Joel Fernandes found that the synchronize_rcu_tasks() was taking a
significant amount of time. He demonstrated it with the following test:
# cd /sys/kernel/tracing
# while [ 1 ]; do x=1; done &
# echo '__schedule_bug:traceon' > set_ftrace_filter
# time echo '!__schedule_bug:traceon' > set_ftrace_filter;
real 0m1.064s
user 0m0.000s
sys 0m0.004s
Where it takes a little over a second to perform the synchronize,
because there's a loop that waits 1 second at a time for tasks to get
through their quiescent points when there's a task that must be waited
for.
After discussion we came up with a simple way to wait for holdouts but
increase the time for each iteration of the loop but no more than a
full second.
With the new patch we have:
# time echo '!__schedule_bug:traceon' > set_ftrace_filter;
real 0m0.131s
user 0m0.000s
sys 0m0.004s
Which drops it down to 13% of what the original wait time was.
Link: http://lkml.kernel.org/r/20180523063815.198302-2-joel@joelfernandes.org
Reported-by: Joel Fernandes (Google) <joel@joelfernandes.org>
Suggested-by: Joel Fernandes (Google) <joel@joelfernandes.org>
Signed-off-by: Steven Rostedt (VMware) <rostedt@goodmis.org>
Signed-off-by: Paul E. McKenney <paulmck@linux.vnet.ibm.com>
struct rcu_head *list;
struct rcu_head *next;
LIST_HEAD(rcu_tasks_holdouts);
+ int fract;
/* Run on housekeeping CPUs by default. Sysadm can move if desired. */
housekeeping_affine(current, HK_FLAG_RCU);
* holdouts. When the list is empty, we are done.
*/
lastreport = jiffies;
- while (!list_empty(&rcu_tasks_holdouts)) {
+
+ /* Start off with HZ/10 wait and slowly back off to 1 HZ wait*/
+ fract = 10;
+
+ for (;;) {
bool firstreport;
bool needreport;
int rtst;
struct task_struct *t1;
- schedule_timeout_interruptible(HZ);
+ if (list_empty(&rcu_tasks_holdouts))
+ break;
+
+ /* Slowly back off waiting for holdouts */
+ schedule_timeout_interruptible(HZ/fract);
+
+ if (fract > 1)
+ fract--;
+
rtst = READ_ONCE(rcu_task_stall_timeout);
needreport = rtst > 0 &&
time_after(jiffies, lastreport + rtst);