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.
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
Looks OK to me.
Thanks for including a comment in the tests. :-)
Mike
On Oct 30 2013, at 02:30 , Paul Sandoz paul.san...@oracle.com wrote:
Hi
Please review:
http://cr.openjdk.java.net/~psandoz/tl/JDK-8027316-distinct-unordered/webrev/
The fix ensures that the distinct operation,
On Oct 30, 2013, at 5:54 PM, Henry Jen henry@oracle.com 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
