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调用签名
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template< std::forward_iterator I, std::sentinel_for<I> S, class Proj = std::identity,
std::indirect_binary_predicate<
std::projected<I,
Proj>,
std::projected<I,
Proj>> Pred =
ranges::equal_to >
constexpr I adjacent_find(
I first, S last, Pred pred = {}, Proj proj = {} );
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(1) |
(C++20 起) |
template< ranges::forward_range R, class Proj
= std::identity,
std::indirect_binary_predicate<
std::projected<ranges::iterator_t<R>, Proj>,
std::projected<ranges::iterator_t<R>, Proj>> Pred =
ranges::equal_to >
constexpr ranges::borrowed_iterator_t<R>
adjacent_find( R&& r, Pred pred =
{}, Proj proj = {} );
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(2) |
(C++20 起) |
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搜索范围 [first, last) 中二个连续的相等元素。
1) 用 pred （在以投影 proj
投影后）比较元素。
2) 同 (1) ，但以 r
为源范围，如同以 ranges::begin(r) 为 first 并以 ranges::end(r) 为 last 。
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
| first, last
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要检验的范围
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| r
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-
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要检验的范围
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| pred
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-
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应用到投影后元素的谓词
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| proj
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-
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应用到元素的投影
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返回值
指向首对等同元素的前一者的迭代器，即首个使得 bool(std::invoke(pred, std::invoke(proj1, *it), std::invoke(proj, *(it + 1)))) 为 true 的 it 。
若找不到这种元素，则返回等于 last 的迭代器。
复杂度
准确应用 min((result-first)+1, (last-first)-1) 次谓词与投影，其中 result 是返回值。
可能的实现
struct adjacent_find_fn {
template< std::forward_iterator I, std::sentinel_for<I> S, class Proj = std::identity,
std::indirect_binary_predicate<
std::projected<I, Proj>,
std::projected<I, Proj>> Pred = ranges::equal_to >
constexpr I operator()( I first, S last, Pred pred = {}, Proj proj = {} ) const
{
if (first == last) {
return first;
}
auto next = ranges::next(first);
for (; next != last; ++next, ++first) {
if (std::invoke(pred, std::invoke(proj, *first), std::invoke(proj, *next))) {
return first;
}
}
return first;
}
template< ranges::forward_range R, class Proj = std::identity,
std::indirect_binary_predicate<
std::projected<ranges::iterator_t<R>, Proj>,
std::projected<ranges::iterator_t<R>, Proj>> Pred = ranges::equal_to >
constexpr ranges::borrowed_iterator_t<R>
operator()( R&& r, Pred pred = {}, Proj proj = {} ) const
{
return (*this)(ranges::begin(r), ranges::end(r), std::ref(pred), std::ref(proj));
}
};
inline constexpr adjacent_find_fn adjacent_find;
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示例
#include <algorithm>
#include <iostream>
#include <vector>
#include <functional>
int main()
{
std::vector<int> v1{0, 1, 2, 3, 40, 40, 41, 41, 5};
// ^^ ^^
namespace ranges = std::ranges;
auto i1 = ranges::adjacent_find(v1.begin(), v1.end());
if (i1 == v1.end()) {
std::cout << "No matching adjacent elements\n";
} else {
std::cout << "The first adjacent pair of equal elements is at ["
<< ranges::distance(v1.begin(), i1) << "] == " << *i1 << '\n';
}
auto i2 = ranges::adjacent_find(v1, ranges::greater());
if (i2 == v1.end()) {
std::cout << "The entire vector is sorted in ascending order\n";
} else {
std::cout << "The last element in the non-decreasing subsequence is at ["
<< ranges::distance(v1.begin(), i2) << "] == " << *i2 << '\n';
}
}
输出：
The first adjacent pair of equal elements is at [4] == 40
The last element in the non-decreasing subsequence is at [7] == 41
参阅
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移除范围中的连续重复元素
(niebloid) |
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查找首对相邻的相同（或满足给定谓词的）元素
(函数模板) |