How can I expose iterators without exposing the co

问题:

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...

回答1:

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

回答2:

I have done the following before so that I got an iterator that was independent of the container. This may have been overkill since I could also have used an API where the caller passes in a vector<T*>& that should be populated with all the elements and then the caller can just iterate from the vector directly.

template <class T>
class IterImpl
{
public:
    virtual T* next() = 0;
};

template <class T>
class Iter
{
public:
    Iter( IterImpl<T>* pImpl ):mpImpl(pImpl) {};
    Iter( Iter<T>& rIter ):mpImpl(pImpl) 
    {
        rIter.mpImpl = 0; // take ownership
    }
    ~Iter() {
        delete mpImpl; // does nothing if it is 0
    }
    T* next() {
    return mpImpl->next(); 
    }
private:
    IterImpl<T>* mpImpl; 
};

template <class C, class T>
class IterImplStl : public IterImpl<T>
{
public:
    IterImplStl( C& rC )
    :mrC( rC ),
    curr( rC.begin() )
    {}
    virtual T* next()
    {
    if ( curr == mrC.end() ) return 0;
    typename T* pResult = &*curr;
    ++curr;
    return pResult;
    }
private:
    C& mrC;
    typename C::iterator curr;
};


class Widget;

// in the base clase we do not need to include widget
class TestBase
{
public:
    virtual Iter<Widget> getIter() = 0;
};


#include <vector>

class Widget
{
public:
    int px;
    int py;
};

class Test : public TestBase
{
public:
    typedef std::vector<Widget> WidgetVec;

    virtual Iter<Widget> getIter() {
        return Iter<Widget>( new IterImplStl<WidgetVec, Widget>( mVec ) ); 
        }

    void add( int px, int py )
    {
        mVec.push_back( Widget() );
        mVec.back().px = px;
        mVec.back().py = py;
    }
private:
    WidgetVec mVec;
};


void testFn()
{
    Test t;
    t.add( 3, 4 );
    t.add( 2, 5 );

    TestBase* tB = &t;
    Iter<Widget> iter = tB->getIter();
    Widget* pW;
    while ( pW = iter.next() )
    {
        std::cout << "px: " << pW->px << " py: " << pW->py << std::endl;
    }
}

回答3:

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.

回答4:

I am unsure about what you mean by "not exposing std::vector publicly" but indeed, you can just define your typedef like that:

typedef typename std::vector<T>::iterator iterator;
typedef typename std::vector<T>::const_iterator const_iterator; // To work with constant references

You will be able to change these typedefs later without the user noticing anything ...

By the way, it is considered good practice to also expose a few other types if you want your class to behave as a container:

typedef typename std::vector<T>::size_type size_type;
typedef typename std::vector<T>::difference_type difference_type;
typedef typename std::vector<T>::pointer pointer;
typedef typename std::vector<T>::reference reference;

And if needed by your class:

 typedef typename std::vector<T>::const_pointer const_pointer;
 typedef typename std::vector<T>::const_reference const_reference;

You'll find the meaning of all these typedef's here: STL documentation on vectors

Edit: Added the typename as suggested in the comments