subscript operator on pointers

Question

If I have a pointer to an object that has an overloaded subscript operator ([]) why can\'t I do this:

 MyClass *a = new MyClass();
 a[1];
         


        

Answer 1

Simply put, with a[1] the a pointer is treated as memory containing array, and you're trying to access the 2nd element in the array (which doesn't exist).

The (*a)[1] forces to first get the actual object at the pointer location, (*a), and then call the [] operator on it.