Re: [Boost-users] variant over base and derived classes

Subject: Re: [Boost-users] variant over base and derived classes
From: Hicham Mouline (hicham_at_[hidden])
Date: 2010-12-09 02:14:34

>> -----Original Message-----
>> From: Hicham Mouline [mailto:hicham_at_[hidden]]
>> Sent: 09 December 2010 00:04
>> To: 'boost-users_at_[hidden]'
>> Subject: variant over base and derived classes
>> 
>> hello,
>> <snip>
>>
>> typedef boost::variant<S,Deriv1, Deriv2, Deriv3> my_variant_t;
>> 
>> Deriv3 d3;
>> 
>> struct print_type: public boost::static_visitor<> {
>>   template <typename T>
>>   void operator()(const T&) const
>>   {
>>     std::cout<< typeid(T).name() <<std::endl;
>>   }
>> };
>> 
>> const S& s = d3;
>> my_variant_t v = s;
>> boost::apply_visitor( print_type(), v );
>> 
>> This prints S.
>> 
>> Is it a truncated copy of d3 that is stored v?
>Yes.
>> How do I construct v polymorphically so that the real object is stored?
>> 
>> regards,
>You have to downcast before creating the variant. 
>boost::variant only uses the static type of the object. 
>In Christ, 
>Steven Watanabe

If I am given a const S* sptr, how do I construct an instance of the above
variant then?

const Deriv1* d1ptr; const Deriv2* d2ptr; const Deriv3* d3ptr;
my_variant_t v = ((d1ptr = dynamic_cast<const Deriv1*>(sptr)) != 0 )?
*d1ptr:
  ((d2ptr = dynamic_cast<const Deriv2*>(sptr)) != 0 )? *d2ptr: 
  ((d3ptr = dynamic_cast<const Deriv3*>(sptr)) != 0 )? *d3ptr: *sptr;

Is there a better way of writing this?

rds,

--------------------------------------------

Trying to do this more generically, I've thought of call mpl::for_each on
the ::types of the variant to see which dynamic_cast works.
The problem is that I need the most derived type to be stored in the
variant.
I can see how to do it by a virtual function but that is completely
intrusive.

regards,