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From: E. Gladyshev (egladysh_at_[hidden])
Date: 2003-10-07 03:44:22

--- Eric Friedman <ebf_at_[hidden]> wrote:
> I disagree. The *only* relevant guarantee made by a variant<T1,...,TN>
> is that at all times it contains exactly one value of one type Ti. Thus,
> in the event of assignment failure, there is no guarantee that the
> contained type Tk after failure is the same type Tj as before.
> There is nothing "unpredictable" about going into assignment with some
> type U and coming out after failure with boost::empty (or some other
> nothrow default-constructible type).

Then why do we need the assignment exception safety that is based
on what the first type is. Sorry it doesn't make sense to me.
I guess I am missing something.


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