Re: [boost] [variant2] documentation request

Subject: Re: [boost] [variant2] documentation request
From: Peter Dimov (lists_at_[hidden])
Date: 2019-03-02 18:19:29

Niall Douglas wrote:
> > It so happens that the strong guarantee is unachievable with variant 
> > (without too much double buffering.) You can either have basic, or 
> > noexcept.
>
> You surprise me, given this is a design capable of double buffering.
>
> For a variant2<A, B>, where both A and B have throwing move constructors 
> and assignment, surely if the variant has state A in buffer1, setting it 
> to state B would use buffer2. If B throws during move, we simply don't 
> change the currently selected buffer to buffer2. The variant's A state 
> remains untouched i.e. strong guarantee.
>
> I guess you haven't written out anywhere how and when the double buffering 
> comes into play.

That is actually the one thing I have written, twice, once in the README and 
one in the Overview section of the documentation. :-)

"To avoid going into a valueless-by-exception state, this implementation 
falls back to using double storage unless

* one of the alternatives is the type monostate,
* one of the alternatives has a nonthrowing default constructor, or
* all the contained types are nothrow move constructible."

So, yes, variant<A, B> will be double-buffered, but variant<monostate, A, B> 
and variant<int, A, B> won't be.