Re: Classification of template typename labels?

From: rwgk (rwgk_at_[hidden])
Date: 2002-01-21 14:40:49

--- In boost_at_y..., "David Abrahams" <david.abrahams_at_r...> wrote:
David Abrahams wrote:
> The library provides the dispatching function:
> 
>     template <class X, class Y>
>     void copy(X x1, X x2, Y y)
>     {
>         copy(typename concept<X>::type(), x1
>             , typename concept<X>::type(), x2
>             , typename concept<Y>::type(), y);
>     }
> 

This looks very interesting in its own right. However, this
does not seem to address Howard Hinnant's problem (correct?):

Howard Hinnant wrote:
> As a library writer I would like to be able to create a restricted
> template foo that would accept only (for example) arithmetic types. 
But
> if client code creates a function foo that takes a user-defined 
type, I
> want it to behave as if there is no template foo, even if he has 
brought
> it into scope.

But I guess smart use of namespaces could help here.

David wrote:
> Things get a lot more interesting when you have return values 
involved. At
> that point we need __typeof__():
> 
>     template <class X, class Y>
> 
>     __typeof__(copy(typename concept<X>::type(), *((X*)())
>             , typename concept<X>::type(), *((X*)())
>             , typename concept<Y>::type(), *((Y*)()))
>     void copy(X x1, X x2, Y y)
>     {
>         return copy(typename concept<X>::type(), x1
>             , typename concept<X>::type(), x2
>             , typename concept<Y>::type(), y);
>     }
> 
> Fortunately, you only need to write one of these ugly dispatching 
functions.

"typeof" does not appear in the ISO standard. How can I learn
more about this?

Thanks,
        Ralf