std::ranges::find_end
定义于头文件 <algorithm>
|
||
调用签名 |
||
template< std::forward_iterator I1, std::sentinel_for<I1> S1, std::forward_iterator I2, std::sentinel_for<I2> S2, |
(1) | (C++20 起) |
template< ranges::forward_range R1, ranges::forward_range R2, class Pred = ranges::equal_to, |
(2) | (C++20 起) |
[first2, last2)
在范围 [first1, last1)
中的最后出现,在分别以proj1 与 proj2 投影后。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
, 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,或以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
[first2, last2)
在范围 [first1, last1)
中的最后一次出现(在以 proj1 与 proj2 投影后)。若 [first2, last2)
为空或找不到该序列,则等效地以 {last1, last1} 初始化返回值。复杂度
至多应用 S·(N-S+1) 次谓词和对应的每次投影,其中对于 (1) S 为 ranges::distance(first2, last2) 而 N 为 ranges::distance(first1, last1) ,或对于 (2) S 为 ranges::distance(r2) 而 N 为 ranges::distance(r1) 。
注解
若输入迭代器实现 std::bidirectional_iterator ,则实现可以通过从末尾到起始搜索改进效率。实现 std::random_access_iterator 可能提升比较速度。然而所有这些不改变最坏情况的理论复杂度。
可能的实现
struct find_end_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 { if (first2 == last2) { auto last_it = ranges::next(first1, last1); return {last_it, last_it}; } auto result = ranges::search( std::move(first1), last1, first2, last2, pred, proj1, proj2); if (result.empty()) return result; for (;;) { auto new_result = ranges::search( std::next(result.begin()), last1, first2, last2, pred, proj1, proj2); if (new_result.empty()) return result; else result = std::move(new_result); } } 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 find_end_fn find_end{}; |
示例
#include <algorithm> #include <array> #include <cctype> #include <iostream> #include <ranges> #include <string_view> void print(const auto haystack, const auto needle) { const auto pos = std::distance(haystack.begin(), needle.begin()); std::cout << "In \""; for (const auto c : haystack) { std::cout << c; } std::cout << "\" found \""; for (const auto c : needle) { std::cout << c; } std::cout << "\" at position [" << pos << ".." << pos + needle.size() << ")\n" << std::string(4 + pos, ' ') << std::string(needle.size(), '^') << '\n'; } int main() { using namespace std::literals; constexpr auto secret{"password password word..."sv}; constexpr auto wanted{"password"sv}; constexpr auto found1 = std::ranges::find_end( secret.cbegin(), secret.cend(), wanted.cbegin(), wanted.cend()); print(secret, found1); constexpr auto found2 = std::ranges::find_end(secret, "word"sv); print(secret, found2); const auto found3 = std::ranges::find_end(secret, "ORD"sv, [](const char x, const char y) { // 用二元谓词 return std::tolower(x) == std::tolower(y); }); print(secret, found3); const auto found4 = std::ranges::find_end(secret, "SWORD"sv, {}, {}, [](char c) { return std::tolower(c); }); // 投影第二范围 print(secret, found4); static_assert(std::ranges::find_end(secret, "PASS"sv).empty()); // => 找不到 }
输出:
In "password password word..." found "password" at position [9..17) ^^^^^^^^ In "password password word..." found "word" at position [18..22) ^^^^ In "password password word..." found "ord" at position [19..22) ^^^ In "password password word..." found "sword" at position [12..17) ^^^^^
参阅
在特定范围中寻找最后出现的元素序列 (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) |
寻找首个满足特定判别标准的元素 (niebloid) |
(C++20) |
搜索元素集合中的任一元素 (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) |