Re: [boost] variant invariant

From: Emil Dotchevski (emil_at_[hidden])
Date: 2008-08-04 01:04:39

On Sun, Aug 3, 2008 at 8:15 PM, Eric Friedman <ebf_at_[hidden]> wrote:
> Hi Emil,
>
> On Wed, Jul 30, 2008 at 6:36 PM, Emil Dotchevski <emil_at_[hidden]>wrote:
>
>> Reading boost::variant's documentation:
>>
>> "If any bounded type is nothrow default-constructible (as indicated by
>> boost::has_nothrow_constructor), the library guarantees variant will
>> use only single storage and in-place construction for every  bounded
>> type in the variant. Note, however, that in the event of assignment
>> failure, an unspecified nothrow default-constructible bounded type
>> will be default-constructed in the left-hand side operand so as to
>> preserve the never-empty guarantee."
>>
>> 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.
>
> This makes it very difficult for the user to enforce some useful
>> invariants for types that have boost::variant members, which in my
>> mind is very important.
>
> I'd like to understand your situation better-- could you elaborate on the
> use case you are facing?

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; therefore
I'm using:

struct value_type
{
    int variant;
    boost::variant<std::string,float> value;
};

For value_type, I want to maintain the invariant that if variant==0,
then value is of type std::string, and if variant is either 1 or 2,
then value is of type float. The problem is that if std::string
operator= throws, boost::variant may break value_type's invariant.

Emil Dotchevski
Reverge Studios, Inc.
http://www.revergestudios.com/reblog/index.php?n=ReCode