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From: tom_usenet@hotmail.com (tom_usenet)
Newsgroups: comp.std.c++
Subject: Re: function template overloading and name lookup
Date: Fri, 4 Jun 2004 16:10:32 +0000 (UTC)
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On Thu, 3 Jun 2004 17:25:13 +0000 (UTC), nesotto@cs.auc.dk ("Thorsten
Ottosen") wrote:

>Hi All,
>
>Consider the small program below. What would you expect the output to be?
>
>#include <iostream>
>#include <deque>
>
>namespace foo
>{
>    template< class T >
>    inline void bar( T )
>    {
>        std::cout << " bar.T";
>    }
>
>    template< class T >
>    inline void use_bar( T t )
>    {
>        foo::bar( t );
>    }
>
>    template<>
>    inline void bar<int>( int i )
>    {
>        std::cout << " bar.Int";
>    }
>
>    template< class T, class A >
>    inline void bar( std::deque<T,A> d )
>    {
>        std::cout << " bar.Deque";
>    }
>}
>
>
>
>int main()
>{
>    using namespace foo;
>    use_bar( char() );

Will print bar.T, for obvious reasons.

>    use_bar( int() );

Will print bar.Int. This is less obvious. But because the explicit
specialization bar<int> has been declared before the point of
instantiation of bar<int>, (which is right after "main") it is fine.
This follows from 14.6.4.1/1 (for point of instantiation) and 14.7.3/6
for information about the declaration of the specialization having to
precede the use of it.

>    use_bar( std::deque<int>() );

Will print bar.T. This is because foo::bar is a non-dependent name,
and therefore is looked up at the point of definition, where the deque
overload is not visible.

(hastily checks with Comeau C++ and GCC 3.4 - both agree!)

Tom
-- 
C++ FAQ: http://www.parashift.com/c++-faq-lite/
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html

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