Re: template copy constructor vs normal copy constructor

From:

James Kanze <james.kanze@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Wed, 29 Sep 2010 09:42:08 -0700 (PDT)

Message-ID:

<a11ab525-3ba7-476c-bd3f-04f19804a0c6@z28g2000yqh.googlegroups.com>

On Sep 29, 2:56 pm, Juha Nieminen <nos...@thanks.invalid> wrote:

James Kanze <james.ka...@gmail.com> wrote:

The only time it makes any sense to speak about a copy
constructor, as opposed to a constructor in general, is when
discussion whether the compiler will generate one automatically.
(If it does, of course, this becomes just one more constructor
to consider in overload resolution.)

What happens in C++0x if you explicitly disable the copy
constructor, and then there's a templated constructor
which would match? Does the disabling take precedence?

Good question. If I understand it correctly, a "deleted"
function is actually defined, so would be considered in
overload resolution. Then, if overload resolution chose it,
you would get an error. This could lead to some interesting
situations:

    struct C
    {
        C(C&) = delete;
        template<typename T>
        C(T const&);
    };

With such a class, you could copy rvalues and const lvalues,
but not non-const lvalues (since the deleted function would
be a better match than the template).

--
James Kanze