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: Andrzej Krzemienski (akrzemi1_at_[hidden])
Date: 2017-06-14 16:13:57

2017-06-14 18:06 GMT+02:00 Peter Dimov via Boost <boost_at_[hidden]>:

> Andrzej Krzemienski wrote:
>
>> > This does not mean that all assignments are strong. In
>> >
>> >    variant<vector<string>> v1, v2;
>> >
>> >    v1 = v2;
>> >
>> > the assignment is basic, because that's the guarantee >
>> vector<string>::operator= gives.
>>
>>
>> Interesting. This satisfies the expectation that a number of people have
>> expressed.
>>
>
> Exactly. For me basic is basic, regardless of whether the type stays the
> same, but apparently this does not match people's intuitive expectations.
>
>  But on the other hand, now that you are doing the double buffering
>> anyway, making the assignment strong would come almost fo free.
>>
>
> Not free here because the above variant doesn't double-buffer (because all
> alternatives are nothrow move constructible).
>

Ok, I see.

>
> If we continue the example with projections: if I change from the drop
>> down menu from one mesh projection to another mesh projection, they are
>> different projections, but can be encoded in the same type; maybe we should
>> get the strong guarantee. If not in the assignment, then perhaps in a
>> dedicated function.
>>
>
> You get the strong guarantee here in move assignment (because vector's
> move assignment doesn't throw (... except for unequal allocators ...)):
>
>    v1 = std::move(v2); // strong
>    v1 = variant<vector<string>>(v2); // strong
>
>    v1 = vector<string>{ "s1", "s2" }; // strong
>
> and in emplace, because it doesn't use assignment:
>
>    v1.emplace<vector<string>>( ... ); // strong too

If you are not doing double buffering, how can you guarantee that `emplace`
is strong? You move the original to the side?

Regards,
&rzej;