scala - Scala future 基础知识

我在Scala中有以下代码：

case class Water(temp: Int)

case class Milk(temp: Int)

def heatWaterFor(minutes: Int, water: Water) = Future {
  Thread.sleep(1000)
  Water(82)
}

def boilMilkFor(minutes: Int, milk: Milk) = Future {
  Thread.sleep(1000)
  Milk(90)
}

def frothMilk(hotwater: Water, hotmilk: Milk) = Future {
  Thread.sleep(1000)
  hotmilk
}

val start = System.currentTimeMillis()

val milkMaker = for {
   water <- heatWaterFor(10, Water(10))
   milk <- boilMilkFor(5, Milk(10))
   frothed = frothMilk(water, milk)
   hotMilk <- frothed
 } yield (hotMilk)

Await.ready(milkMaker, Duration.Inf)
val end = System.currentTimeMillis() - start
println(milkMaker.value + " , Time taken: "+((end/1000))+" seconds.")

我的目的是并行化heatWaterFor(...)和boilMilkFor(...)，因为它们是独立的。但是我觉得上面的代码是顺序的，根本没有利用期货的力量。显然，这需要3000毫秒才能运行（这是一个额外的证明）。

我在这里想念的最根本的东西是什么？

最佳答案

for表达式可简化为一系列map，flatMap和withFilter操作。您要表达的内容简化为以下形式：

heatWaterFor(10, Water(10))
        .flatMap(water => boilMilkFor(5, Milk(10))
                             .flatMap(milk => frothMilk(water, milk))

如您在此处看到的，下一个future的执行在上一个完成时开始。因此，如果要并行执行它们，则需要执行以下操作：

val heatWater = heatWaterFor(10, Water(10))
val boilMilk = boilMilkFor(5, Milk(10))

val milkMaker = for {
   water <- heatWater
   milk <- boilMilk
   hotMilk <- frothMilk(water, milk)
} yield (hotMilk)

这将同时启动heatWaterFor和boilMilkFor，当这两个都完成时将启动frothMilk（因为这取决于其他两个期货的结果）。