reverse_iterator适配器 [英] reverse_iterator adapter

问题描述

我正在尝试为我的迭代器和const_iterator类实现一个反向迭代器适配器，但有点麻烦。如果有人可以指导我完成这个，那将非常感谢！

I'm trying to implement a reverse-iterator adaptor for my iterator and const_iterator classes with a little bit of trouble. If anyone could guide me through this, that would be greatly appreciated!

我的想法是我应该能够从我的rbegin（）和rend创建一个反向迭代器（）函数调用

The idea is that I should be able to create a reverse-iterator from my rbegin() and rend() function calls

reverse_iterator rbegin();
reverse_iterator rend();
const_reverse_iterator rbegin() const;
const_reverse_iterator rend() const;

我在课堂上使用以下typedef：

I'm using the following typedef's in the class:

typedef btree_iterator<T> iterator;
typedef const_btree_iterator<T> const_iterator;
typedef reverse_btree_iterator<iterator> reverse_iterator;
typedef reverse_btree_iterator<const_iterator> const_reverse_iterator;

如您所见，我希望能够使用模板创建反向迭代器， reverse_iterator类是迭代器还是const_iterator。

As you can see, I would like to be able to create reverse-iterators using templates, giving the reverse_iterator class either an iterator or const_iterator.

不幸的是，这是我坚持的......

Unfortunately, it is this bit I'm stuck on...

下面是我目前的类定义，但有错误。

Below is the class definition that I currently have, with errors.

template <typename I> class reverse_btree_iterator {

  typedef ptrdiff_t                     difference_type;
  typedef bidirectional_iterator_tag    iterator_category;

public:

  reverse_btree_iterator() : base_(I()) {}
  template <typename T> reverse_btree_iterator(const btree_iterator<T>& rhs) : base_(rhs) {}

  I base() { return base_; }

  I::reference operator*() const;
  I::pointer operator->() const;
  I& operator++();
  I operator++(int);
  I& operator--();
  I operator--(int);
  bool operator==(const I& other) const;
  bool operator!=(const I& other) const;

private:

  I base_;

};

我以前从未使用过这样的模板，所以我很可能完全误解了它们可以使用...

I've never used templates like this before, so it is very likely I'm completely misunderstanding how they can be used...

由于我可以是迭代器或const_iterator，因此引用和指针的typedef在两个类之间有所不同。不编译的行是：

Since I can be an iterator or a const_iterator, the typedef of reference and pointer vary between the two classes. The lines that aren't compiling are these:

I::reference operator*() const;
I::pointer operator->() const;

我不知道如何让一个reverse_iterator类同时用于iterator和const_iterator我不能做I :: reference和I :: pointer。我也尝试在前面添加模板，因为它们在迭代器类（例如）中被定义为：

I'm not sure how else I can make the one reverse_iterator class work for both iterator and const_iterator if I'm not able to do I::reference and I::pointer. I also tried adding template in front of those, since they are defined in the iterator class (for example) as:

typedef T*                            pointer;
typedef T&                            reference;

推荐答案

引用和指针是从属名称，所以你必须使用

reference and pointer are dependent names, so you have to use

typename I::reference operator*() const;
typename I::pointer operator->() const;

此外，构造函数应该只接受 I 。

In addition, the constructor should accept just the I.

但是，根本不需要写这个类。标准库具有 reverse_iterator 。或者如果您对此不满意，还可以 Boost。 ReverseIterator 。

However, there is no need to write this class at all. The standard library has reverse_iterator for this. Or if you are not happy with that, there's also Boost.ReverseIterator.

所需要的只是

typedef std::reverse_iterator<iterator> reverse_iterator;
typedef std::reverse_iterator<const_iterator> const_reverse_iterator;

这篇关于reverse_iterator适配器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！