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    Bug 
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    Resolution: Fixed
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     P3 P3
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    8
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        b115
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        Verified
                    A pipeline such as:
ThreadLocalRandom.current().ints().boxed().parallel().distinct().findAny();
will not terminate since distinct() is a barrier for an unordered input.
In such cases it is possible for the distinct operation to be lazy and use a distinct wrapping spliterator, which internally uses a ConcurrentHashMap to track distinct elements.
ThreadLocalRandom.current().ints().boxed().parallel().distinct().findAny();
will not terminate since distinct() is a barrier for an unordered input.
In such cases it is possible for the distinct operation to be lazy and use a distinct wrapping spliterator, which internally uses a ConcurrentHashMap to track distinct elements.