On Oct 30, 2013, at 5:54 PM, Henry Jen wrote:
> Looks good to me.
>
> In the test,
>
>> Integer one = Stream.iterate(1, i -> i +
>> 1).unordered().parallel().distinct().findAny().get();
>> assertEquals(one.intValue(), 1);
>
> The implementation is probably make sure this will return 1, but
On Oct 30, 2013, at 5:54 PM, Henry Jen wrote:
> Looks good to me.
>
Thanks.
> In the test,
>
>> Integer one = Stream.iterate(1, i -> i +
>> 1).unordered().parallel().distinct().findAny().get();
>> assertEquals(one.intValue(), 1);
>
> The implementation is probably make sure this will ret
Looks OK to me.
Thanks for including a comment in the tests. :-)
Mike
On Oct 30 2013, at 02:30 , Paul Sandoz wrote:
> Hi
>
> Please review:
>
> http://cr.openjdk.java.net/~psandoz/tl/JDK-8027316-distinct-unordered/webrev/
>
> The fix ensures that the distinct operation, for a parallel st
Looks good to me.
In the test,
> Integer one = Stream.iterate(1, i -> i +
> 1).unordered().parallel().distinct().findAny().get();
> assertEquals(one.intValue(), 1);
The implementation is probably make sure this will return 1, but is that what
we spec to do? I sort of think it can have various
Hi
Please review:
http://cr.openjdk.java.net/~psandoz/tl/JDK-8027316-distinct-unordered/webrev/
The fix ensures that the distinct operation, for a parallel stream, is not a
full barrier if the upstream is unordered (and not already distinct, otherwise
it is a no-op).
Thanks,
Paul.
