Alex Rivera | Logout

How can I expose iterators without exposing the container used?

Asked 2008-10-01T10:02:35.870
26

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.

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;
};

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...

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1 Answer

1

This should do what you want:

typedef typename std::vector<T>::iterator MyIterator;

From Accelerated C++:

Whenever you have a type, such as vector<T>, that depends on a template parameter, and you want to use a member of that type, such as size_type, that is itself a type, you must precede the entire name by typename to let the implementation know to treat the name as a type.

answered 2008-10-01T10:23:07.990

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