检测均匀和奇数C 矢量迭代器

Detecting even and odd C++ vector iterators

本文关键字：迭代器 检测      更新时间：2023-10-16

我搜索了，我很惊讶这尚未问。我知道如何使用简单的循环进行操作，使用矢量迭代器呢？

for( std::vector<int>::iterator it = somevector.begin(); it != somevector.end(); ++it )
{
    //Conditions stating a certain vector has an even or odd index.
}

对不起，我的意思是要检测到向量的索引是否奇怪甚至是。

这将是一种简单的方法：

{
  bool is_even = true;
  for (const auto& v: somevector) {
    if (is_even) even_handler(v);
    else         odd_handler(v);
    is_even = !is_even;
  }
}

想要一个更复杂的解决方案吗？没问题：

#include <iostream>
#include <string>
#include <utility>
#include <vector>
using std::next;
template<typename Iter, typename Func, typename...Funcs>
void RotateHandlers(Iter b, Iter e, Func f, Funcs...fs) {
    if (b != e) {
        f(*b);
        RotateHandlers(next(b), e, fs..., f);
    }
}
int main() {
    std::vector<std::string> v({"Hello", "world", "it's", "really", "great", "to", "be", "here"});
    RotateHandlers(v.begin(), v.end(),
      [](const std::string& s){std::cout << "First|" << s << std::endl;},
      [](const std::string& s){std::cout << "Then |" << s << std::endl;},
      [](const std::string& s){std::cout << "And  |" << s << std::endl
                                         << "     |" << std::string(s.size(), '-') << std::endl;}
    );
    return 0;
}

在此处查看：http：//ideone.com/jmlv5f

我会猜测您的意思是您要检测当前索引是否均匀或奇怪：

#include <iostream>
#include <iterator>
#include <vector>
int main()
{
    std::vector<int> somevector;
    somevector.push_back(1);
    somevector.push_back(2);
    somevector.push_back(4);
    somevector.push_back(8);
    somevector.push_back(111605);
    for (auto it = somevector.begin(); it != somevector.end(); ++it)
    {
        // current index
        const auto index = std::distance(somevector.begin(), it);
        if ((index % 2) == 0) // even
        {
            std::cout << "Index " << index << " (even) is: " << *it;
        }
        else
        {
            std::cout << "Index " << index << " (odd) is: " << *it;
        }
        std::cout << std::endl;
    }
}

您可以通过std::distance获得迭代器之间的距离。（索引与开始距离。）

如果我正确理解[在上次编辑之前]的问题，则选项是：

bool is_odd(const std::vector<int> &somevector) {
    for( std::vector<int>::iterator it = somevector.begin(); it != somevector.end(); ++it ) {
        //Conditions stating a certain vector is even or odd.
        if (*it % 2 == 0) {
          return false; 
        }
    }
    return true;
}

分别用于"偶数向量"。

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