scala – 如何将隐式参数导入匿名函数

发布时间：2020-12-16 10:03:16 所属栏目：安全来源：网络整理

导读：如何将隐式val myConnection置于execute(true)函数的范围内 def execute[T](active: Boolean)(blockOfCode: = T): Either[Exception,T] = { implicit val myConnection = "how to get this implicit val into scope" Right(blockOfCode)}execute(true){// my

如何将隐式val myConnection置于execute(true)函数的范围内

def execute[T](active: Boolean)(blockOfCode: => T): Either[Exception,T] = {
  implicit val myConnection = "how to get this implicit val into scope"
  Right(blockOfCode)
}



execute(true){
// myConnection is not in scope
  useMyConnection()   // <- needs implicit value
}

解决方法

你不能直接这样做.在调用execute之前,myConnection的值是否真的没有确定？在这种情况下,您可以这样做：

def execute[T](active: Boolean)(blockOfCode: String => T): Either[Exception,T] = {
  val myConnection = "how to get this implicit val into scope"
  Right(blockOfCode(myConnection))
}

execute(true) { implicit connection =>
  useMyConnection() 
}

基本上,您将参数传递给evaluate函数,但是您必须记住在调用站点上隐式标记它.

如果您有几个这样的含义,您可能希望将它们放在专用的“隐式提供者”类中.例如.：

class PassedImplicits(implicit val myConnection: String)

def execute[T](active: Boolean)(blockOfCode: PassedImplicits => T): Either[Exception,T] = {
  val myConnection = "how to get this implicit val into scope"
  Right(blockOfCode(new PassedImplicits()(myConnection)))
}

execute(true) { impl =>
  import impl._
  useMyConnection() 
}

如果你想避免导入,你可以为PassedImplicits中的每个字段提供“隐式getter”并写下这样的东西,然后：

implicit def getMyConnection(implicit impl: PassedImplicits) = impl.myConnection

execute(true) { implicit impl =>
  useMyConnection() 
}

（编辑：李大同）

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