Re: [boost] [variant2] documentation request

Subject: Re: [boost] [variant2] documentation request
From: Niall Douglas (s_sourceforge_at_[hidden])
Date: 2019-03-02 22:11:51

> 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.

Could we have instead then the following more elemental variants:

1. single_buffered_variant<...>: Never enters a trap state (where
valueless_on_exception() = true). Always single buffered. Requires at
least one state to have a nothrow default constructor, or all states to
have nothrow move constructors.

2. double_buffered_variant<...>: Never enters a trap state (where
valueless_on_exception() = true). Always double buffered. Always
implements the strong exception guarantee.

I would prefer this design because Boost users are more than capable of
writing a std::conditional_t<> which chooses what the variant
implementation is, based on input types e.g.

template<class... Args> using mylocalvariant =
std::conditional_t<is_trivially_copyable_for_all<Args...>,
single_buffered_variant<Args...>, double_buffered_variant<Args...>>;

I personally don't think that you need to choose a hard coded mix of
single and double buffered variant. Rather, let the Boost user choose
the mix.

(But if you're dead set on there being a boost::variant2::variant<...>,
let it be a template alias to your current variant mix)

Niall