Dear all, while I appreciate boost::fusion very much, I've recently stumbled across a point that I don't understand. It appears to me that - const is transitive, i.e. dereferencing an iterator into a const vector yields a reference to a const element and - transformation algorithms work on const sequences. Combined, these two imply that an in-place modification of a sequence through a view generated by a transformation algorithm is impossible. While this is perfectly reasonable for transform, it seems quite restrictive for filter or erase. After all, filter_view accepts a non-const input sequence. My point is probably best illustrated by the following code snippet: #include <iostream> #include <boost/type_traits/is_same.hpp> #include <boost/fusion/algorithm/iteration/for_each.hpp> #include <boost/fusion/algorithm/transformation/filter.hpp> #include <boost/fusion/sequence/container/vector.hpp> #include <boost/fusion/sequence/io/out.hpp> #include <boost/fusion/sequence/view/filter_view.hpp> struct SetToZero { template <typename X> void operator()(X& x) const { x = 0; } }; int main(void) { using namespace boost; using namespace boost::fusion; typedef vector<int,char,double,int> Vec; Vec vec(3,'.',1.0,4); std::cout << vec << '\n'; // This does not compile (error: assignment of read-only reference »x«) for_each(filter<int>(vec),SetToZero()); // This works... for_each(filter_view<Vec,is_same<mpl::_,int> >(vec), SetToZero()); std::cout << vec << '\n'; return 0; } So I'm wondering if this is intended? If yes, how do I best modify (part of) sequences defined by transformation algorithms? Thanks, Martin -- Dr. Martin Weiser web: www.zib.de/weiser Zuse Institute Berlin mail: weiser@zib.de Numerical Analysis and Modelling pgp key: www.zib.de/weiser/weiser.asc