Re: [boost] variant2 never empty guarantees (was: Re: Outcome/expected/etc/etc/etc)

Subject: Re: [boost] variant2 never empty guarantees (was: Re: Outcome/expected/etc/etc/etc)
From: Andrzej Krzemienski (akrzemi1_at_[hidden])
Date: 2017-06-13 07:31:37

2017-06-12 21:00 GMT+02:00 Gottlob Frege via Boost <boost_at_[hidden]>:

> On Mon, Jun 12, 2017 at 1:29 PM, Peter Dimov via Boost
> <boost_at_[hidden]> wrote:
> > Gottlob Frege wrote:
> >>
> >> On Wed, Jun 7, 2017 at 10:28 AM, Peter Dimov via Boost
> >> <boost_at_[hidden]> wrote:
> >> > Gottlob Frege wrote:
> >> >
> >> >> Agreed.  But I don't see much value in the never-empty guarantee if
> it
> >> >> >> doesn't give you the strong guarantee.
> >> >
> >> >
> >> > I'm not sure I understand this fully; could you please explain from
> what
> >> > > expressions, and under what conditions, you expect the strong
> guarantee?
> >> >
> >> >    variant<X, Y> v1, v2;
> >> >    X x;
> >> >
> >> >    v1= v2; // do you expect strong guarantee here?
> >> >    v1 = std::move(v2); // here?
> >> >    v1 = x; // here?
> >> >    v1 = std::move(x); // here?
> >> >    v1.emplace<X>(); // here?
> >>
> >>
> >> How about "all of the above"?
> >> At least when X and Y each offer the strong guarantee?
> >
> >
> > I'm interested in a practical answer, not a theoretically sound one
> which is
> > of no use. Suppose that X is something that occurs in practice, such as
> > std::vector, not some hypothetical X with a strong assignment, which
> > doesn't.
> >
> > Unless of course you only put types with strong assignment operators into
> > your variants, which in practice confines you to built-ins, in which case
> > all of the above will indeed be not just strong, but nonthrowing as well.
> >
>
> I think I'm going with Niall's comment some time earlier - to have
> variant give as strong a guarantee as the types it holds.  If X = X
> (assignment) is only basic, I don't really need variant to somehow
> magically make variant<X> = variant<X> strong.
>
> Maybe I just don't have enough time to think about this all the way
> through - your comments about how strong doesn't compose, and using
> the swap idiom at the correct level are interesting.  Are you saying
> your variant2 fixes swap, but doesn't go full strong, and that that is
> all we really need?
>

First of all. I haven't seen anyone give an example where they need a
strong guarantee for variant's assignment. There was one attempt with a
state machine, but even there it seamed only theoretical. I think the
questions if we can implement it are secondary to if anyone really needs it.

If whoever has a real (as opposed to theoretical) need for a strong
guarantee, please respond.

Regards,
&rzej;