How do I call "operator->()" directly?

For some strange reason, I need to call the operator->() method directly. For example:

class A {
    public:
        void foo() { printf("Foo"); }
};

class ARef {
    public:
        A* operator->() { return a; }
    protected:
        A* a;
}; 

If I have an ARef obje开发者_StackOverflow社区ct, I can call foo() by writing:

aref->foo();

However, I want to get the pointer to the protected member 'a'. How can I do this?

aref.operator->(); // Returns A*

Note that this syntax works for all other operators as well:

// Assuming A overloads these operators
A* aref1 = // ...
A* aref2 = // ...
aref1.operator*();
aref1.operator==(aref2);
// ...

For a cleaner syntax, you can a implement a Get() function or overload the * operator to allow for &*aref as suggested by James McNellis.

You can call the operator directly using the following syntax:

aref.operator->()

You should probably overload both -> and *, so that usage is consistent with the usage of a pointer:

class ARef {
    public:
        A* operator->() { return a; }
        A& operator*() { return *a; }
    protected:
        A* a;
}; 

Then you can use the following to get the pointer value:

&*aref

You can also implement a get() member function that returns the A* directly, which is much, much cleaner than either of these solutions (most smart pointer classes provide a get() member function).

You can call operator->() directly.

A* result = a.operator->();

I would probably not call it together, but instead create a new member which does the same work since it would read better:

class ARef {
public:
    A* operator->() { return get(); }
    A* get() const { return a; }
protected:
    A* a;
};

Absent any friends, you cannot get a "pointer to the protected member a" outside class A or one of its descendants. Operator -> will not help you here, regardless of how you call it.

Given the current interface of ARef, the only way to get a pointer to a is to do something like A **pa = &a inside one of the methods of class A or some descendant class.

I don't know why you need to get a pointer to a, and frankly it doesn't seem to make much practical sense to me, but if you really need it, you can add a separate method that does exactly that

class ARef {
public:
  ...
  A** get_ptr_to_a() { return &a; }
  ...
};