Pointer dereference operator ( (*) vs -> )

Question

Is there a general difference between doing

(*ptr).method()

vs

ptr->method()

I saw this question in a comment on another question and thought I would ask it here. Although I just remembered that pretty much every operator in C++ can be overloaded, so I guess the answer will depend. But in general, is there a difference between doing one versus the other?

Answer 1

As "jamesdlin" already noted, the * and -> operators can be overloaded for class types.

And then the two expressions (*ptr).method() and ptr->method() can have different effect.

However, for the built-in operators the two expressions are equivalent.

The -> operator is more convenient when you're following a chain of pointers, because . has higher precedence than *, thus requiring a lot of ungrokkable parentheses.

Consider:

pBook->author->snailMail->zip 

versus

(*(*(*pBook).author).snailMail).zip 

Answer 2

For raw pointer types, they are the equivalent.

And yes, for general types, the answer is indeed "it depends", as classes might overload operator* and operator-> to have different behaviors.

Answer 3

Yes. ptr->method() is two characters shorter than (*ptr).method().

It is also prettier.