Scala

Question

可能标题不是很清楚。这是我的问题。

假设我有一个使用一系列配置参数定义应用程序的特征。这些参数包含在一个Map中，其中一些有默认值。

trait ConfApp {
  val dbName: String
  lazy val conf: scala.collection.mutable.Map[String, Any] = scala.collection.mutable.Map("db" -> dbName, "foo" -> "bar")
}

所以我可以创建一个自定义应用程序如下：

class MyApp extends ConfApp {
  override val dbName = "my_app_db"

  // print app configuration parameters
  println(conf)

  def add() = {...}
  ...
}

val M1 = new Myapp    // Map(db -> my_app_db, foo -> bar)

我想创建其他特征来设置一些其他配置参数。换句话说，我希望能够做类似的事情：

class MyApp2 extends ConfApp with LogEnabled {
  override val dbName = "my_app2_db"
  // print app configuration parameters
  println(conf)

  def add() = {...}
  ...
}

val M2 = new Myapp2    // Map(db -> my_app_db, foo -> bar, log -> true)

到目前为止，我已经设法做到了以下几点：

trait LogEnabled {
  val conf: scala.collection.mutable.Map[String, Any]
  conf("log") = true
}

trait LogDisabled {
  val conf: scala.collection.mutable.Map[String, Any]
  conf("log") = false
}

trait ConfApp {
  val dbName: String
  lazy val conf: scala.collection.mutable.Map[String, Any] = scala.collection.mutable.Map("db" -> dbName, "foo" -> "bar")
}

class MyApp extends ConfApp {
  val dbName = "my_app_db"
  println(conf)
}

class MyApp2 extends ConfApp with LogDisabled {
  val dbName = "my_app_db"
  println(conf)
}

val M = new MyApp         // Map(db -> my_app_db, foo -> bar)
val M2 = new MyApp2       // Map(log -> false, foo -> bar, db -> null)

但是正如您在 M2 中看到的那样，db 参数是 null。我不明白我做错了什么。

真诚地，我一点也不喜欢这种使用可变 Map 的方法，但我没能做得更好。

Answer 1

如果我是你，我不会为此使用 vals，因为初始化顺序存在很多问题。一方面，您还可以在特征中使用自引用来指定它们必须与 ConfApp 混合在一起。那就是说我会做这样的事情：

trait LogEnabled { self: ConfApp =>
  self.conf("log") = true
}

trait LogDisabled { self: ConfApp =>
  self.conf("log") = false
}

trait ConfApp {
  def dbName: String
  lazy val conf: scala.collection.mutable.Map[String, Any] = scala.collection.mutable.Map("db" -> dbName, "foo" -> "bar")
}

class MyApp extends ConfApp {
  override def dbName = "my_app_db"
  println(conf)
}

class MyApp2 extends ConfApp with LogDisabled {
  override def dbName = "my_app_db"
  println(conf)
}

这似乎工作正常。

Answer 2

您仍然可以这样使用不可变 Map：

scala> trait ConfApp {
     |   val dbName: String
     |   def conf: Map[String, Any] = Map("db" -> dbName, "foo" -> "bar")
     | }
defined trait ConfApp

scala> trait LogEnabled extends ConfApp {
     |   override def conf = super.conf.updated("log", true)
     | }
defined trait LogEnabled

scala> trait LogDisabled extends ConfApp {
     |   override def conf = super.conf.updated("log", false)
     | }
defined trait LogDisabled

scala> class MyApp extends ConfApp {
     |   val dbName = "my_app_db"
     |   println(conf)
     | }
defined class MyApp

scala> class MyApp2 extends ConfApp with LogDisabled {
     |   val dbName = "my_app_db2"
     |   println(conf)
     | }
defined class MyApp2

scala> new MyApp
Map(db -> my_app_db, foo -> bar)
res0: MyApp = MyApp@ccc268e

scala> new MyApp2
Map(db -> my_app_db2, foo -> bar, log -> false)
res1: MyApp2 = MyApp2@59d91aca

scala> new ConfApp with LogDisabled with LogEnabled {
     |   val dbName = "test1"
     |   println(conf)
     | }
Map(db -> test1, foo -> bar, log -> true)
res2: ConfApp with LogDisabled with LogEnabled = $anon@16dfdeda

scala> new ConfApp with LogEnabled with LogDisabled  {
     |   val dbName = "test2"
     |   println(conf)
     | }
Map(db -> test2, foo -> bar, log -> false)
res3: ConfApp with LogEnabled with LogDisabled = $anon@420c2f4a

如果您需要 val conf 而不是 def conf，您可以这样做：

scala> class MyApp extends ConfApp {
     |   val dbName = "my_app_db"
     |   override val conf = super.conf
     |   println(conf)
     | }
defined class MyApp

scala> new MyApp
Map(db -> my_app_db, foo -> bar)
res4: MyApp = MyApp@17ebbd2a

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