在Scala中处理future [Ething]类型

Question

I'm a bit struggling to get this structured. Here is what I'm trying to do:

def checkResultAndFetchUser(result: WriteResult, encryptedEmail: String): Future[Either[ServiceError, User]] = Future {
  if (result.code contains 11000)
    Left(ServiceError("Email already exists"))
  else if (result.hasErrors)
    Left(ServiceError(result.writeErrors.map(_.errmsg).toString))
  else
    userByEmail(encryptedEmail).map(user =>
      user
    ).recover {
      case NonFatal(ex) => Left(ServiceError(ex.getMessage))
    }
}

checkResultAndFetchUser(
  await(userCollection.insert(encryptedUser)), encryptedUser.email
)

I'm expecting that the checkResultAndFetchUser returns a Future[Either[ServiceError, User]] , but I get to see the following compiler failures:

Error:(155, 28) type mismatch;
 found   : scala.concurrent.Future[Either[DBService.this.ServiceError,com.inland.model.User]]
 required: Either[DBService.this.ServiceError,com.inland.model.User]
Error occurred in an application involving default arguments.
    checkResultAndFetchUser(
                           ^
Error:(150, 19) type mismatch;
 found   : scala.concurrent.Future[Either[DBService.this.ServiceError,com.inland.model.User]]
 required: Either[DBService.this.ServiceError,com.inland.model.User]
        ).recover {
                  ^

The userByEmail(encryptedEmail) method gives me a Future[Either[ServiceError, User]] as I would expect it to, but why and where is the problem?

EDIT: I've found a solution:

def checkResultAndFetchUser(result: WriteResult, encryptedEmail: String): Future[Either[ServiceError, User]] = {
  if (result.code contains 11000)
    Future(Left(ServiceError("Email already exists")))
  else if (result.hasErrors)
    Future(Left(ServiceError(result.writeErrors.map(_.errmsg).toString)))
  else
    userByEmail(encryptedEmail)
}

await(checkResultAndFetchUser(
  await(userCollection.insert(encryptedUser)), encryptedUser.email
))

Is that Okay? I mean, the implementation is safe as I'm using local variables to return a Future !

Answer 1

Your code is ok in the sense that it produce the expected result. However as @Łukasz mentioned in the comment, doing it this way is a little bit wasteful.

The reason is that whenever you instantiate a Future like that, a new task is spawned that needs to be scheduled on some ExecutionContext. Usually whenever you just need to wrap an already computed result in a Future (or if the computation is really quick) is better to use Future.successful so to avoid overhead.

Here's how I would modify the checkResultAndFetchUser function:

def checkResultAndFetchUser(result: WriteResult, encryptedEmail: String): Future[Either[ServiceError, User]] = {
  if (result.code contains 11000)
    Future.successful(Left(ServiceError("Email already exists")))
  else if (result.hasErrors)
    Future.successful(Left(ServiceError(result.writeErrors.map(_.errmsg).toString)))
  else
    userByEmail(encryptedEmail)
}