Question

在未等待复合的情况下调用.toString未来会导致不确定的结果。我的问题是为什么在未完成的期货上调用.toString在scala 2.10.x和2.11.x中返回"List()"？ The implementation似乎并不明确。

可以从REPL中看到此行为：

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

scala> Future(1).toString
res0: scala.concurrent.Future[Int] = Success(1)

scala> Future(1).toString
res1: scala.concurrent.Future[Int] = List()

请注意，Scala 2.12.x可能会明确实现Future#toString而不是返回"Future(<not completed>)"（source）。

编辑：证明这不是来自REPL或“隐藏隐藏”的人工制品（-Yno-predef删除所有默认隐含）：

Future.scala：

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

object Main extends App {
  System.out.println(Future(1).toString)
}

build.sbt：

scalaVersion := "2.11.8"

scalacOptions := Seq(
  "-deprecation",
  "-encoding", "UTF-8",
  "-feature",
  "-unchecked",
  "-Yno-predef",
  "-Xfatal-warnings",
  "-Xlint",
  "-Yinline-warnings",
  "-Yno-adapted-args",
  "-Ywarn-dead-code",
  "-Ywarn-unused-import",
  "-Ywarn-numeric-widen",
  "-Ywarn-value-discard",
  "-Xfuture")

Answer 1

消除sun.misc.Unsafe的依赖是一个不幸的副作用。它在Scala 2.12和更新版本的Scala 2.11 IIRC中得到了纠正。

为什么`Future＃toString`返回`“List（）”`？

1 个答案: