Overloading the indirection operator in c++ [duplicate]

Question

This question already has answers here:

How to overload the indirection operator? (C++) (2 answers)

Closed 7 years ago.

my problem is a simple one. I have a class template that holds a pointer to a dynamically allocated type. I want to overload the indirection operator so that referring to the class template instance with the -> operator I get redirected as if I use the pointer contained within directly.

template<class T>
class MyClass
 {
  T *ptr;
  ...
  // created dynamic resource for ptr in the constructor
 };

Create myclass of some type:

MyClass<SomeObject> instance;

So what I want is instead of having to type:

instance.ptr->someMemberMethod();

I simply type:

intance->someMemberMethod();

Even thou instance is not a pointer it behaves as if it is the pointer instance contains. How to bridge that gap by overloading the operator?

Answer 1

You can just overload operator-> and operator*:

template<class T>
class MyClass
{
    T* ptr;

public:
    T* operator->() {
        return ptr;
    }

    // const version, returns a pointer-to-const instead of just a pointer to
    // enforce the idea of the logical constness of this object 
    const T* operator->() const {
        return ptr;
    }

    T& operator*() {
        return *ptr;
    }

    // const version, returns a const reference instead of just a reference
    // to enforce the idea of the logical constness of this object
    const T& operator*() const {
        return *ptr;
    }
};

Note that, due to a design decision by the creator of the language, you can't overload the . operator.

Also, you might think that operator* would overload the multiplication operator instead of the dereference operator. However, this is not the case, because the multiplication operator takes a single argument (while the dereference operator takes no arguments), and because of this, the compiler can tell which one is which.

Finally, note that operator-> returns a pointer but operator* returns a reference. It's easy to accidentally confuse them.

Answer 2

Overload the -> operator:

template <typename T> class MyClass
{
    T * p;
public:
    T       * operator->()       { return p; }  // #1
    T const * operator->() const { return p; }
};

Note that both overloads don't mutate the object; nonetheless we decide to make #1 non-const, so that we bequeath constness of the object onto the pointee. This is sometimes called "deep constness propagation" or something of this sort. The language D takes this much further.

Answer 3

The member access operator can be overloaded to return a pointer to the object to be accessed:

T * operator->() {return ptr;}
T const * operator->() const {return ptr;}

You might also want the deference operator, to make it feel even more like a pointer; this returns a reference instead:

T & operator*() {return *ptr;}
T const & operator*() const {return *ptr;}