Reputation:
I have the following problem:
template <int N, typename T>
/*what is the return type*/ nviewgetter( T const& t )
{
typename T::const_iterator it(t.begin());
typedef BOOST_TYPEOF_TPL(*it) etype;
typedef typename boost::fusion::result_of::as_nview<etype, N>::type netype;
std::vector<netype> r;
while(it!=t.end()){
r.push_back( boost::fusion::as_nview<N>(*it) );
it++;
}
//return r;
}
What is expected is that T is a sequence of Forward Sequences (e.g. boost::fusion::vector) and I want to get a view of the N-th element in each element of T . However, I do not know from beforehand the type of boost::fusion::vector
, e.g. boost::fusion::vector<int, double>
or boost::fusion::vector<int, double, std::string>
. In the code I can figure out the correct type but I cannot figure out this in the function declaration.
Thanks !
Any suggestions for code improvement are also welcome. :)
Upvotes: 4
Views: 262
Reputation: 254621
If you don't want to write the full type out, you could move the type definitions into a separate template, so that they are available when you declare the function template; something like:
template <int N, typename T>
struct nviewgetter_traits
{
typedef BOOST_TYPEOF_TPL(typename T::value_type) etype;
typedef typename boost::fusion::result_of::as_nview<etype, N>::type netype;
typedef std::vector<netype> result_type;
// Or combine it into a single monstrosity if you prefer:
// typedef std::vector<
// typename boost::fusion::result_of::as_nview<
// BOOST_TYPEOF_TPL(typename T::value_type), N
// >::type> result_type;
};
template <int N, typename T>
typename nviewgetter_traits<N,T>::result_type nviewgetter(T const & t)
{
typename nviewgetter_traits<N,T>::result_type r;
for (auto it = t.begin(); it != t.end(); ++it) {
r.push_back( boost::fusion::as_nview<N>(*it) );
}
return r;
};
Upvotes: 4