Boost logo

Boost Users :

Subject: [Boost-users] [phoenix] Extending actors
From: paul Fultz (pfultz2_at_[hidden])
Date: 2012-02-22 19:57:11

Hi all, I am wanting to extend actors to access a member variable from the class. In the examples they show extending an acotr by using a lazy function to call the function. I was thinking I could use a lazy function to access the member, something like this: template <typename Expr> struct point_actor     : actor<Expr> {     typedef actor<Expr> base_type;     typedef point_actor<Expr> that_type;         point_actor( base_type const& base )         : base_type( base ) {}             typename expression::function<x_impl, that_type>::type x;     typename expression::function<y_impl, that_type>::type y; }; Where x_impl and y_impl are lazy functions that access the x and y variable. Would this work as member variables, rather than member functions? Also, why wouldnt the second parameter to expression::function take just actor instead of just point_actor<Expr>? Does that make sense? The reason I ask is I would like to control the actor that is returned and do something like this(using the above actor): template <typename Expr> struct rect_actor     : actor<Expr> {     typedef actor<Expr> base_type;     typedef rect_actor<Expr> that_type;         rect_actor( base_type const& base )         : base_type( base ) {}             typename expression::function<top_impl, point_actor >::type top;     typename expression::function<bottom_impl, point_actor >::type bottom; }; expression::terminal<phoenix::argument<1>, rect_actor > arg1; (cout <<; Does that make sense? I can't seem to find any reference on the predefined expressions, and I also can't find the header file for expression::function. Finally, the is_actor trait doesn't seem work for extended actors.Is there a workaround for that? Thanks, Paul

Boost-users list run by williamkempf at, kalb at, bjorn.karlsson at, gregod at, wekempf at