search(haystack, none); print(3, haystack, none, found3); // 'needle' 将不会被找到: constexpr auto awl {"efg"sv}; constexpr auto found4 = std::ranges::search(haystack, awl); print(4, haystack, awl, found4); // 使用自定义比较器和投影进行搜索: constexpr auto bodkin {"234"sv}; auto found5 = std::ranges::search(haystack, bodkin, [](const int x, const int y) { return x == y; }, // pred [](const int x) { return std::toupper(x); }, // proj1 [](const int y) { return y + 'A' - '1'; }); // proj2 print(5, haystack, bodkin, found5); }
输出:
1) search("abcd abcd", "bcd"); found: "bcd"; subrange: {1, 4}
2) search("abcd abcd", "bcd"); found: "bcd"; subrange: {1, 4}
3) search("abcd abcd", ""); not found; subrange: {0, 0}
4) search("abcd abcd", "efg"); not found; subrange: {9, 9}
5) search("abcd abcd", "234"); found: "bcd"; subrange: {1, 4}
参见
(C++20) |
查找首对相邻的相等(或满足给定谓词)的元素 (算法函数对象) |
(C++20)(C++20)(C++20) |
查找满足特定条件的首个元素 (算法函数对象) |
(C++20) |
查找特定范围内元素的最后一个序列 (算法函数对象) |
(C++20) |
搜索元素集合中的任意一个元素 (算法函数对象) |
(C++23)(C++23) |
检查范围是否包含给定元素或子范围 (算法函数对象) |
(C++20) |
确定一个序列是否为另一个序列的子序列 (算法函数对象) |
(C++20) |
查找两个范围第一个出现差异的位置 (算法函数对象) |
(C++20) |
搜索范围内元素连续出现指定次数的首次出现 (算法函数对象) |
| 搜索元素集合的首次出现 (函数模板) |