不带参数继承

Question

我是Scala的新手并尝试使用以下继承:

abstract class Command(arg: Any)

case object Help extends Command // <---- Ok

但如果我用这个代替它:

abstract class Command(arg: Option[Any])

case object Help extends Command // <---- Fail

为什么？为什么它在第一个例子中工作？

Answer 1

第一种情况是编译器接受的,因为它使构造函数参数适应(),即Unit类型的单个实例:

$ scala -Xlint:adapted-args
Welcome to Scala 2.12.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.8.0_112).
Type in expressions for evaluation. Or try :help.

scala> abstract class Command(arg: Any)
defined class Command

scala> case object Help extends Command
<console>:12: warning: Adaptation of argument list by inserting () is deprecated: leaky (Object-receiving) target makes this especially dangerous.
        signature: Command(arg: Any): Command
  given arguments: <none>
 after adaptation: new Command((): Unit)
       case object Help extends Command
                                ^
defined object Help

如你所见,这一行:

case object Help extends Command

已被编译器重写为:

case object Help extends Command(())

也就是说,它()作为构造函数参数传递,这很好,因为参数应该是类型Any,并Unit满足该约束.

但是,在第二种情况下,构造函数参数是类型的Option,并且Unit不满足该约束,因此编译器不能神奇地填充构造函数参数.

无论如何,最好通过使用scalac选项完全禁用这种魔术行为-Yno-adapted-args.