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未来方法参数的默认值

scala

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Omnipresent · 技术社区 · 7 年前

是否有方法为类型为的方法参数提供默认值 Future ?

我们可以做到:

def myMethod(test: String = "") {...}

但如果我们有未来呢

def myMethod(test: Future[List[Color]]) {...} //How can I set default for this?

1 回复 | 直到 7 年前

1

4

prayagupa soxunyi 7 年前

Future[A] 将包含

类型的成功结果 A

例子:

scala> import scala.concurrent.Future
import scala.concurrent.Future

scala> val success = Future.successful("some data")
success: scala.concurrent.Future[String] = Future(Success(some data))

scala> val failed = Future.failed(throw new Exception("failed"))
java.lang.Exception: failed
  ... 30 elided

这意味着您可以将示例中的默认值指定为 Future.successful ,

def myMethod(test: Future[List[Color]] = Future.successful(List.empty[Color])) {...}

例子:

scala> implicit val ec = scala.concurrent.ExecutionContext.global
ec: scala.concurrent.ExecutionContextExecutor = scala.concurrent.impl.ExecutionContextImpl$$anon$3@3c8802cf[Running, parallelism = 8, size = 0, active = 0, running = 0, steals = 0, tasks = 0, submissions = 0]

scala> def addLoyaltyPoints(previousLoyalty: Future[Int] = Future.successful(0)): Future[Int] = previousLoyalty.map(_ + 2)
addLoyaltyPoints: (previousLoyalty: scala.concurrent.Future[Int])scala.concurrent.Future[Int]

scala> addLoyaltyPoints()
res16: scala.concurrent.Future[Int] = Future(<not completed>)

scala> res16
res17: scala.concurrent.Future[Int] = Future(Success(2))

Future.successful(a) 而不是 Future(a) 因为 未来(a) 将在线程上运行。为预先计算的结果分配线程是没有意义的。