🚀 go-pugleaf

Could someone possibly tell me why there is a difference in the 
functionality of user-defined operators "->" and "*"?

from:  13.5.6  Class member access                                 [over.ref]
>1 operator-> shall be a non-static member function taking no parameters.
>  It implements class member access using ->
>          postfix-expression -> id-expression
>  An  expression  x->m is interpreted as (x.operator->())->m for a class
>  object x of type T if T::operator->() exists and if  the  operator  is
>  selected  as the best match function by the overload resolution mechaÐ
>  nism (_over.match_).

Wheras *x is not interpreted as *(x.operator*()) as far as I can tell 
(the april WP does not seem to talk about unary operator*() much).

Why is there this difference?

Also, the ARM (13.4.6) added:
> It follows that operator->() must return either a pointer to a class or 
> an object of or a reference to a class for which operator->() is defined.

Does this still hold under the draft standard?

If so, does it follow given:
T* f();

that:
class X { T* operator->() { return f(); } };
is always does the same as:
class X { T& operator-:() { return *f(); } };

assuming that T does not have user-defined operator*

Thanks for any help on this matter,
Atman Binstock
ajb@panix.com


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