Open GoogleCodeExporter opened 9 years ago
Well I think it's a matter of taste. Why do you want to remove something that
is not
present? It seems like an error to me. Of course, adding a method for checking
if a
widget is in a container or not is a good idea. I guess std::remove has
combined the
two methods in one operatotion.
Original comment by olof.nae...@gmail.com
on 13 Feb 2010 at 5:10
I've found that, in several instances, I'm not sure if the widget is present or
not
(it may or may not be, depending on a number of other variables). I agree,
though,
that this is a matter of taste and the manner in which I've been programming.
I
feel that a hasWidget() method could allow for such flexibility in both cases.
I've
attached a patch.
Original comment by holy.smo...@gmail.com
on 13 Feb 2010 at 5:30
Attachments:
Crud, the hpp patch should have been 'virtual bool', not 'virtual void'. Sorry
:-/
Original comment by holy.smo...@gmail.com
on 13 Feb 2010 at 5:52
Original issue reported on code.google.com by
holy.smo...@gmail.com
on 13 Feb 2010 at 4:20