operator ->

//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 through chained -> operators.

//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