Re: [Boost-users] [variant] make_recursive_variant containing a variant ?

26 Feb 2008

      Steven Watanabe wrote:
...
AMDG
Alban Linard wrote:
...
Hi,
i am trying to build a recursive variant containing another variant, as 
described in the following source code:
#include <cstdlib>
#include <vector>
#include <string>
#include 
typedef ::boost::variant
        < int
        , bool
        >
        first_variant;
typedef ::boost::make_recursive_variant
        < first_variant
        , ::std::string
        , ::std::vector< ::boost::recursive_variant_ >
        >::type
        second_variant;
int main()
{
  first_variant f(true);
  second_variant s(f);
  return EXIT_SUCCESS;
}
It does not compile (g++-4.0 and g++-4.3). It seems that the 
"first_variant" part of second_variant is not used.
Why this behavior ? Is there a workaround ?
This is a bug in boost::variant.  You can fix it by changing line 101 of 
boost/variant/recursive_variant.hpp
from
    typedef ::boost::variant< types > type;
to
    typedef ::boost::variant< over_sequence<types> > type;
In Christ,
Steven Watanabe
Thank you very much!

Next step, i would like to store a container of first_variant instead of 
   a single one.

#include <cstdlib>
#include <vector>
#include 

typedef ::boost::variant
         < int
         , bool
         >
         first_variant;

typedef ::boost::make_recursive_variant
         < ::std::vector< first_variant >
         , ::std::vector< ::boost::recursive_variant_ >
         >::type
         second_variant;

int main()
{
   ::std::vector< first_variant > f;
   second_variant s(f);
   return EXIT_SUCCESS;
}

This crashes, too. Again, did i use it wrong or is it another problem ?

Thanks,
Alban Linard