提高融合和const正确性

Question

I am learning boost fusion and am trying to take a view of a std::vector<boost::fusion::vector<int,double,double> > . The code appears simple but I appear to be running into some problems with const. I clearly misunderstand something about const and would love for someone to explain where I am going wrong.

The code:

template<int N, class T>
struct viewTraits{
    typedef typename T::value_type etype;
    typedef typename boost::fusion::result_of::as_nview<etype, N>::type netype;
    typedef std::vector<netype> result_type;
};

template <int N, typename T>
typename viewTraits<N,T>::result_type c( T const &t ) 
{
   typename viewTraits<N, T>::result_type retVal;  
    for (typename T::const_iterator it(t.begin());it<t.end();++it){
        retVal.push_back(fusion::as_nview<N>(*it));
    } 
    return retVal;
}

template <typename Container>
typename Container::value_type sum( Container const &container )
{
    typedef typename Container::value_type value_type;
    return std::accumulate( container.begin(), container.end(), value_type() );
}

int main(){
    typedef fusion::vector<int, double, double>             row;
    typedef std::vector<row>                                                container;

    container x;
    b( x, 200 );
    std::cout << sum(c<1>(x)) << std::endl;
}

The code fails to compile at retVal.push_back() because of an issue with casting const. I tried several permutations of removing and adding some const keywords but have not been successful at programming by permutation and would rather understand what I am doing.

Anyone have any thoughts? BTW the const in the function definition of sum and c must stay.

EDIT: I forgot to mention that b fills x which is a std::vector<fusion::vector<int, double, double> >

EDIT2: The corrected code:

template<int N, class T>
struct viewTraits{
    typedef typename T::value_type etype;
    typedef typename fusion::result_of::as_nview<etype, N>::type netype;
    typedef typename fusion::result_of::at_c<netype,0>::type reference;
    typedef typename boost::decay<reference>::type value_type;
    typedef std::vector<value_type> result_type;
};

template <int N, typename T>
typename viewTraits<N,T>::result_type c( T const &t )
{
    typename viewTraits<N,T>::result_type retVal;

    for(typename T::const_iterator it(t.begin()); it<t.end();++it){
        retVal.push_back(fusion::deref(fusion::begin(fusion::as_nview<N>(*it))));
    }
    return retVal;
}

Answer 1

There are two immediate problems with your code:

• you compute fusion::as_nview<N>(*it) and try to store it as if it were a fusion::result_of::as_nview<T::value_type, N>::type ; in truth it is a fusion::result_of::as_nview<T::value_type const, N>::type ( *it has type T::const_reference ); this is the source of your const -related errors.

• you try to accumulate the resulting views, but, as far I can tell from the Fusion docs , valid operations on such a view are mostly those of a Fusion Random Access Sequence. In particular you can't add two views together, or default construct one. Here's a lead to a possible fix (by no means the only way though), not actually tested:

   typedef typename Container::value_type view_type; typedef typename fusion::result_of::at_c<view_type, 0>::type reference; // boost::decay is from Boost.TypeTraits typedef typename boost::decay<reference>::type value_type; value_type zero = 0; return std::accumulate(container.begin(), container.end(), zero, add_views()); 

  where add_views is a functor that returns something in the spirit of lhs + at<0>(rhs) . Note that this solution only really makes sense if you intend for c to make views of length exactly one.

Answer 2

Your Sequence type isn't T::value_type , it's const T::value_type . Try writing typedef const typename T::value_type etype; .