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

Subject: Re: [Boost-users] variant over base and derived classes
From: Krzysztof Czainski (1czajnik_at_[hidden])
Date: 2010-12-09 10:49:47

2010/12/9 Hicham Mouline <hicham_at_[hidden]>

> struct S {
>  virtual ~S()
>  {}
> };
> struct Deriv1 : public S {
>  virtual ~Deriv1()
>  {}
> };
> struct Deriv2 : public S {
>  virtual ~Deriv2()
>  {}
> };
> struct Deriv3 : public Deriv2 {
>  virtual ~Deriv3()
>  {}
> };
>
> typedef boost::variant<S,Deriv1, Deriv2, Deriv3> my_variant_t;
> typedef boost::mpl::sort<my_variant_t::types,
> boost::mpl::not_<boost::is_base_and_derived<boost::mpl::_,boost::mpl::_> >
> >::type v0_sorted_t;
>
>
> template <typename Variant, typename Base>
> struct assign {
>  assign(Variant& v, const Base& b)
>    : v_(v), assigned_(false), b_(b)
>  {}
>
>  template <typename T>
>  void operator()(const T&)
>   {
>    std::cout<< typeid(T).name() <<std::endl;
>     if (!assigned_) {
>      const T* t = dynamic_cast<const T*>(&b_);
>      if (t!=0) {
>        v_ = (*t);
>        assigned_ = true;
>      }
>    }
>  }
> private:
>  Variant& v_;
>  bool assigned_;
>  const Base& b_;
> };
>
>
> int main()
> {
>  S s; Deriv1 d1; Deriv2 d2; Deriv3 d3;
>
>  const S& base = s;
>
>  my_variant_t vvv;
>  boost::mpl::for_each<v0_sorted_t>( assign<my_variant_t,S>(vvv, base) );
>
>  std::cout<< vvv.which() <<std::endl;
>  return 0;
> }
>

A thought: won't the following minor design change simplify things? I think
You can avoid sorting the types by introducing struct Deriv0 : S to be used
wherever objects of class S were used. While You're at it, You can make S an
abstract base ;-)

Regards, Kris