Re: [boost] Strong guarantee variant (Was: Outcome/expected/etc/etc/etc)

2017-06-08 14:59 GMT+02:00 Peter Dimov via Boost :

Andrzej Krzemienski wrote:

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2017-06-06 14:20 GMT+02:00 Peter Dimov via Boost :

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Types don't generally provide the strong guarantee on assignment as it doesn't compose. If type T has strong assign, type U has strong assign, > struct { T t; U u; } no longer does. So it's more efficient to just do > basic everywhere, which does compose, and then at the specific point you > want the strong guarantee, use f.ex. the copy and swap trick to get it.

I think the reasoning behind Niall's position is that a type `X` can have a custom guarantee: upon its assignment it guarantees that after a throw the type is either unchanged or goes to fallback state `0`. This guarantee is

stronger than basic, and weaker than strong.

Types _could_ provide such a guarantee, but don't, for the same reason they don't provide strong - as outlined in the quoted paragraph above, if types X and Y provide it, struct { X x; Y y; } no longer does. So it's additional work, and buys nothing much.

Yes, such guarantee is not composable when you put X into product types. But it still could work for certain cases of assignments of variant (those cases where both variants hold an X). Personally, I have never found such guarantee useful in practice, but the demand seems theoretically sound. Regards, &rzej;