Joel Falcou wrote:

How can I write a transform that matches expression like

a0+a1+a2+ ... +an (with N +)

You write a grammar.

and turn into a custom node like

sum(a0,a1,...,an)

You use proto::functional::flatten to turn the binary + expression tree into a flat Fusion sequence, and then you use proto::functional::unpack_expr to turn the Fusion sequence into an expression node with any tag you want. See below: #include <iostream> #include #include #include namespace mpl = boost::mpl; namespace proto = boost::proto; namespace fusion = boost::fusion; using proto::_; struct Plus : proto::or_< proto::terminal<_> , proto::plus > {}; struct as_vector : proto::callable { template<typename Sig> struct result; template struct result : fusion::result_of::as_vector< typename boost::remove_reference<Seq>::type > {}; template<typename Seq> typename fusion::result_of::as_vector<Seq const>::type const operator()(Seq const &seq) const { return fusion::as_vector(seq); } }; struct FlattenPlus : proto::when< proto::plus , proto::functional::unpack_exprproto::tag::plus( as_vector(proto::functional::flatten(_)) ) > {}; int main() { proto::literal<int> i(0), j(1), k(2), l(3), m(4); proto::display_expr(FlattenPlus()(i+j+k+l+m)); return 0; } This displays: plus( terminal(0) , terminal(1) , terminal(2) , terminal(3) , terminal(4) ) The only reason I need as_vector above is because currently unpack_expr requires a RandomAccessSequence, and a flattened expression is a ForwardSequence. Fixing unpack_expr to handle ForwardSequences is on my TODO list. HTH, -- Eric Niebler BoostPro Computing http://www.boostpro.com