Re: [boost] variant invariant

From: Mathias Gaunard (mathias.gaunard_at_[hidden])
Date: 2008-08-04 10:00:24

Emil Dotchevski wrote:

>>> Do I read correctly that in an attempt to assign one variant object to
>>> another, the left-hand object _might_ silently change its type?
>> If an exception is thrown during assignment to a boost::variant<T0, T1,
>> ...>, and one of the Ti is nothrow default-constructible, then the
>> boost::variant will end up with a default-constructed value of one of those
>> Ti. So even if the type happens already to be that Ti, the value will still
>> change to the default value.

> I have 3 variants of values, let's call them V0, V1 and V2. V0 values
> are of type std::string, and V1 and V2 values are of type float.
> However, boost::variant<std::string,float,float> is illegal;

Then you should use tags, and not ugly hacks.

Variant can be implemented in three (four?) different techniques
(temporary heap backup, nothrow copy or move, nothrow default constructor).
Only the last one (which is, unfortunately, the one invoked by your
code), does not satisfy strong exception-safety. It's fairly funny how
an optimization diminishes the safety of variant.