Re: [Boost-users] enable-if overloading out the wazoo

Subject: Re: [Boost-users] enable-if overloading out the wazoo
From: John Dlugosz (JDlugosz_at_[hidden])
Date: 2010-04-26 19:40:10

> >   template <typename T1, typename T2>
> >   dim_s foo (const T1& left, const T2& right);
> >
> > Then, in a header that gets more concrete, provide its own form that
> is meant to be used when objects of those types are involved:
> >
> >    template<typename T1, typename T2>
> >    typename enable_if_c<
> >       is_base_of<genericbase,T1>::value &&
> >       is_base_of<genericbase,T2>::value,
> >    dim_s>::type foo (const T1& left, const T2& right);
> >
> Isn't that causing ODR violation ?
> the enable_if'ed foo should also present in the header IMHO
> 
> --
> ___________________________________________
> Joel Falcou - Assistant Professor


I'm assuming that you won't have any objects of the types being enabled unless that header is included.  I know that the compiler needs to be aware of explicit specializations before making a call that would use one.  But this isn't a specialization anyway, but an overloaded foo.

I see that overloading would not be allowed if it differs in the return type only.  But I was under the impression that for templates they are indeed different templates.  Now there is no way to distinguish between them as written, when is an extension of the issue I was asking about.  I could put the enable_if in a dummy extra argument that has a default value -- that would prevent the signatures from being the same, and it would not affect my question.


--John






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