runtime: leave cleantimers early if G is being preempted

The cleantimers can run for a while in some unlikely cases.
If the GC is trying to preempt the G, it is forced to wait as the
G is holding timersLock. To avoid introducing a GC delay,
return from cleantimers if the G has a preemption request.

Fixes #37779

Change-Id: Id9a567f991e26668e2292eefc39e2edc56efa4e0
Reviewed-on: https://go-review.googlesource.com/c/go/+/223122
Run-TryBot: Ian Lance Taylor <iant@golang.org>
TryBot-Result: Gobot Gobot <gobot@golang.org>
Reviewed-by: Michael Knyszek <mknyszek@google.com>
Reviewed-by: Cherry Zhang <cherryyz@google.com>

1 files changed, 10 insertions, 0 deletions

diff --git a/src/runtime/time.go b/src/runtime/time.go
index 155e0501fe..c62207d7f7 100644
--- a/src/runtime/time.go
+++ b/src/runtime/time.go
@@ -499,10 +499,20 @@ func resettimer(t *timer, when int64) {
 // slows down addtimer. Reports whether no timer problems were found.
 // The caller must have locked the timers for pp.
 func cleantimers(pp *p) {
+	gp := getg()
 	for {
 		if len(pp.timers) == 0 {
 			return
 		}
+
+		// This loop can theoretically run for a while, and because
+		// it is holding timersLock it cannot be preempted.
+		// If someone is trying to preempt us, just return.
+		// We can clean the timers later.
+		if gp.preemptStop {
+			return
+		}
+
 		t := pp.timers[0]
 		if t.pp.ptr() != pp {
 			throw("cleantimers: bad p")