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From: jamiller@hpcupt1.cup.hp.com (Jim Miller)
Newsgroups: comp.std.c++
Subject: "->" == "." proposal
Message-ID: <52390002@hpcupt1.cup.hp.com>
Date: 30 Oct 90 18:15:22 GMT
Organization: Hewlett Packard, Cupertino
Lines: 41


I'd like suggest a modification for C++:
    Make "." and "->" equivalent.
       "a.b",  if "a" is a pointer, becomes "(*a).b".
       "x->y", if "x" is a structrue/class, becomes "x.y" --
	     unless "->" has been overloaded for x.

Any problems?


Commentary:

A while ago, due to a design error, I had to change a pointer to a
structure to a local structure, and similarly a local structure to a
pointer to a structure.  The compiler told me in each case that the
"." or "->" was wrong. 

Good, but if it can tell me that a.b is an error because a is a pointer,
why can't it turn it into "a->b" or "(*a).b)"?

Similarly, it tells me that x->y is an error because x is not a pointer
but a structure, why can't it turn the "x->y" into a "x.b"?

From my point of view, what I'm doing is getting an element out of
a structure, and having to decide if the variable is the structure or
a pointer to a structure is a silly thing for me to have to know,
especially since the compiler knows explicitly, since I told it so
earlier.  No ambiguity.

In the case of an overloaded "x->", "x->y" would not become "x.y" if
"x" was a structure instead of a pointer.


Responses? Problems? Opinions?


   jim miller
   jamiller@hpmpeb7.cup.hp.com
   (a.k.a James A. Miller; Jim the JAM; stupid; @!?$$!; ... )
   Anything I say will be used against me ...
   But my company doesn't know or approve or condone anything of mine here.