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adamw avatar adamw commented on May 24, 2024 1

You're right, the example in the docs is misleading.

What happens there is that because .tell is non-blocking, the invocation is scheduled, and the scope ends (there's not other code to run - the main body is complete). When this happens, all threads that are still running (which includes actors) are interrupted. Hence, the println is never invoked.

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adamw avatar adamw commented on May 24, 2024 1

Interesting! This surprised me as well, but it seems println is not interruptible!

You can observe the interruption e.g. as follows:

    def increment(delta: Int): Int =
      counter += delta
      try Thread.sleep(100)
      catch
        case e: Exception =>
          println("Exception: " + e)
          throw e
      println(counter)
      counter

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adamw avatar adamw commented on May 24, 2024 1

I've expanded the example (4542e47), I'll close this for now as I think the code is fine, but please reopen if there are some more problems with tell

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windbird123 avatar windbird123 commented on May 24, 2024

Thank you for your kind explanation.

When I replace Thread.sleep(100) with (1 to 10000000).foreach(println), it seemed like there was no interruption. (It printed all numbers)

 def increment(delta: Int): Int =
    counter += delta
    // Thread.sleep(100)
    (1 to 10000000).foreach(println)
    println(counter)
    counter

I thought that (1 to 10000000).foreach(println) to be interrupted since the scope already ends (and main body is complete).
In this case, could we assume that ox tried to interrupt but was not interrupted.

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