如果我在内部抛出异常，以下代码如何打印成功（4）？

Question

object MyRealMainObj extends App {

  println(
    Try(1)
      .map(doOne)
      .map(doTwo)

  )

  def doOne(i: Int): Int = i + 1; throw new RuntimeException("failed in one")
  def doTwo(i: Int): Int = i + 2
}

结果：

Success(4)
Exception in thread "main" java.lang.RuntimeException: failed in one
    at MyRealMainObj$.delayedEndpoint$MyRealMainObj$1(TestMainArgs.scala:16)
    at MyRealMainObj$delayedInit$body.apply(TestMainArgs.scala:7)
    at scala.Function0$class.apply$mcV$sp(Function0.scala:40)
    at scala.runtime.AbstractFunction0.apply$mcV$sp(AbstractFunction0.scala:12)
    at scala.App$$anonfun$main$1.apply(App.scala:76)
    at scala.App$$anonfun$main$1.apply(App.scala:76)

怎么打印Success(4) doOne不应该失败，因为我在那里抛出异常？

Answer 1

将代码包装在大括号内，如

def doOne(i: Int): Int = {i + 1; throw new RuntimeException("failed in one")}

抛出新的RuntimeException（＆＃34;在一个＆＃34;中失败）不在函数内，它是主流内的一个单独的行。