Hi all, let _o_ be any proto terminal and v a vector. I would like to extract the vector v from the expression _o_(v) to do some modification on this vector. For this purpose i defined a grammar Rhs to extract the most right hand side of the tree. But unfortunately the compiler is asking me for constness for this code: boost::result_of<Rhs(Expr)>::type & result = Rhs()(expr); Whereas the expression _o_ + v compiles fine without asking me for constness. Am I missing something? Best regards, Kim Here is the code to demonstrate the problem. # include <vector> # include <iostream> # include <algorithm> # include <boost/proto/proto.hpp> namespace proto = boost::proto; namespace mpl = boost::mpl; using proto::_; template<typename Expr> struct MixedExpr; struct MixedDomain : proto::domain<proto::generator<MixedExpr> > {}; template<typename Expr> struct MixedExpr : proto::extends<Expr, MixedExpr<Expr>, MixedDomain> { explicit MixedExpr(Expr const &expr) : proto::extends<Expr, MixedExpr<Expr>, MixedDomain>(expr) {} }; struct oo { friend std::ostream &operator<<(std::ostream &s, oo const&) { return s << "oo"; } }; proto::terminal<oo>::type const _o_ = {{}}; template<typename T> struct IsMixed : mpl::false_ {}; template<typename T, typename A> struct IsMixed<std::vector<T, A> > : mpl::true_ {}; namespace MixedOps { BOOST_PROTO_DEFINE_OPERATORS(IsMixed,MixedDomain) }; namespace boost { namespace proto { template<typename T, typename A> std::ostream &operator<<(std::ostream &s, std::vector<T, A> const& ) { return s << "std::vector"; } }} struct Rhs : proto::or_< proto::when<proto::terminal<_>, proto::_value > ,proto::when<_, Rhs(proto::_child1)>
{};
template<typename Expr> void o_o(Expr& expr) { proto::display_expr(expr); boost::result_of<Rhs(Expr)>::type & result = Rhs()(expr); std::fill(result.begin(), result.end(),2); }; int main() { using namespace MixedOps; std::vector<int> u(10),v(10); //o_o( _o_(v) ); //uncomment this won't compile o_o( u+v ); std::copy(v.begin(),v.end(),std::ostream_iterator<int>(std::cout,",") ); std::cout<<std::endl; return 1; }