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