AnyIterator 和 boost 迭代器外观
是否可以使用 boost 迭代器外观实现任何迭代器？ 我不想在我的基类中定义实现细节
class Base
{
public:
typedef std::vector<int>::iterator iterator;//implementation detail
...
virtual iterator begin()=0;
virtual iterator end()=0;
};
，或者我是否必须完全从头开始编写一个；
Is it possible to implement an any iterator with boost iterator facade?
I don't want to define implementation details in my baseclass
class Base
{
public:
typedef std::vector<int>::iterator iterator;//implementation detail
...
virtual iterator begin()=0;
virtual iterator end()=0;
};
or do i have to write one completely from scratch;
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评论（2）
您发布的代码已修复从
Base返回的迭代器类型及其对std::vector::iterator 的所有实现，这可能不是您想要的想。 Jeremiah 的建议是一种方法，但有一个缺点：你失去了与 STL 的兼容性...我知道多态迭代器包装器的三种实现：any_iterator（它实现boost::iterator_facade< /code>）any_iterator的层次结构。这个问题比看起来更难......我自己尝试了一下，主要是因为我需要
any_iterators类型参数中的协变（any_iterator应该自动转换为 < code>any_iteratorEnumerator这样的 AC# 更容易实现（*）（恕我直言，通常是比类似 STL 的迭代器对更清晰的概念），但同样，您“失去”了 STL。(*) = 当然没有“产量”:-)
The code you've posted has fixed the type of iterators returned from
Baseand all it's implementantions tostd::vector<int>::iteratorwhich is probably not what you want. Jeremiah's suggestion is one way to go with one drawback: you loose compatibility with STL... I know of three implementations of a polymorphic iterator wrapper:any_iterator(which implementsboost::iterator_facade)opaque_iteratorlibrary (google for it), orany_iterators.The problem is harder than it might seem... I made an attempt myself mainly because I needed covariance in
any_iteratorstype argument (any_iterator<Derived>should be automatically convertible toany_iterator<Base>) which is difficult to implement cleanly with STL like iterators. A C# likeEnumerator<T>is easier to implement(*) (and imho generally a cleaner concept than STL-like pairs of iterators) but again, you "loose" the STL.(*) = without 'yield' of course :-)
我想这可能就是您正在寻找的：
以下是该页面的片段：
I think this may be what you're looking for:
Here's a snippet from that page::