scala - 如何将隐式参数放入匿名函数中

Question

如何将隐式 val myConnection 放入 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
}

Answer 1

你不能直接这样做。是myConnection的值真的没有确定之前你打电话execute ?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，您将一个参数传递给评估函数，但是您必须记住在调用站点将其标记为隐式。

如果您有多个这样的隐式，您可能希望将它们放在一个专用的“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 import ，您可以为 PassedImplicits 中的每个字段提供“隐式 setter/getter ”。并写下这样的东西，然后:

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

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