//The arrow operator has no inputs. Technically, it can return whatever you want, but it should return something that either is a pointer or can become a pointer .

//The -> operator automatically dereferences its return value before calling its argument using the built-in pointer dereference, not operator*, so you could have the following class:

class PointerToString{    string a;public:    class PtPtS    {    public:        PtPtS(PointerToString &s) : r(s) {}        string* operator->()        {            std::cout << "indirect arrow\n";            return &*r;        }    private:        PointerToString & r;    };    PointerToString(const string &s) : a(s) {}    PtPtS operator->()    {        std::cout << "arrow dereference\n";        return *this;    }    string &operator*()    {        std::cout << "dereference\n";        return a;    }};

这个类可以这么用 

PointerToString ptr(string("hello"));string::size_type size = ptr->size();

上面语句 也可以转换为

string::size_type size = (*ptr.operator->().operator->()).size();

分析这个问题 主要是因为 看stl代码时发现了 auto_ptr 中对其的重载