What is the difference between the -> operator and .(dot) operator?

Question

C++ / C#

I have a simple question, why is the -> operator used when referring to a dynamic class object, and not . ? Yes, the IDE (In my case, Visual Studio 2019) itself replaces the "dot" with an "arrow", but why is this so? What's the Difference?

Answer 1

a -> b is a shorter notation for (* a ). b
That is, an arrow combines a pointer dereference and a method call.
Well, if a is a class, then the arrow operator can be defined in it at the request of the class author.

Answer 2

The compiler does not replace anything - apparently this is what your IDE does.
The difference is in the purpose. You will find in any textbook on pluses.

Answer 3

EMNIP
a->b≡(*a).b

Answer 4

The arrow is used with pointers to a class/struct, while the dot is used with references or directly with a class/struct instance. Those. if you try to use a dot with a pointer, you will get a syntax error.