Topic: can the '.' operator be overloaded ??
Author: b91926@fsgm01.fnal.gov (David Sachs)
Date: 20 Jan 1995 18:09:54 -0600 Raw View
rajs1@ix.netcom.com (Raj Sumangali) writes:
>Hi,
> can anyone tell me wheteher the '.' operator can be
>overloaded ?
It cannot. The following operators also may not be overloaded.
.*
::
?:
sizeof
# (preprocessing token)
## (preprocessing token)
Author: fjh@munta.cs.mu.OZ.AU (Fergus Henderson)
Date: Sun, 22 Jan 1995 18:16:06 GMT Raw View
rajs1@ix.netcom.com (Raj Sumangali) writes:
> can anyone tell me wheteher the '.' operator can be
>overloaded ?
No, it can't.
--
Fergus Henderson - fjh@munta.cs.mu.oz.au
all [L] (programming_language(L), L \= "Mercury") => better("Mercury", L) ;-)
Author: rajs1@ix.netcom.com (Raj Sumangali)
Date: 19 Jan 1995 17:58:03 GMT Raw View
Hi,
can anyone tell me wheteher the '.' operator can be
overloaded ?
thanks,
raj
Author: euamts@eua.ericsson.se (Mats Henricson)
Date: 20 Jan 1995 14:46:21 GMT Raw View
In article k4@ixnews2.ix.netcom.com, rajs1@ix.netcom.com (Raj Sumangali) writes:
>Hi,
> can anyone tell me wheteher the '.' operator can be
>overloaded ?
No. There has been very long discussions on this topic, and if I
am not mistaken, most experts agree it should not be possible to
do so, or that the impact on the language of this overloadibility
is not completely understood by anyone.
Mats Henricson
Sweden