Re: c++/10649: is_pointer<T> reports pointer to member function is not a pointer

[Robert Schwartz <robertallanschwartz@yahoo.com>]

The following reply was made to PR c++/10649; it has been noted by GNATS.

From: Robert Schwartz <robertallanschwartz@yahoo.com>
To: Gabriel Dos Reis <gdr@integrable-solutions.net>
Cc: giovannibajo@libero.it, gcc-bugs@gcc.gnu.org, gcc-gnats@gcc.gnu.org
Subject: Re: c++/10649: is_pointer<T> reports pointer to member function is not a pointer
Date: Thu, 8 May 2003 15:20:41 -0700 (PDT)

 --- Gabriel Dos Reis <gdr@integrable-solutions.net> wrote:
 > Robert Schwartz <robertallanschwartz@yahoo.com> writes:
 > 
 > | --- giovannibajo@libero.it wrote:
 > | > Synopsis: is_pointer<T> reports pointer to member function is not a pointer
 > | > 
 > | > State-Changed-From-To: open->closed
 > | > State-Changed-By: bajo
 > | > State-Changed-When: Wed May  7 00:40:10 2003
 > | > State-Changed-Why:
 > | >     No, a member function pointer cannot match "T*". What would
 > | >     be T, in fact?
 > | 
 > | is_pointer<int foo::*>::value (i.e. pointer to member variable) evaluates to true.
 > | If I remove the * from "int foo::*", then I get "int foo::".
 > | I know this isn't a valid C++ type, but a pointer to a member
 > | variable is still a pointer, no? 
 > 
 > In which sense?  A pointer to member is a different kind of beast.  It
 > is much more close to a struct than to a usual pointer.
 > 
 > -- Gaby
 
 I'm not saying anything about how pointer-to-member is *implemented*; I'm
 saying something about its syntax and its semantics. A pointer to a member
 is a pointer. That's all I'm saying.
 
 Robert
 
 __________________________________
 Do you Yahoo!?
 The New Yahoo! Search - Faster. Easier. Bingo.
 http://search.yahoo.com