CODEWITHSUNDEEP
scala
user830818
user830818

Reputation: 417

Option[Serializable] does not conform to Try[String]

The following code generates the compilation error Option[serializable] does not conform to Try[String]

In all the places where the method can return, I am returning either a Try(String) or Failure(Exception). Given that why does compiler think I am returning Option[Serializable]?

def m1(arg1 : String, arg2 : String): Try[String] = {
    for {
      x1 <- getM1(arg1)
      x2 <- getM2(arg2)
    } yield {
      val id : Either[Exception, String]=  weirdFunction(x1, x2)
      id match {
        case Left(e) => Failure(Exception("")))
        case Right(id) => Success(id)
      }
   } getOrElse(Failure(Exception("")))
 }

Upvotes: 1

Views: 1310

Answers (1)

Ben Reich
Ben Reich

Reputation: 16324

It's just an issue of parentheses:

def m1(arg1 : String, arg2 : String): Try[String] = {
    for {
      x1 <- getM1(arg1)
      x2 <- getM2(arg2)
    } yield {
      val id : Either[Exception, String]=  weirdFunction(x1, x2)
      id match {
        case Left(e) => Failure(Exception("")))
        case Right(id) => Success(id)
      }
   } /* getOrElse used to be here */
}.getOrElse(Failure(Exception("")))

The compiler thought you were trying to call getOrElse on the Try returned in the yield statement (how could it know otherwise?). Since there is a getOrElse method on Try as well as Option, this is a particularly confusing error message. So, it thought you were trying to either return the id in the Success of the yield statement or the Failure afterwards. Their least common super type is Serializable, and since it thought we were still wrapped in the for comprehension, it deduced the return type to be Option[Serializable.

Upvotes: 3

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