在Scala中将Option转换为Either

问题描述

假设我需要在Scala中将Option[Int]转换为Either[String, Int].我想这样做:

Suppose I need to convert Option[Int] to Either[String, Int] in Scala. I'd like to do it like this:

def foo(ox: Option[Int]): Either[String, Int] =
  ox.fold(Left("No number")) {x => Right(x)}

不幸的是，上面的代码无法编译，我需要显式添加类型Either[String, Int]:

Unfortunately the code above doesn't compile and I need to add type Either[String, Int] explicitly:

ox.fold(Left("No number"): Either[String, Int]) { x => Right(x) }

是否可以通过这种方式将Option转换为Either而无需添加类型?
您如何建议将Option转换为Either?

Is it possible to convert Option to Either this way without adding the type ?
How would you suggest convert Option to Either ?

推荐答案

否，如果您采用这种方式，就不能忽略该类型.

No, if you do it this way, you can't leave out the type.

推断Left("No number")的类型为Either[String, Nothing].仅从Left("No number")编译器就无法知道您是否希望Either的第二种类型为Int，并且类型推断的步伐还很遥远，以至于编译器将查看整个方法并决定应采用哪种方法是Either[String, Int].

The type of Left("No number") is inferred to be Either[String, Nothing]. From just Left("No number") the compiler can't know that you want the second type of the Either to be Int, and type inference doesn't go so far that the compiler will look at the whole method and decide it should be Either[String, Int].

您可以通过多种不同的方式来执行此操作.例如，使用模式匹配:

You could do this in a number of different ways. For example with pattern matching:

def foo(ox: Option[Int]): Either[String, Int] = ox match {
  case Some(x) => Right(x)
  case None    => Left("No number")
}

或带有if表达式:

def foo(ox: Option[Int]): Either[String, Int] =
  if (ox.isDefined) Right(ox.get) else Left("No number")

或使用Either.cond:

def foo(ox: Option[Int]): Either[String, Int] =
  Either.cond(ox.isDefined, ox.get, "No number")

这篇关于在Scala中将Option转换为Either的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！