为什么Scala编译器说该副本不是我的case类的成员？

Question

首先,这是在Scala 2.8中,所以它应该在那里!=)

我正在研究Lift的Javascript对象,我希望得到以下内容:

case class JsVar(varName: String, andThen: String*) extends JsExp {
  // ...
  def -&(right: String) = copy(andThen=(right :: andThen.toList.reverse).reverse :_*)
}

不幸的是,我收到以下编译器错误:

[error] Lift/framework/web/webkit/src/main/scala/net/liftweb/http/js/JsCommands.scala:452: not found: value copy
[error]     def -&(right: String) = copy(andThen=(right :: andThen.toList.reverse).reverse :_*)
[error]

case类有属性,所以应该有一个copy方法,对吗？

如果我尝试this.copy我几乎得到相同的错误:

[error] Lift/framework/web/webkit/src/main/scala/net/liftweb/http/js/JsCommands.scala:452: value copy is not a member of net.liftweb.http.js.JE.JsVar
[error]     def -&(right: String) = this.copy(andThen=(right :: andThen.toList.reverse).reverse :_*)
[error]

为什么这样,我如何copy在我的case类方法中使用？或者是copy在声明我的方法之后编译器添加的想法？

我应该这样做吗？

case class JsVar(varName: String, andThen: String*) extends JsExp {
  // ...
  def -&(right: String) = JsVar(varName, (right :: andThen.toList.reverse).reverse :_*)
}

Answer 1

规范在这方面没有提及,但这实际上是预期的.该copy方法取决于默认参数,并且重复参数(varargs)不允许使用默认参数:

不允许使用重复参数在参数部分中定义任何默认参数.

(Scala参考,第4.6.2节 - 重复参数)

scala> def f(xs: Int*) = xs
f: (xs: Int*)Int*

scala> def f(xs: Int* = List(1, 2, 3)) = xs
<console>:24: error: type mismatch;
 found   : List[Int]
 required: Int*
       def f(xs: Int* = List(1, 2, 3)) = xs
                            ^
<console>:24: error: a parameter section with a `*'-parameter is not allowed to have default arguments
       def f(xs: Int* = List(1, 2, 3)) = xs
           ^