Edward, Thanks a lot. See below:
On Fri, Aug 16, 2019 at 10:10 AM Edward K. Ream <[email protected]> wrote: [...] > As Vitalije says" > > - Adding (or deleting) the last sibling will not change the position of > any existing node. This includes adding/deleting the last top-level node. > - If you can recast your algorithm in terms of vnodes, it will be > impervious to changes in the outline. > The way I read Vitalije's explanation, it sounded like the last child of a subtree. But are you saying it is only the last node in the entire outline? [...] > > Another possible pattern: > > 1. Create, say, an @dates node (somewhere) *first*, if it doesn't already > exist. If you do create an @dates node, then call c.redraw, which will, in > effect, recompute all positions. > Maybe I should start my journey on position enlightenment with c.redraw(). If I see how positions are recomputed, maybe I will better understand the behavior. > 2. Now that you have an @dates node, create your @date nodes as children > of the @dates node. This will preserve all positions. > Hmm, are you saying no matter where I create the @dates node all previously create positions will still be valid? Even after making more modifications like inserting @date nodes? Brian -- You received this message because you are subscribed to the Google Groups "leo-editor" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/leo-editor/CAO5X8CzE%3DyWvfuTr_oXDF_jeGwNAGCRbdeL%2BuzZa91TgESmB1A%40mail.gmail.com.
