Question:
Which arguments do you receive when you overload . and -> , and what do you do with the arguments? I know that for overloading [] you receive the index. How about . and -> ?

Answer:
First, the . operator cannot be overloaded. If it could be, there would be no way to access members of the objectdirectly.

Here are the rules for operator -> :

1. Operator -> must be defined as a member of a class.
2. The operator takes a second argument but the type is unspecified.
3. The return value must be a type for which operator -> is meaningful.

Here is an example:

struct X{   Y * operator ->() { return yPtr_; }   struct Y { void foo (); }   Y * pPtr;}X x; x->foo () ; // calls X::Y::foo

The operand to operator -> (in this case foo) is applied to its return value. If the return value is something for which -> does not make sense, that is an error.