overloading -> operator in c++

Question

I saw this code but I couldn't understand what it does:

inline S* O::operator->() const
{
    return ses; //ses is a private member of Type S*
}

so what happens now if I used ->?

Answer 1

Now if you have

O object;
object->whatever()

first the overloaded operator-> will be called, which will return ses stored inside the object, then operator-> (built-in in case of S*) will be called again for the returned pointer.

So

object->whatever();

is equivalent to pseudocode:

object.ses->whatever();

the latter would be of course impossible since O::ses is private - that's why I call it pseudocode.

With such overload you can create a wrapper around a pointer - such wrapper is typically called smart pointer.