C++ - Difference between (*). and ->?

Question

Is there any difference in performance - or otherwise - between:

ptr->a();

and

(*ptr).a(); 

?

Answer 1

[Edit]

If the variable is defined as T* (where T is some type) then both -> and * are the same (unless ptr is null).

If the variable is an instance of a class (by value or by reference) then -> and * should behave the same (per best practice) but this requires the class to overload them the same way.