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: Peter Dimov (lists_at_[hidden])
Date: 2017-06-13 22:21:47

Gottlob Frege wrote:

> I'm happy with std::variant.  It makes some trade-offs, but I can live 
> with them.
> But once someone tries to make a variant with less trade-offs, it seems to 
> me you should just go all the way - no trade-offs.  I can and will live 
> with trade-offs, but who wouldn't like no trade-offs?
>
> Basically, I was surprised that there was something between std::variant 
> and no-compromises-variant.  It seems you want to explore that area - 
> _different_ trade-offs, or just _less_ trade-offs.  But you are 
> approaching zero, I think you should just do zero.  I could easily be 
> wrong.

All right.

My current iteration is "zero" unless you specifically opt into nonzero by 
placing the special type "valueless" as a first alternative.

For the example use cases in my previous message,

>     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?

where X is not the type 'valueless', this would translate to:

    v1 = v2; // as strong as X::op=(X const&)
    v1 = std::move(v2); // as strong as X::op=(X&&)

    v1 = x; // as strong as X::op=(X const&)
    v1 = std::move(x); // as strong as X::op=(X&&)

    v1.emplace<X>(); // always strong