我有以下功能,我想返回A
,但IDE会提示我返回Sub ArrayDemo()
Dim A(10 ^ 6) As Integer
Dim B(10) As Integer
For i = 0 To 10 ^ 6
A(i) = Round(Rnd * 10, 0)
Next
Start = GetTickCount
k = 0
For i = 0 To 10 ^ 6
For j = 0 To k
If B(j) = A(i) Then GoTo nxt
Next
B(k) = A(i)
Debug.Print B(k)
k = k + 1
nxt:
Next
Debug.Print GetTickCount - Start
End Sub
。我是Scala的新手。有人能指出我做错了吗?
Future[Boolean]
答案 0 :(得分:2)
将onSuccess
替换为flatMap
。假设您的remove(x: String)
方法也返回Future
,那么还需要flatMap
ped:
def remove(loginInfo: LoginInfo): Future[Boolean] = {
val result = findObject(loginInfo)
result.flatMap {
case Some(persistentPasswordInfo) =>
val removeResult = remove(persistentPasswordInfo._id.toString)
removeResult.flatMap {
case Left(ex) => Future.successful(false)
case Right(b) => Future.successful(b)
}
case None => Future.successful(false)
}
}