Structure pointer operator conjecture (theory)

The below expressions are equivalent:?

(*x).y x->y 

Yes, Both are two different ways to access structure member y.

(*x).y operator is . DOT that works with value variable Element selection by reference. that the reason * used. means x is pointer to struct.

x->y operator -> is used called Element selection through pointer. This work with pointer to struct. that is the reason * not used this time.

Both works same.

Would it be fair to think of this operator simply as preprocessor macro defined as such:

#define (x)-> (*(x).) 

No First it give an error: macro names must be identifiers. This error is because we can't have -> operator as macro name.

a valid macro name can be:

Macro names should only consist of alphanumeric characters and underscores, i.e. 'a-z', 'A-Z', '0-9', and '_', and the first character should not be a digit. Some preprocessors also permit the dollar sign character '$', but you shouldn't use it; unfortunately I can't quote the C standard since I don't have a copy of it.

Also, note -> and . are differences operators as I state above. also their precedence are different so its bad idea to replace one operator with other.

Valid macros to access struct elements:

Additionally I would like to share today only i came to know that most C header files. Defined macros like:

#define S(x) ((x).y) 

for specific strcut element.

Notice (x) parenthesis around x is to overwrite precedence of * over . . By default . DOT has higher precedence over * So that it can be use for pointer and simple variable. Below my example will be helpful I think.

#define S(x) ((x).y) typedef struct { int y; }X; X *x; X b; int main(){ S(*x); S(b); } 

EDIT:
Better Option

I am extending my idea to access strcut elements, I defined new macro:

#define S(x,y) ((x).y) typedef struct { int a; }X; X *x; X b; int main(){ S(*x,a); S(b,a); } 

Not its not more for specif elements via macros.

Hope at-least OP love it :)