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

-----Original Message----- 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 v(abc);

without deep copying. The Di objects are big.

Then I would apply the visitation.

Maybe storing variant...

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(&abc)) // do something ... if (dynamic_cast(&abc)) // do something } // no warnings here If I miss a case Maybe what I want is just not possible,