scala - 从异常中恢复，scala Future

Question

为什么这个scala代码可以编译，恢复的签名是，

def recover[U >: T](pf: PartialFunction[Throwable, U])(implicit executor: ExecutionContext): Future[U]

那为什么下面的代码可以编译。您可以看到下面代码中的 recover 没有返回一个 Unit。

 object TestRecover {

    import scala.concurrent.ExecutionContext.Implicits.global
    import scala.concurrent.Future

    def failingFunction(input: Seq[String]): Future[Unit] = {
      Future {
        if (input.isEmpty) throw new Exception("Cannot be empty")
        else ()
      }
    }

    def callFailingFunc(input: Seq[String]): Future[Unit] = {
      failingFunction(input).recover {
        case _ =>
          //Not returning Unit here, but Future[Unit]. Shouldn't type checker fail this ?
          callFailingFunc(input.reverse)
      }
    }
  }

另外，为什么 Await.result(TestRecover.callFailingFunc(Seq()), 20.seconds) 由于无限递归而不会产生 stackoverflow？

Answer 1

then why does below code compile. You can see that recover in below code doesn't return a Unit.

编译器会稍微“帮助”您，并执行从 Future[Unit] 到 Unit 的隐式转换。像这样有效地编译它。

def callFailingFunc(input: Seq[String]): Future[Unit] = {
  failingFunction(input).recover {
    case _ =>
      //Not returning Unit here, but Future[Unit]. Shouldn't type checker fail this ?
      callFailingFunc(input.reverse)
      ()
  }
}

这个博客解释的很好:http://blog.bruchez.name/2012/10/implicit-conversion-to-unit-type-in.html

Also why Await.result(TestRecover.callFailingFunc(Seq()), 20.seconds) doesn't produce stackoverflow because of infinite recursion ?

您不会出现堆栈溢出，因为每次调用 failingFunction 都会创建一个带有新堆栈的新 Future。