[Boost-users] [fusion] transform a vector of values to refs

7 Sep 2006

      Hi,

    I am trying to transform from a fusion vector of values to a fusion vector 
of non-const references to those values, the ultimate aim being to assign the 
original sequence from another sequence.

The code below works ok with make_vector, but I want to be able to automate it 
for any fusion sequence. How can I do that? Current efforts follows:

regards
Andy Little

#include <quan/out/current.hpp>
#include <quan/matrix/out/matrix.hpp>
#include <quan/operators/binary_operator_functors.hpp>
#include <boost/fusion/sequence/io.hpp>
#include <boost/fusion/sequence.hpp>
#include <boost/ref.hpp>

// try to make a function object to wrap references
// in fusion style, with boost::ref
struct add_reference_wrapper{
    template <typename T>
    struct result{
        typedef boost::reference_wrapper<T> type;
    };
    template <typename T>
    typename result<T>::type
    operator()(T & t)const
    {
        return boost::ref(t);
    }
};

int main()
{
    using boost::fusion::at_c;

    typedef boost::fusion::vector4<
        quan::current::A, quan::current::mA,
        double,double
    > fusion_sequence_type;
    fusion_sequence_type sequence(
        quan::current::A(3), quan::current::mA(2),
        1,1
    );
    std::cout << "before : " << sequence <<'\n';

    // This works OK, and and assign the sequence to itself;
    std::cout << "after : " << boost::fusion::as_vector(
        boost::fusion::transform(
            boost::fusion::make_vector(
                boost::ref(at_c<0>(sequence)), boost::ref(at_c<1>(sequence)),
                boost::ref(at_c<2>(sequence)), boost::ref(at_c<3>(sequence))
            ),sequence,
            quan::operator_plus_equals()
        )
    ) <<'\n';

//#############################################
    // But cant make a vector of non const references
    boost::fusion::as_vector(
        boost::fusion::transform(
            sequence,add_reference_wrapper()
        )
    );
//#############################################
}