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runtime: leave cleantimers early if G is being preempted

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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>
This commit is contained in:
Ian Lance Taylor 2020-03-11 22:03:50 -07:00
parent 29b36a88ab
commit 5d70cb0667
1 changed files with 10 additions and 0 deletions
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10
src/runtime/time.go
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@ -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")