TITLE: type deduction in template arguments

(Newsgroups: comp.lang.c++.moderated, 27 Apr 99)

VANDEVOORDE: David Vandevoorde

>> Most standard conversions aren't considered during the deduction
>> process either, but a few are: in particular, derived-to-base
>> conversions and the corresponding contravariant conversions on
>> pointer-to-members apply.


BONNARD: Valentin Bonnard <Bonnard.V@wanadoo.fr>

>Hum... AFAIK, that doesn't make the following valid:
>
>    template <typename T>
>    void foo (T, T);
>
>    struct B {};
>    struct D : B {};
>
>    foo (B(), D()); // ill-formed, contradicting deduction
>    foo<B> (B(), D()); // ok, standard conversion (derived-to-base)
>
>only the following
>
>    template <typename T>
>    struct Y {};
>
>    template <typename T>
>    void bar (Y<T>, Y<T>);
>
>    struct Z : Y<int> {};
>
>    bar (Z(), Y<int>()); // ok, Y<Y> is a template-id
>    bar (Z(), Z()); // ok, same reason
>
>[ BTW, egcs agrees with my interpretation. ]
>

VANDEVOORDE:

Exactly: even though deduction can see derived-to-base
conversions in some cases, it still looks for a best
match on a argument-by-argument basis (and the "no
disagreement" rule remains in effect after that).