Re: dynamic_cast is ugly!
On Mar 15, 10:29 am, Jerry Coffin <jcof...@taeus.com> wrote:
danie...@earthlink.net says...
There are any number of ways to do it, Dave Mikes mentioned one,
another would be to have a virtual function in Shape that lets Shapes
know that a "change rectangles to yellow" request has been made.
I would posit that (at least as stated) this would constitute quite a
poor design.
In many cases, yes it would. There are almost as many ways to design
the system as there are systems. :-)
Instead, if we want to be able to highlight all the interfaces (or
whatever) we'd share the "highlighted" vs. "normal" state (or perhaps
the current color) among all objects of that type:
struct UML_object {
int x_, y_;
virtual void draw(surface_t &) = 0;
};
class UML_interface : public UML_object {
static color_t color;
public:
static void highlight(bool state=true) {
static color_t colors[] = {
RGB(0,0,0),
RGB(255, 0,0)
};
color = colors[state];
// code to force re-draw goes here.
}
square(int x, int y) : x_(x), y_(y) {}
virtual void draw(surface_t &s) {
// draw "interface" on specified surface
}
};
class UML_class : public UML_object {
public:
UML_class(int x, int y) : x_(x), y_(y) {}
virtual void draw(surface_t &s) {
// draw "class" on specified surface
}
};
color_t UML_interface::color;
This way we don't need separate containers OR a dynamic_cast to
highlight all your UML_interface objects -- instead, you call:
UML_interface::highlight();
and they all highlight. To change them all back to normal, you call:
UML_interface::highlight(false);
IMO, if you want shared state, it's better to create real shared state
than to force all objects of that type to the same state, independently
of each other.
Excellent post! Another way would be to have some sort of state object
that each of the elements literally share, kind of like a style sheet.
That way it would be easy to support exceptions to the rule, simply
give them a different/non-standard style sheet.