Question

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 object, 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?

Answer 1

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.