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.

Answer 2

Since you are asking for it in the comments. What you are probably looking for can be found in the Standard (5.2.5 Class member access):

3 If E1 has the type “pointer to class X,” then the expression E1->E2 is converted to the equivalent form (*(E1)).E2;

The compiler will produce the exact same instructions and it will be just as efficient. Your machine will not know if you wrote "->" or "*.".