actually, i think the race does not exist. up() is perfectly safely done
on the on-stack semaphore, because both the wake_up() done by __up() and
the __down() path takes the waitqueue spinlock, so i cannot see where the
up() touches the semaphore after the down()-ed task has been woken up.
the second argument still holds though - a completion is probably slightly
cheaper in this case.
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