如何在不暴露所使用的容器的情况下公开迭代器？ [英] How can I expose iterators without exposing the container used?

问题描述

我一直在使用C＃一段时间，回到C ++是一个头痛。我试图从C＃中获取一些我的C ++实践，但是我发现了一些阻力，我很乐意接受你的帮助。

I have been using C# for a while now, and going back to C++ is a headache. I am trying to get some of my practices from C# with me to C++, but I am finding some resistance and I would be glad to accept your help.

我想为这个类展示一个迭代器：

I would like to expose an iterator for a class like this:

template <class T>
class MyContainer
{
public:
    // Here is the problem:
    // typedef for MyIterator without exposing std::vector publicly?

    MyIterator Begin() { return mHiddenContainerImpl.begin(); }
    MyIterator End() { return mHiddenContainerImpl.end(); }

private:
    std::vector<T> mHiddenContainerImpl;
};

我在尝试不是问题的东西吗？我应该只是typedef std :: vector< T> :: iterator？我希望只是依赖于迭代器，而不是实现容器...

Am I trying at something that isn't a problem? Should I just typedef std::vector< T >::iterator? I am hoping on just depending on the iterator, not the implementing container...

推荐答案

您可能会发现以下文章有趣它解决了您发布的问题：关于面向对象与C ++中的通用编程之间的紧张关系类型擦除可以做到这一点

You may find the following article interesting as it addresses exactly the problem you have posted: On the Tension Between Object-Oriented and Generic Programming in C++ and What Type Erasure Can Do About It

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