C++ - Difference between (*). and ->?
c++, dereference, pointers
Solution
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 "*.".
Problem
Is there any difference in performance - or otherwise - between: ``` ptr->a(); ``` and ``` (*ptr).a(); ``` ?