> Regarding the "This is a lot faster than iterating over all the > incidences in the calendar" comment - is that based on measurements or > the observation in the KCalCore code that one operation is O(n) and the > other O(log(n)), or something like it?
Strictly based on complexity. If I had doubts I would have written a performance test for it but in this case, I just don't see how it can be slower than iterating over all the incidences (i.e. O(n)). Fetching the incidences instances is O(log n) because the memory calendar uses a QMap to store the incidences, then we iterate only over the incidence instances. The worst case complexity is the same but it is significantly better on average. Kr, -- Dr. Christophe Dumez Linux Software Engineer Intel Finland Oy - Open Source Technology Center _______________________________________________ MeeGo-dev mailing list [email protected] http://lists.meego.com/listinfo/meego-dev http://wiki.meego.com/Mailing_list_guidelines
