std::ranges::adjacent_find_C++中文网

C++ 参考手册

std::ranges::adjacent_find

          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 = {} );
(1) (C++20 èµ·)
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 = {} );
(2) (C++20 èµ·)

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，或以某些特殊编译器扩展实现。

参数

first, last - 要检验的范围
r - 要检验的范围
pred - 应用到投影后元素的谓词
proj - 应用到元素的投影

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;

示例

运行此代码
#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 at: "
                  << ranges::distance(v1.begin(), 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) << '\n';
    }
}

输出： 

The first adjacent pair of equal elements at: 4
The last element in the non-decreasing subsequence is at: 7

参阅