Hi!
The only solution I can come up with is the declaration of the visitor in your ABCBase as virtual function and visitor has a function template to dispatch different types:
class ABCVisitor
{
public:
template<class T>
void visit(T& t)
{
// no visit implemented ...
}
};
class ABCBase
{
//... dtor, copy ctor, assignment operator etc.
public:
void accept_visitor(ABCVisitor& v)
{
do_accept_visitor(v);
}
private:
virtual void accept_visitor(ABCVisitor& v)=0;
};
class ABCDerived1
{
//... dtor, copy ctor, assignment operator etc.
void accept_visitor(ABCVisitor& v)
{
v.visit(*this);
}
};
template<>
void ABCVisitor::visit<ABCDerived1>(ABCDerived1& derived)
{
// handle derived here...
}
Hope that helps,
Ovanes
> -----Original Message-----Then that's not what I need.
> From: boost-users-bounces@lists.boost.org [mailto:boost-users-
> bounces@lists.boost.org] On Behalf Of Igor R
> Sent: 03 August 2009 14:50
> To: boost-users@lists.boost.org
> 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.
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.
Switch (abc.getTag())
void f( const ABC& abc )
{
{
case D1tag: // do something
...
Case Dntag:
}
} // I get the warning from g++ if I miss 1 case
Or
if (dynamic_cast<D1*>(&abc))
void f( const ABC& 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,
_______________________________________________
Boost-users mailing list
Boost-users@lists.boost.org
http://lists.boost.org/mailman/listinfo.cgi/boost-users