scala.util.control.Exception Problem

Question
1

Can anyone tell me some knowledge and usage of ‘scala.util.control.Exception._’?
Such as
val rollbackIfException = handling(classOf[Throwable]) by { //???
t =>
{
tx.rollback()
throw t
}
}
rollbackIfException[A] { //??
val session = new Session(conn, Some(tx))
val result: A = execution(session)
tx.commit()
result
}

2

Is there something specific you want to know that isn’t covered in the Scaladoc?

fwiw, I don’t think scala.util.control.Exception is widely known or widely used.

3

Hi @yangmingsen,

You didn’t explain the problem you’re having, but I’m guessing it’s to do with the [A] type parameter?

You need to define rollbackIfException as a function with a type parameter for it to work, and I also passed it the tx:

imporWelcome to Scala 2.13.1 (OpenJDK 64-Bit Server VM, Java 11.0.6).
Type in expressions for evaluation. Or try :help.

scala> import scala.util.control.Exception._
import scala.util.control.Exception._

scala> trait Tx {
     |   def commit(): Unit
     |   def rollback(): Unit
     | }
defined trait Tx

scala> class FakeTx extends Tx {
     |   def commit()   = println("committing")
     |   def rollback() = println("rolling back")
     | }
defined class FakeTx

scala> def rollbackIfException[A](tx: Tx) =
     |   handling[A](classOf[Throwable]).by { t => tx.rollback(); throw t }
rollbackIfException: [A](tx: Tx)util.control.Exception.Catch[A]

scala> val tx: Tx = new FakeTx
tx: Tx = FakeTx@531bec12

scala> rollbackIfException[String](tx) {
     |   val result = "woohoo"
     |   tx.commit()
     |   result
     | }
committing
res1: String = woohoo

scala> rollbackIfException[String](tx) {
     |   val result = throw new Exception("boom")
     |   tx.commit()
     |   result
     | }
rolling back
java.lang.Exception: boom
  at .$anonfun$res2$1(<console>:2)
  at scala.util.control.Exception$Catch.apply(Exception.scala:227)
  ... 43 elided