Re: [boost] variant2 never empty guarantees (was: Re: Outcome/expected/etc/etc/etc)

2017-06-14 0:21 GMT+02:00 Peter Dimov via Boost :

Gottlob Frege wrote:

I'm happy with std::variant. It makes some trade-offs, but I can live

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with them. But once someone tries to make a variant with less trade-offs, it seems to me you should just go all the way - no trade-offs. I can and will live with trade-offs, but who wouldn't like no trade-offs?

Basically, I was surprised that there was something between std::variant and no-compromises-variant. It seems you want to explore that area - _different_ trade-offs, or just _less_ trade-offs. But you are approaching zero, I think you should just do zero. I could easily be wrong.

All right.

My current iteration is "zero" unless you specifically opt into nonzero by placing the special type "valueless" as a first alternative.

For the example use cases in my previous message,

variant v1, v2;

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X x;

v1 = v2; // do you expect strong guarantee here? v1 = std::move(v2); // here? v1 = x; // here? v1 = std::move(x); // here? v1.emplace<X>(); // here?

where X is not the type 'valueless', this would translate to:

v1 = v2; // as strong as X::op=(X const&) v1 = std::move(v2); // as strong as X::op=(X&&)

v1 = x; // as strong as X::op=(X const&) v1 = std::move(x); // as strong as X::op=(X&&)

v1.emplace<X>(); // always strong

So, you mean -- unless I am using the special "valueless" type -- all assignments/emplacements/swaps are strong (possibly at the expense of double buffering in some cases)? That is, there is no "get C upon unsuccessful assignment of A to B"? Regards, &rzej;