在STL容器中查找迭代器索引 - 需要模板功能 [英] Find index of iterator in STL container - need template function

问题描述

  template< typename T，typename R> int find_index（const T& list，const R& value）; 
  

据我所知，有 find（）在STL中返回迭代器。我需要返回迭代器的索引（即使对于非索引容器，如 std :: list ）。我试过这个代码：

  template< typename T，typename R> 
 int find_index（const T& list，const R& value）
 {
 int index = 0; 
 for（T :: const_iterator it = list.begin（）; it！= list.end（）; it ++，index ++）
 if（（* it）== value）
 return指数; 
返回-1; 
 
 $ / code> 

但编译器在 it - 似乎不允许从模板化类型名称中获取 const_iterator 。我可以绕过吗？



在最坏的情况下，我可以传递开始和结束迭代器find_index参数，但它看起来不那么好。感谢优雅的解决方案。
解决方案

  for（typename T :: const_iterator it = list.begin（）; it！= list.end（）; ++ it，++ index）
  



<当使用依赖类型（取决于模板参数的类型）时，编译器不知道 const_iterator c $ c>是一个类型，直到它用一个具体类型实例化模板，它也可以是一个静态变量或其他。使用 typename 关键字，告诉他 const_iterator 确实是一个类型。

在C ++ 11中，还可以使用 auto 关键字绕过整个 typename 问题。

  for（auto it = list.begin（）; it！= list.end（）; ++ it ++ index）
  

---

如果您已经有了迭代器从另一个操作开始），你也可以计算从列表的开始到这个迭代器的距离：
$ b $ $ $ $ $ $ $ $ #include< iterator> ;

int index = std :: distance（list.begin（），it）;


但是由于这对 std :: list ，使用自制的 find_index 函数比 std :: find 后面跟着 std :: distance ，至少在表现方面。

I want to have a function with interface like this:

template<typename T, typename R> int find_index (const T& list, const R& value);

As I know, there is find() in STL that returns iterator. I need to return index of iterator (even for non-indexed containers such as std::list). I tried this code:

template<typename T, typename R>
int find_index (const T& list, const R& value)
{
    int index = 0;
    for (T::const_iterator it = list.begin(); it != list.end(); it++, index++)
        if ((*it) == value)
            return index;
    return -1;
}

But compiler shows error on it - seems like it is not allowed to get const_iterator from templated typename. Can I go around it?

At the worst case I can pass begin and end iterators to find_index arguments, but it looks not so fine. Would be thankful for elegant solution.

解决方案

for (typename T::const_iterator it = list.begin(); it != list.end(); ++it, ++index)

should solve your problem.

When using dependent types (types depending on template parameters), the compiler does not know that const_iterator is a type until it instantiates the template with a concrete type, it could also just be a static variable or whatever. Using the typename keyword, you tell him that const_iterator is really a type.

In C++11 you can also circumvent the whole typename issue using the auto keyword:

for (auto it = list.begin(); it != list.end(); ++it, ++index)

---

If you already have the iterator (maybe from some other operation), you can also just compute the distance from the list's begin to this iterator:

#include <iterator>

int index = std::distance(list.begin(), it);

But since this has linear complexity for a std::list, using your self-made find_index function is a better idea than std::find followed by std::distance, at least performance-wise.

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