CODEWITHSUNDEEP
c++operator-overloading
zer0Id0l
zer0Id0l

Reputation: 1444

Overloading dereferencing operators

noGiven a private Member pData

private:
    T*    pData; // Generic pointer to be stored

Below is the overloaded implementation of * and ->

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

T* operator-> ()
{
    return pData;
}

We call the same from the main as shown below:

void main(){
    SP<PERSON> p(new Person("Scott", 25));
    p->Display();
}

I cannot understand how -> and "*" operator overloading will work here? or to be more clear how p->Display(); will be interpreted?

Upvotes: 1

Views: 123

Answers (1)

Daniel Frey
Daniel Frey

Reputation: 56863

The -> operator is special. When it returns an object, it is automatically applied again. If it returns another object it is also applied again, until finally a plain pointer is returned. This is called chaining, the plain pointer is finally dereferenced and the chain stops.

p->Display() is therefore interpreted like this:

p->Display(); // Compiler sees this
T* tmp = p.operator->(); // First applied operator-> (the one you provided)
tmp->Display(); // since T* is a pointer itself, operator-> (the built-in one for pointers) is applied

Upvotes: 2

Related Questions