On Wed, Jun 12, 2019 at 11:57:29AM +0200, Daniel Bristot de Oliveira wrote:
> When a static key has more than one entry, these steps are called once for > each entry. The number of IPIs then is linear with regard to the number 'n' of > entries of a key: O(n*3), which is O(n). > Doing the update in this way, the number of IPI becomes O(3) with regard > to the number of keys, which is O(1). That's not quite true, what you're doing is n/X, which, in the end, is still O(n). It just so happens your X is 128, and so any n smaller than that ends up being 1.
