Hi everybody ,
I'd like to know if there are functions that could help you to set up a kind of
a log for a widget : for an option menu , which menu item was 'activate' last ,
for a combo the last strings entered in the entry , for a clist which rows were
selected last , etc ?
In the same idea , how do you know the number of rows in a clist ?
Of course , you create the clist initially , you allow the user to add or
remove rows , so you might be able to monitor what's in it . But forcibly the
clist knows what's composing itself , so is there no fonction to easily
retrieve the numer of rows , and so forth knowing everything that's in it
without keeping a track , callback after callback , of all the alterations ?
Continuing on this track , if this is not implemented natively and that i
really want this kind of feature , i'll need a place to store the informations
. I saw on the C documentation that there are two methods of the Gtk::Object
class : set_data , set_user_data ( the former doesn't seem to be ported in
Gtk-Perl ) . Can i use this , or can it just hold a string ( and not , for
instance a hash , itself containing references and hashes according to my needs
) ?
If i can't , it seems that Gtk-Perl uses a hash ( in regard of the informations
given when you make a "print" on a widget ) , could i create ( cautiously )
just one entry ( a hash reference ) in it that would hold afterwards all the
informations needed ? What's its little name ?
Last question leading to a , truly ( i swear ! ;-) , ultimate one : aren't this
kind of features implemented in Gnome ( as it seems that Gnome is , among other
things , another layer intended to automate a few things for the programmer ) ?
Hence , if i ask this question , it's because i can't seem to find a good
tutorial or documentation on Gnome-Perl ( and i reaaally don't understand a
thing in C , i'm not sure i know enough of it to code a "Hello world" ...;-) ,
could you indicate an URL ?
Thanks for those who are still here reading this , and moreover to those who
will take time to answer ...
D@vid.
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