在 Scala 中，是否有用于将异常转换为选项的预先存在的库函数? [英] In Scala, is there a pre-existing library function for converting exceptions to Options?

问题描述

这基本上是为了包装 java 工厂方法，如果无法根据输入创建项目，则会抛出异常.我正在基础库中寻找类似的东西:

This is basically to wrap java factory methods which throw exceptions if the item can't be created based on the inputs. I'm looking for something in the base library like:

 def exceptionToOption[A](f: => A):Option[A] ={
    try{
      Some(f)}
    catch{
      case e:Exception => None}
  }

用法:

val id:Option[UUID] = exceptionToOption(UUID.fromString("this will produce None"))

我知道我可以自己写，但我想检查一下我没有重新发明轮子.

I know I can write my own but I want to check I am not re-inventing the wheel.

推荐答案

使用 scala.util.control.Exception:

import scala.util.control.Exception._

allCatch opt f

你可以让它更复杂.例如，只捕获算术异常并检索异常:

And you can make it more sophisticated. For example, to catch only arithmetic exceptions and retrieve the exception:

scala> catching(classOf[ArithmeticException]) either (2 / 0)
res5: Either[Throwable,Int] = Left(java.lang.ArithmeticException: / by zero)

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