Scala Try[Unit] confusion

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滥情空心 2021-01-21 00:23

I have this piece of code

import scala.util.Try
val t: Try[Unit] = Try(Try(1))

and 2 questions:

  • What is happening here? How can t
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2条回答
  • 2021-01-21 00:35

    You are basically forcing compiler to assign Try as Unit.

    For exmple doSomething method below is supposed to return Int as that is last statement, but return type Unit is forcing it to return ().

    scala> def doSomething: Unit = 1 + 1
    doSomething: Unit
    

    In your example val t: Try[Unit] = Try(Try(1 / 0)), you asking compiler to treat inner Try(1 / 0) as Unit; which means

    scala> val innerTry: Unit = Try(1 / 0)
    innerTry: Unit = ()
    

    Which means even if Try fails, its Unit which is always Success for another Try that you have.

    scala> val t: Try[Unit] = Try(someOperation)
    t: scala.util.Try[Unit] = Success(())
    

    Better remove the specific type you are providing and let the compiler figure it out,

    scala> val t = Try(Try(1 / 0))
    t: scala.util.Try[scala.util.Try[Int]] = Success(Failure(java.lang.ArithmeticException: / by zero))
    

    Also read: Scala: Why can I convert Int to Unit?

    final abstract class Unit private extends AnyVal {
      // Provide a more specific return type for Scaladoc
      override def getClass(): Class[Unit] = ???
    }
    
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  • 2021-01-21 00:44

    Try(1) returns Success(1). Try(Try(1) returns Success(())because of its assignment to Type Try[Unit] Similar to the following commands in Scala REPL where x is forced to take Unit type:

    scala> val x = 5
    x: Int = 5
    
    scala> val x:Unit = 5
    <console>:23: warning: a pure expression does nothing in statement position; you may be omitting necessary parentheses
           val x:Unit = 5
                        ^
    x: Unit = ()
    
    scala>
    

    So obviously the compiler is taking the return value and is forced to type () which is Unit. Because it is Unit may be it makes sense the way compiler interprets it.

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