Re: [boost] [variant2] documentation request

Subject: Re: [boost] [variant2] documentation request
From: Peter Dimov (lists_at_[hidden])
Date: 2019-03-02 14:41:32

Andrzej Krzemienski wrote:

> For instance, the variant's move assignment will sometimes use T's move 
> assignment rather than move constructor when changing the variant's type, 
> as in the following example:
>
> https://wandbox.org/permlink/GDRyS54bpdLP7GVa

Interesting.

Since the destructors and the move assignments are trivial, this variant has 
a pseudo-trivial move constructor, that is, it does the equivalent of 
memcpy.

This however doesn't quite match the specification.

I need to look into the latest changes to std::variant to see how it handles 
triviality in this case.

> Describe the algorithm for achieving the [strong] guarantee.

It so happens that the strong guarantee is unachievable with variant 
(without too much double buffering.) You can either have basic, or noexcept.

If all contained types have noexcept move constructors and noexcept move 
assignments, the move assignment of variant is noexcept.

If all types have noexcept move constructors but not all have noexcept move 
assignments (this can be the case for f.ex. std::vector with a 
not-always-equal allocator), variant's move assignment is only as strong as 
the move assignments of the contained types, but emplace is noexcept.

If not all types have noexcept move constructors, I don't know of a good way 
to achieve the strong guarantee.