On Thu, Sep 10, 2020 at 08:18:01PM -0700, Alexei Starovoitov wrote: > On Thu, Sep 10, 2020 at 1:20 PM <paulmck@xxxxxxxxxx> wrote: > > > > From: "Paul E. McKenney" <paulmck@xxxxxxxxxx> > > > > The various RCU tasks flavors currently wait 100 milliseconds between each > > grace period in order to prevent CPU-bound loops and to favor efficiency > > over latency. However, RCU Tasks Trace needs to have a grace-period > > latency of roughly 25 milliseconds, which is completely infeasible given > > the 100-millisecond per-grace-period sleep. This commit therefore reduces > > this sleep duration to 5 milliseconds (or one jiffy, whichever is longer) > > in kernels built with CONFIG_TASKS_TRACE_RCU_READ_MB=y. > > The commit log is either misleading or wrong? > If I read the code correctly in CONFIG_TASKS_TRACE_RCU_READ_MB=y > case the existing HZ/10 "paranoid sleep" is preserved. Yes, for CONFIG_TASKS_TRACE_RCU_READ_MB=y, the previous 100-millisecond "paranoid sleep" is preserved. Preserving previous behavior is of course especially important for rcupdate.rcu_task_ipi_delay, given that real-time applications are degraded by IPIs. And given that we are avoiding IPIs in this case, speeding up the polling is not all that helpful. > It's for the MB=n case it is reduced to HZ/200. Yes, that is, to roughly 5 milliseconds for large HZ or to one jiffy for HZ<200. Here, we send IPIs much more aggressively, so polling more frequently does help a lot. > Also I don't understand why you're talking about milliseconds but > all numbers are HZ based. HZ/10 gives different number of > milliseconds depending on HZ. As long as HZ is 10 or greater, HZ/10 jiffies is roughly 100 milliseconds. In the unlikely event that HZ is less than 10, the code clamps to one jiffy. Since schedule_timeout_idle() sleep time is specified in jiffies, it all works out. Thanx, Paul
