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From: "Gabor Greif" <gabor@no.netopia.com>
Subject: Re: [Q] function member template in a template class
Date: 1999/05/10
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Approved: Valentin Bonnard <bonnard@clipper.ens.fr>
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On my question , how to externally define a member template function of a
templated class

On Fri, May 7, 1999 18:43 Uhr, Howard Hinnant
<mailto:hinnant@_anti-spam_metrowerks.com> wrote:
>This ought to do it:
>
>template <int I, typename T>
>struct Res { };
>
>
>template <int I, typename T>
>struct Bar
>{
>        template <int J>
>        typename Res<J, T> foo(void);
>};
>
>template <int I, typename T>
>template <int J>
>typename Res<J, T>
>Bar<I, T>::foo(void)
>{
>   return Res<J, T>();
>}
>
>int main()
>{
>   Bar<1, int> b;
>   b.foo<2>();
>}
>
>-Howard
>


Thanks to Howard and also Siemel, who responded privately. The latter has
pointed out, that my original code contained a non-conformant line:

>        typename Res<J, T> foo(void);

Here typename is not allowed, since it is inferrable that Res<J, T> is a
type. The typename would be necessary to disambiguate the access to some
member in Res<J, T>, such as a typedef or a nested class.

He has also brought to my attention that calling foo as in

>   b.foo<2>();

will not work, since the "<" could be parsed as a smaller operator.

The correct version would be:

>  b.template foo<2>();

It seems that Howard and me are using the same compiler ;-)

	Gabor
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