模板组合和友谊传递性

Question

我有包含私有数据的容器和访问私有数据的朋友 class：

template<class T>
class Container
{
    friend typename T::MyAccessor;
    vector<T> _data;
};

template<class T>
class Accessor
{
public:
    void doSomething(Container<T> *c)
    {
        cout << c->_data.size() << endl;
    }
};

template<class T, template<typename> class CustomAccessor>
struct MyBase
{
    typedef Container<T> MyContainer;
    typedef CustomAccessor<T> MyAccessor;
};

现在我想这样组合访问器：

template<
    template<typename> class First
    ,template<typename> class Second 
    ,class T
>
class Composite
{
public:
    typedef First<T> MyFirst;
    typedef Second<T> MySecond;

    void doSomething(Container<T> *c)
    {
        MyFirst a;
        a.doSomething(c);

        MySecond b;
        b.doSomething(c);
    }
};

template<class T>
class DoubleAccessor : public Composite<Accessor, Accessor, T> {};

但是友谊是不可传递的，组合访问者不能访问容器私有数据。有没有办法在不向所有人公开容器私有数据的情况下解决这个问题？

Answer 1

如果您只是将 class Accessor 声明为 class Container 的 friend 怎么办：

template<class T>
class Container {
    template<class T1> friend class Accessor;
    std::vector<T> _data;
};

template<class T>
class Accessor {
public:
    void doSomething(Container<T> *c) {
        std::cout << c->_data.size() << std::endl;
    }
};

template<class T, template<typename> class CustomAccessor = Accessor>
struct MyBase {
    typedef Container<T> MyContainer;
    typedef CustomAccessor<T> MyAccessor;
};

template<class T, template<typename> class First = Accessor, template<typename> class Second = Accessor>
class Composite {
public:
    typedef First<T> MyFirst;
    typedef Second<T> MySecond;

    void doSomething(Container<T> *c) {
        MyFirst a;
        a.doSomething(c);

        MySecond b;
        b.doSomething(c);
    }
};

LIVE DEMO

模板组合和友谊传递性

Template composition and friendship transitivity

c++

refactoring

templates

friend