Hi Martin,
Thanks for explanation, now I can understand why you set the DELAY_MS to
100 seconds, it is true that it prevents failure on a slow host,
however, i still have some concerns.
Because the test tests to schedule tasks at the time in the past, so all
13 tasks should be executed immediately and finished within a short time.
If set the elapsed time limitation to 50s (DELAY_MS/2), it seems that
the timer have plenty of time to finish tasks, so whether it causes
above test point lost.
Back to the original test, i think it should be a test stabilization
issue, because the original test assumes that the timer should be
cancelled within < 1 second before the 14th task is called. this
assumption may not be guaranteed due to 2 reasons:
1. if test is executed in jtreg concurrent mode on a slow host.
2. the system clock of virtual machine may not be accurate (maybe faster
than physical).
To support the point, i changed the test as attached to print the
execution time to see whether the timer behaves expected as the API
document described. the result is as expected.
The unrepeated task executed immediately: [1401855509336]
The repeated task executed immediately and repeated per 1 second:
[1401855509337, 1401855510337, 1401855511338]
The fixed-rate task executed immediately and catch up the delay:
[1401855509338, 1401855509338, 1401855509338, 1401855509338,
1401855509338, 1401855509338, 1401855509338, 1401855509338,
1401855509338, 1401855509338, 1401855509338, 1401855509836, 1401855510836]
Thanks,
Eric
On 2014/6/4 9:16, Martin Buchholz wrote:
On Tue, Jun 3, 2014 at 6:12 PM, Eric Wang <yiming.w...@oracle.com
<mailto:yiming.w...@oracle.com>> wrote:
Hi Martin,
To sleep(1000) is not enough to reproduce the failure, because it
is much lower than the period DELAY_MS (10*1000) of the repeated
task created by "scheduleAtFixedRate(t, counter(y3), past, DELAY_MS)".
Try sleep(DELAY_MS), the failure can be reproduced.
Well sure, then the task is rescheduled, so I expect it to fail in
this case.
But in my version I had set DELAY_MS to 100 seconds. The point of
extending the DELAY_MS is to prevent flaky failure on a slow machine.
Again, how do we know that this test hasn't found a Timer bug?
I still can't reproduce it.
diff -r a02731d3f739 test/java/util/Timer/Args.java
--- a/test/java/util/Timer/Args.java Thu May 29 22:32:11 2014 -0700
+++ b/test/java/util/Timer/Args.java Wed Jun 04 14:17:57 2014 +0800
@@ -49,9 +49,9 @@
new F(){void f(){ t.scheduleAtFixedRate(task, d, period); }});
}
- TimerTask counter(final CountDownLatch latch) {
+ TimerTask counter(final CountDownLatch latch, final ArrayList<Long> list) {
return new TimerTask() { public void run() {
- check(latch.getCount() > 0);
+ list.add(System.currentTimeMillis());
latch.countDown();
}};
}
@@ -97,10 +97,12 @@
final CountDownLatch y3 = new CountDownLatch(11);
final long start = System.currentTimeMillis();
final Date past = new Date(start - 10500);
-
- schedule( t, counter(y1), past);
- schedule( t, counter(y2), past, 1000);
- scheduleAtFixedRate(t, counter(y3), past, 1000);
+ ArrayList<Long> l1 = new ArrayList<>();
+ ArrayList<Long> l2 = new ArrayList<>();
+ ArrayList<Long> l3 = new ArrayList<>();
+ schedule( t, counter(y1, l1), past);
+ schedule( t, counter(y2, l2), past, 1000);
+ scheduleAtFixedRate(t, counter(y3, l3), past, 1000);
y3.await();
y1.await();
y2.await();
@@ -108,9 +110,11 @@
final long elapsed = System.currentTimeMillis() - start;
System.out.printf("elapsed=%d%n", elapsed);
check(elapsed < 500);
-
+ Thread.sleep(2000);
t.cancel();
-
+ System.out.println("l1 " + Objects.toString(l1));
+ System.out.println("l2 " + Objects.toString(l2));
+ System.out.println("l3 " + Objects.toString(l3));
THROWS(IllegalStateException.class,
new F(){void f(){ t.schedule(x, 42); }},
new F(){void f(){ t.schedule(x, past); }},
