std::ranges::find_first_of
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, | (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。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:
目录 |
[编辑]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 |
[编辑]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。
[编辑]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)。
[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>constexprranges::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));}}; inlineconstexpr find_first_of_fn find_first_of {}; |
[编辑]示例
#include <algorithm>#include <iostream>#include <iterator> int main(){namespace rng = std::ranges; constexprstaticauto haystack ={1, 2, 3, 4};constexprstaticauto needles ={0, 3, 4, 3}; constexprauto found1 = rng::find_first_of(haystack.begin(), haystack.end(), needles.begin(), needles.end()); static_assert(std::distance(haystack.begin(), found1)==2); constexprauto found2 = rng::find_first_of(haystack, needles); static_assert(std::distance(haystack.begin(), found2)==2); constexprstaticauto negatives ={-6, -3, -4, -3};constexprauto not_found = rng::find_first_of(haystack, negatives); static_assert(not_found == haystack.end()); constexprauto 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;};constexprstaticauto p1 ={P{1, -1}, P{2, -2}, P{3, -3}, P{4, -4}};constexprstaticauto 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:constauto 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}。
[编辑]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) |