Re: [Boost-users] variant: bounded types derive from same base class

Subject: Re: [Boost-users] variant: bounded types derive from same base class
From: Hicham Mouline (hicham_at_[hidden])
Date: 2009-08-03 10:22:39

> -----Original Message-----
> From: boost-users-bounces_at_[hidden] [mailto:boost-users-
> bounces_at_[hidden]] On Behalf Of Igor R
> Sent: 03 August 2009 14:50
> To: boost-users_at_[hidden]
> Subject: Re: [Boost-users] variant: bounded types derive from same base class
> 
> > My restriction is that f takes a ABC const-ref as an input.
> > Can I construct the variant from the    const ABC& abc
> > Maybe like
> >    variant<D1, D2, ... Dn> v(abc);
> >
> > without deep copying. The Di objects are big.
> >
> > Then I would apply the visitation.
> >
> > Maybe storing variant<D1*, D2*, Dn*>...
> 
> Yes, you can store ptrs in the variant.
> 
> > So f would look like
> >
> > void f( const ABC& abc )
> > {
> >   const VariantTypeDefProvidedByLib v(&abc);
> >
> >    v.apply_visitor( YourVisitor );
> > }
> 
> You cannot construct the variant in f(), because you already lost the
> compile-time type of the object - it's ABC now.
> If you'd like to work with ABC's and virtual functions, then why do
> you need variant? Variant is about preserving the compile-time type.

Then that's not what I need.
I need a runtime visitor pattern.

Parts of the user code deal with ABCs, and the user also writes functions for specific derived types,
that use non common public functions of the derived types.

The f function still needs to dispatch to the appropriate piece of code depending on the ABC type.

void f( const ABC& abc )
{

 Switch (abc.getTag())
  { 
   case D1tag:  // do something
   ...
    Case Dntag:
   }
}  // I get the warning from g++ if I miss 1 case

Or 

void f( const ABC& abc )
{ 
 if (dynamic_cast<D1*>(&abc))
   // do something
...
if (dynamic_cast<Dn*>(&abc))
   // do something
}  // no warnings here If I miss a case

Maybe what I want is just not possible,