4 Jul
2014
4 Jul
'14
10:54 a.m.
2014-07-04 11:52 GMT+02:00 Benedek Thaler <thalerbenedek@gmail.com>:
What if the second element cannot be pushed to the downstream queue because it's full? The transformation should yield (and not spin).
that problem could be solved by fibers - if a fiber would enqueue an item in a bounded-queue and the queue is full, the fiber is suspended and the thread can resume another fiber (executing another transformation). if items are removed from the bounded-queue, the waiting fiber gets notified an will be resumed, e.g. is able to enqueue its item.