在C ++ 11中找到元素的位置基于范围的for循环？ [英] Find position of element in C++11 range-based for loop?

问题描述

假设我有以下代码：

  vector< int&列表; 
 for（auto& elem：list）{
 int i = elem; 
} 
  

我可以找到 elem

解决方案

是的，只需要一些按摩; / p>

诀窍是使用composition：而不是直接在容器上迭代，你可以使用索引来压缩它。

专用拉链代码：

  template< typename T& 
 struct iterator_extractor {typedef typename T :: iterator type; }; 
 
 template< typename T> 
 struct iterator_extractor< T const> {typedef typename T :: const_iterator type; }; 
 
 
 template< typename T> 
 class Indexer {
 public：
 class iterator {
 typedef typename iterator_extractor< T> :: type inner_iterator; 
 
 typedef typename std :: iterator_traits< inner_iterator> :: reference inner_reference; 
 public：
 typedef std :: pair< size_t，inner_reference>参考; 
 
迭代器（inner_iterator it）：_pos（0），_it（it）{} 
 
引用运算符*（）const {return reference（_pos，* _it）; } 
 
迭代器& operator ++（）{++ _ pos; ++ _ it; return * this; } 
 iterator operator ++（int）{iterator tmp（* this）; ++ * this; return tmp; } 
 
 bool operator ==（iterator const& it）const {return _it == it._it; } 
 bool operator！=（iterator const& it）const {return！（* ​​this == it）; } 
 
 private：
 size_t _pos; 
 inner_iterator _it; 
}; 
 
索引器（T& t）：_container（t）{} 
 
迭代器begin（）const {return iterator（_container.begin（））; } 
 iterator end（）const {return iterator（_container.end（））; } 
 
 private：
 T& _容器; 
}; // class Indexer 
 
 template< typename T> 
 Indexer< T>索引（T& t）{return Indexer T（t）; } 
  

并使用它：

  #include< iostream> 
 #include< iterator> 
 #include< limits> 
 #include< vector> 
 
 //这里的拉链代码
 
 int main（）{
 std :: vector< int> v {1，2，3，4，5，6，7，8，9}; 
 
 for（auto p：index（v））{
 std :: cout< p.first<< ：<< p.second<< \\\
; 
} 
} 
  

您可以在 ideone ，但它缺少范围循环支持，因此不太漂亮。

编辑：

记住我应该经常检查Boost.Range。很遗憾，没有 zip 范围，但我发现了一个perl： boost :: adapters :: indexed 。但是它需要访问迭代器来拉取索引。耻辱：x

否则使用 counting_range 和一个通用的 zip 可以做一些有趣的事情...

在理想的世界里，我会想象：

  int main（）{
 std :: vector< int> v {1，2，3，4，5，6，7，8，9}; 
 
 for（auto tuple：zip（iota（0），v））{
 std :: cout< tuple.at 0（）<< ：<< tuple.at 1（）<< \\\
; 
} 
} 
  

使用 zip 自动创建一个作为引用元组范围的视图， iota（0）只需创建一个从 0开始的 ，并且只计入无穷大（或者好的，其类型的最大值...）。

Assume I have the following code:

vector<int> list;
for(auto& elem:list) {
    int i = elem;
}

Can I find the position of elem in the vector without maintaining a separate iterator?

解决方案

Yes you can, it just take some massaging ;)

The trick is to use composition: instead of iterating over the container directly, you "zip" it with an index along the way.

Specialized zipper code:

template <typename T>
struct iterator_extractor { typedef typename T::iterator type; };

template <typename T>
struct iterator_extractor<T const> { typedef typename T::const_iterator type; };


template <typename T>
class Indexer {
public:
    class iterator {
        typedef typename iterator_extractor<T>::type inner_iterator;

        typedef typename std::iterator_traits<inner_iterator>::reference inner_reference;
    public:
        typedef std::pair<size_t, inner_reference> reference;

        iterator(inner_iterator it): _pos(0), _it(it) {}

        reference operator*() const { return reference(_pos, *_it); }

        iterator& operator++() { ++_pos; ++_it; return *this; }
        iterator operator++(int) { iterator tmp(*this); ++*this; return tmp; }

        bool operator==(iterator const& it) const { return _it == it._it; }
        bool operator!=(iterator const& it) const { return !(*this == it); }

    private:
        size_t _pos;
        inner_iterator _it;
    };

    Indexer(T& t): _container(t) {}

    iterator begin() const { return iterator(_container.begin()); }
    iterator end() const { return iterator(_container.end()); }

private:
    T& _container;
}; // class Indexer

template <typename T>
Indexer<T> index(T& t) { return Indexer<T>(t); }

And using it:

#include <iostream>
#include <iterator>
#include <limits>
#include <vector>

// Zipper code here

int main() {
    std::vector<int> v{1, 2, 3, 4, 5, 6, 7, 8, 9};

    for (auto p: index(v)) {
        std::cout << p.first << ": " << p.second << "\n";
    }
}

You can see it at ideone, though it lacks the for-range loop support so it's less pretty.

EDIT:

Just remembered that I should check Boost.Range more often. Unfortunately no zip range, but I did found a perl: boost::adaptors::indexed. However it requires access to the iterator to pull of the index. Shame :x

Otherwise with the counting_range and a generic zip I am sure it could be possible to do something interesting...

In the ideal world I would imagine:

int main() {
    std::vector<int> v{1, 2, 3, 4, 5, 6, 7, 8, 9};

    for (auto tuple: zip(iota(0), v)) {
        std::cout << tuple.at<0>() << ": " << tuple.at<1>() << "\n";
    }
}

With zip automatically creating a view as a range of tuples of references and iota(0) simply creating a "false" range that starts from 0 and just counts toward infinity (or well, the maximum of its type...).

这篇关于在C ++ 11中找到元素的位置基于范围的for循环？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！