[Boost-users] [phoenix] Extending actors

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 << arg1.top.x)(my_rect());

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