>> I thought I proved by testing that is_visible() would merely check >> whether show() had been called - not if the widget or window was >> actually visible to the user, but maybe I got it wrong... >> > > The signal responds to things done on the desktop rather than by you. You > (the programmer) can keep your own flags about what you have done > programmatically. For that reason if hide() is called the signal is not > emitted (but Gtk::Widget::is_visible() will correctly state that it is not > visible). I do not really understand what you mean about show(). If you > have called hide() and it remains hidden (that is, you have not called > Gtk::Window::present() in the case of a window or Gtk::Widget::show() in the > case of a widget) then Gtk::Window::is_visible() will return the correct > result, at least when I use it. > I suspect that you did not understand my original question. I want to know if the user can actually see the widget, not just if it is theoretically visible in the sense that it has been shown via show() and not subsequently hidden via hide(). In other words, I'm looking for a call that also returns "false" if for instance the widget is covered by a different window.
- Toralf _______________________________________________ gtkmm-list mailing list [email protected] http://mail.gnome.org/mailman/listinfo/gtkmm-list
