Re: 14/8 says my code is right, compiler says it's wrong

From:
Alberto Ganesh Barbati <AlbertoBarbati@libero.it>
Newsgroups:
comp.lang.c++.moderated
Date:
10 Oct 2006 08:27:08 -0400
Message-ID:
<zcJWg.141771$_J1.915401@twister2.libero.it>
Matthias Hofmann ha scritto:

"Alberto Ganesh Barbati" <AlbertoBarbati@libero.it> schrieb im Newsbeitrag

?14.7.1.2/6... did you read my post?


Yes, I did. But your quote of 14.7.1.2/6 only *requires* a definition,
without saying *where* it should be made.


Well, maybe that statement does not say it explicitly, but the intent is
clear: a definition is required to be available at the point of implicit
instantiation. The example following that statement in the standard
clarifies that.

template <class T> struct item_traits;

template <class T> class Item; // forward declaration

// definition of item_traits< Item<int> >
template <> struct item_traits< Item<int> >
{
     static int foo(); // declaration of foo()
};

template <class T> class Item
{
public:
     typedef item_traits< Item<T> > traits_type;
     int foo() { return traits_type::foo(); }
};

---- item.cpp

#include "item.h"

int item_traits< Item<int> >::foo() { return 42; }


Oh, I see. That's good news. But doesn't this require the notoriously
unsupported 'export' feature? I vaguely remember that there is an exception
for template *specializations*, though...


What makes you think so? I didn't use the export keyword... so, no: you
don't need that feature. And it's not an "exception" from the rule.
Notice that the definition of item_traits< Item<int> >::foo does not
even have a the template keyword, because it's no longer a template:
it's simply a definition of a regular member function of a regular class
(albeit with a funny name) that have been declared elsewhere.

Ganesh

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