`On 12/06/2019 16:52, Peter Zijlstra wrote:> 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.> Correct! In the v1, when I was using a (dynamic) linked list of keys, it wasO(1), now it is O(n).Using an academic hat of easy assumptions, I could argue that:"Doing the update in this way, the number of IPI becomes O(3) with regard to thenumber of keys*, which is O(1).* Given that the number of elements in the vector is larger than or equals tothe numbers of entries of a given key, O(n) otherwise."Life is so easy when we can do such assumptions, like infinity memory :-)So, yeah, with a fixed size vector, it is O(n) in the worst case, but still"O(1)" in the vast majority of cases.-- Daniel`