如何在不暴露使用的容器的情况下暴露迭代器? [英] How can I expose iterators without exposing the container used?

问题描述

我使用 C# 已经有一段时间了，回到 C++ 是一件很头疼的事情.我正在尝试将我的一些实践从 C# 带到 C++，但我发现了一些阻力，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.

我想为这样的类公开一个迭代器:

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;
};

我是否正在尝试一些不成问题的事情?我应该只是 typedef std::vector<T >::迭代器?我希望只依赖于迭代器，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很有趣，因为它完全解决了您发布的问题:关于 C++ 中面向对象和泛型编程之间的张力以及类型擦除可以做什么

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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