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Subject: Re: [boost] variant2 never empty guarantees (was: Re: Outcome/expected/etc/etc/etc)
From: Peter Dimov (lists_at_[hidden])
Date: 20170612 17:29:19
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 neverempty 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 builtins, in which case
all of the above will indeed be not just strong, but nonthrowing as well.
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