On Wed, Jul 18, 2018 at 02:03:18PM +0200, Peter Zijlstra wrote: > On Thu, Jul 12, 2018 at 01:29:40PM -0400, Johannes Weiner wrote: > > + for (to = 0; (bo = ffs(set)); to += bo, set >>= bo) > > + tasks[to + (bo - 1)]++; > > You want to benchmark this, but since it's only 3 consecutive bits, it > might actually be faster to not use ffs() and simply test all 3 bits: > > for (to = set, bo = 0; to; to &= ~(1 << bo), bo++) if (to & (1 << bo)) > tasks[bo]++;