Scala Future - 用于理解、混契约(Contract)步和异步

Question

在我的method1中，我需要异步调用另一个method2，它返回Option(result1)。比，如果 result1 为空，我需要异步调用另一个 method3，但如果 result1 不为空，我只需要返回它。

这是方法:

  def signIn(username: String): Future[User] = {
    for {
      foundUser <- userService.findByUsername(username) // this method returns Future[Option[User]], 
      // foundUser is Option[User]
      user <- if (foundUser.isEmpty) {
        val newUser = User(username = "User123")
        userService.create(newUser).map(Some(_)) // this method returns Future[Option[User]]
      }
      else
        // Here I want to return just foundUser, of course, it is not possible. 
        // IS THIS APPROACH CORRECT?? DOES THIS LINE CREATE ASYNCHRONOUS CALL?          
        Future.successful(foundUser)
    } yield user 
  }

问题是:
Future.successful(foundUser) - 这种方法在上面的代码中是否正确？此行是否创建异步调用？如果是，如何避免？我已经取了 找到用户 异步，我不想进行额外的异步调用只是为了返回已经获取的值。

Answer 1

Future.successful不会在提供的 ExecutionContext 上排队附加功能.它只是使用 Promise[T] 创建一个完整的Future[T] :

/** Creates an already completed Future with the specified result.
   *
   *  @tparam T       the type of the value in the future
   *  @param result   the given successful value
   *  @return         the newly created `Future` instance
   */
  def successful[T](result: T): Future[T] = Promise.successful(result).future

作为旁注，您可以使用 Option.fold 减少样板的数量。 :

def signIn(username: String): Future[User] = 
  userService
    .findByUsername(username)
    .flatMap(_.fold(userService.create(User(username = "User123")))(Future.successful(_))