给出以下两种方法:
def f: Future[Int] = Future { 10 }
def g: Future[Int] = Future { 5 }
我想把它们写成:
scala> import scala.concurrent.Future
import scala.concurrent.Future
scala> import scala.concurrent.Future._
import scala.concurrent.Future._
scala> import scala.concurrent.ExecutionContext.Implicits.global
import scala.concurrent.ExecutionContext.Implicits.global
scala> for {
| a <- f
| b <- g
| } yield (a+b)
res2: scala.concurrent.Future[Int] = scala.concurrent.impl.Promise$DefaultPromise@34f5090e
现在,我会打电话Await.result
阻止直到完成。
scala> import scala.concurrent.duration._
import scala.concurrent.duration._
正如预期的那样,我得到15
, since Await.result
took a Future[Int]
并返回一个Int
.
scala> Await.result(res2, 5.seconds)
res6: Int = 15
定义一个recoverFn
对于失败的未来:
scala> val recoverFn: PartialFunction[Throwable, Future[Int]] =
{ case _ => Future{0} }
recoverFn: PartialFunction[Throwable,scala.concurrent.Future[Int]] = <function1>
我尝试定义一个 failedFuture:
scala> def failedFuture: Future[Int] = Future { 666 }.failed.recoverWith{ recoverFn }
<console>:20: error: type mismatch;
found : scala.concurrent.Future[Any]
required: scala.concurrent.Future[Int]
def failedFuture: Future[Int] = Future { 666 }.failed.recoverWith{ recoverFn }
^
但是,我收到上述编译时错误。
具体来说,我该如何修复这个错误?一般来说,是Future#recoverWith
通常是如何失败的Future
的处理了吗?