Scalaz流括号提前关闭资源

考虑以下代码：

import java.time.Instant
import scalaz.concurrent.Task
import scalaz.stream._

  class MyResource {

    println("resource obtained")

    @volatile var v = 1

    def read() = v

    def close() {
      v = 0
      println("resource closed")
    }
  }

  val expensiveCalculation = (i: Int, rc: MyResource) => {
    Thread.sleep(1000)
    val r = rc.read()
    println(s"calculating ($i, $r) on ${Thread.currentThread} at ${Instant.now}")
    i / r
  }

我想使用Process.bracket来管理资源：

val resource = Process.bracket(Task.delay(new MyResource))(r => Process.eval_(Task.now(r.close())))(rc => Process.repeatEval(Task.now(rc)))

并对一系列数字进行计算：

val range = Process.range(0, 5).toSource

val s1 = range.tee(resource)(tee.zip).map {
    case (i, rc) => expensiveCalculation(i, rc)
}

println(s"start s1 at ${Instant.now}")
val res1 = s1.runLog.attemptRun
println(s"done at ${Instant.now}: $res1")

到目前为止还不错。。。我得到这个结果：

start s1 at 2016-04-04T13:44:26.450Z
resource obtained
calculating (0, 1) on Thread[main,5,main] at 2016-04-04T13:44:27.565Z
calculating (1, 1) on Thread[main,5,main] at 2016-04-04T13:44:28.573Z
calculating (2, 1) on Thread[main,5,main] at 2016-04-04T13:44:29.575Z
calculating (3, 1) on Thread[main,5,main] at 2016-04-04T13:44:30.577Z
calculating (4, 1) on Thread[main,5,main] at 2016-04-04T13:44:31.578Z
resource closed
done at 2016-04-04T13:44:31.597Z: \/-(Vector(0, 1, 2, 3, 4))

start s2 at 2016-04-04T13:44:31.601Z
resource obtained
resource closed
calculating (0, 0) on Thread[pool-1-thread-3,5,main] at 2016-04-04T13:44:32.717Z
calculating (1, 0) on Thread[pool-1-thread-8,5,main] at 2016-04-04T13:44:32.717Z
calculating (4, 0) on Thread[pool-1-thread-7,5,main] at 2016-04-04T13:44:32.717Z
calculating (2, 0) on Thread[pool-1-thread-2,5,main] at 2016-04-04T13:44:32.717Z
calculating (3, 0) on Thread[pool-1-thread-6,5,main] at 2016-04-04T13:44:32.717Z
done at 2016-04-04T13:44:32.779Z: -\/(java.lang.ArithmeticException: / by zero)

现在，我想并行运行昂贵的计算：

val s2 = merge.mergeN(range.zipWith(resource){
    case (i, rc) => Process.eval(Task.delay(expensiveCalculation(i, rc)))
})

println(s"start s2 at ${Instant.now}")
val res2 = s2.runLog.attemptRun
println(s"done at ${Instant.now}: $res2")

我得到了这个结果：

start s1 at 2016-04-04T13:44:26.450Z
resource obtained
calculating (0, 1) on Thread[main,5,main] at 2016-04-04T13:44:27.565Z
calculating (1, 1) on Thread[main,5,main] at 2016-04-04T13:44:28.573Z
calculating (2, 1) on Thread[main,5,main] at 2016-04-04T13:44:29.575Z
calculating (3, 1) on Thread[main,5,main] at 2016-04-04T13:44:30.577Z
calculating (4, 1) on Thread[main,5,main] at 2016-04-04T13:44:31.578Z
resource closed
done at 2016-04-04T13:44:31.597Z: \/-(Vector(0, 1, 2, 3, 4))

start s2 at 2016-04-04T13:44:31.601Z
resource obtained
resource closed
calculating (0, 0) on Thread[pool-1-thread-3,5,main] at 2016-04-04T13:44:32.717Z
calculating (1, 0) on Thread[pool-1-thread-8,5,main] at 2016-04-04T13:44:32.717Z
calculating (4, 0) on Thread[pool-1-thread-7,5,main] at 2016-04-04T13:44:32.717Z
calculating (2, 0) on Thread[pool-1-thread-2,5,main] at 2016-04-04T13:44:32.717Z
calculating (3, 0) on Thread[pool-1-thread-6,5,main] at 2016-04-04T13:44:32.717Z
done at 2016-04-04T13:44:32.779Z: -\/(java.lang.ArithmeticException: / by zero)

显然，调用括号的释放函数太早，过早地关闭了资源。我做错了什么？有没有办法在并行计算完成后关闭资源

---

谢谢。

在阅读了更多的文章并试用了scalaz stream之后，我找到了这个解决方案，介绍了如何在并行执行后释放资源：

val p: Process[Task, Int] = Process.await(Task.delay(new MyResource)) { rc =>

    val range = Process.range(0, 5).toSource

    merge.mergeN(range.map { i =>
      Process.eval(Task.delay(expensiveCalculation(i, rc)))
    }).onComplete(Process eval_ Task.delay {
      rc.close()
    })
}

p.runLog.attemptRun