Hi Ivan,
The updated code looks fine except for style.
Please put the return and the if's on different lines (by the style
conventions).
Line 210 and 211, and 216.
Thanks, Roger
On 03/15/2019 04:08 AM, Ivan Gerasimov wrote:
Hi David!
The code could be written like following:
long start = System.nanoTime();
do {
...
long elapsed = System.nanoTime() - start;
remainingNanos = timeoutNanos - elapsed;
} while (remainingNanos > 0);
but then one realizes that this always calculates (start +
timeoutNanos), both of which are effectively final.
So, the sum can be calculated in advance, and named, say, "deadline" :)
With kind regards,
Ivan
On 3/15/19 12:16 AM, David Holmes wrote:
On 15/03/2019 5:03 pm, Ivan Gerasimov wrote:
Thank you David!
On 3/14/19 11:01 PM, David Holmes wrote:
Hi Ivan,
On 15/03/2019 12:02 pm, Ivan Gerasimov wrote:
Hi David!
On 3/14/19 5:28 PM, David Holmes wrote:
Hi Ivan,
On 15/03/2019 5:49 am, Ivan Gerasimov wrote:
Hello!
The default implementation of Process.waitFor(long, TimeUnit)
does not check if the process has exited after the last portion
of the timeout has expired.
Please clarify. There is always a race between detecting a
timeout and the process actually terminating. If the process
actually terminates while you're deciding to report a timeout
that seems just an acceptable possibility. No matter what you do
the process could terminate just after you decided to report the
timeout.
Current code for waitFor(...) is
212 do {
213 try {
214 exitValue();
215 return true;
216 } catch(IllegalThreadStateException ex) {
217 if (rem > 0)
218 Thread.sleep(
219 Math.min(TimeUnit.NANOSECONDS.toMillis(rem) + 1, 100));
220 }
221 rem = unit.toNanos(timeout) - (System.nanoTime()
- startTime);
222 } while (rem > 0);
223 return false;
So, if the loop has just processed the last 100 ms portion of the
timeout, it ends right away, without checking if the process has
exited during this period.
Ah I see. Yes there should be a final check after what will be the
last sleep.
Not a big deal of course, but it is more accurate to check if the
process has exited at the *end* of the specified timeout, and not
100 milliseconds before the end.
Agreed.
I share Pavel's concern about the implicit overflow in calculating
a deadline, rather than comparing elapsed time with the timeout.
There has been some effort in core-libs to remove assumptions about
how overflows are handled.
We only check sign of the remainingNanos, which is initially
strictly positive timeout. Over time it gets decreased by elapsed
time.
The elapsed time is the difference between System.nanoTime()
measured at line 217 and 213, so it is always non-negative (well,
strictly positive, as there's sleep in between).
I don't really see a possibility of overflow for remainingNanos here.
The deadline may overflow, and it is totally fine, as we don't care
about its sign.
It's deadline that I was flagging.
Am I missing something?
Just a code cleanliness issue. I had thought, but now can't find,
these kinds of overflowing calculations were being cleaned up in
core-libs. But perhaps I'm just mis-remembering.
Cheers,
David
With kind regards,
Ivan
Thanks,
David
-----
With kind regards,
Ivan
David
-----
JDK has two implementations of Process (for Unix and Windows)
and they both override waitFor(), so it's not an issue for them.
Still, it is better to provide a more accurate default
implementation.
I'm not quite certain the regression test needs to be included
in the fix. The test does demonstrate the issue with the
unfixed JDK and passed Okay on all tested platforms in Mach5.
Yet, I suspect the test can still show false negative results,
as there are no guaranties that even such simple application as
`true` will finish in 100 ms.
I can tag the test as @ignored with a comment, or simply remove
it from the fix.
BUGURL: https://bugs.openjdk.java.net/browse/JDK-8220684
WEBREV: http://cr.openjdk.java.net/~igerasim/8220684/00/webrev/
Thanks in advance for reviewing!
