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 was O(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 the number of keys*, which is O(1). * Given that the number of elements in the vector is larger than or equals to the 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
