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From: goo...@vandevoorde.com (Daveed Vandevoorde)
Newsgroups: comp.lang.c++.moderated
Subject: Re: Template argument deduction fails on EDG - but why?
Date: 15 May 2003 19:40:12 -0400
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"Rani Sharoni" <rani_shar...@hotmail.com> wrote:
> Stefan Slapeta wrote:
[...]
> > I think that's not really true, especially as there are strict rules
> > for type deduction in function template calls. Take a look at
> > [14.8.2.1] where it says:
> 
> I looked at 14.8.2 (especially 14.8.2.1/3) and I didn't see any
> treatment in
> the case (conversion) you presented. Notice that the following actually
> works (EDG and GCC):
> template<typename T> int test(T const&);
> 
> In this case T==int[10] and test<int[10]> is instantiated (per
> 14.8.2.1/3).
> 
> > There are no DRs on these lines, so I think EDG behaves incorrectly
> > here.
> 
> I'm really not authorized to say that but maybe Daveed Vandevoorde can
> help us with this issue.

OK.  Just work your way through 14.8.2.1.

Para 1: Our A is "int[10]" and our P is "T const(&)[S]".
Para 2:
   P is a reference type, so skip the 3 bullets.
   P is not a cv-qualified type (references never are),
     so skip the first sentence after the bullets.
   P is a reference type, so we use P' = "T const [S]"
     for deduction purposes.
Para 3:
   We try to find T, S such that P' becomes identical
   to A.  That's not possible, so we looks at the
   three bullets.  The 2nd and 3rd don't apply, but
   the first implies that T=int, S=10 is OK because
   "int const[10]" is more cv-qualified that "int[10]"
   (see 3.9/5).

So deduction succeeds.  Now you've got to check that
the resulting specialization is viable (it is; left
as an exercise to the reader <grin>), and selected
by overload resolution (trivial; there is but one
viable candidate).

So yes, this is valid code and an EDG bug (known).

        Daveed

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