How do I forward repeated arguments in Scala?

Question

在Scala(2.7)中,如果我有这个功能:

def foo(args: Array[String]) =
  for (arg <- args) println(arg)

如果我现在尝试定义以下内容:

def bar(args: String*) = foo(args)

然后编译器抱怨:

<console>:5: error: type mismatch;
 found   : String*
 required: Array[String]
       def bar(args: String*) = foo(args)
                                ^

我不明白这个错误,因为编程Scala书中指出args内部函数的类型bar实际上是Array[String].我怎么能用重复的参数写这样的包装函数？

Answer 1

scala> def foo(args: Array[String]) = for(arg <- args) println(arg)
foo: (args: Array[String])Unit

scala> def bar(args: String*) = foo(args.toArray)
bar: (args: String*)Unit

scala> bar("hello", "world")
hello
world

您需要执行上述转换,因为Scala中的varargs实现为Seq,而不是Array.

以下是通常在Scala中转发varargs的方法:

scala> def fooV(args: String*) = args foreach println
fooV: (args: String*)Unit

scala> def fooS(args: Seq[String]) = fooV(args: _*)
fooS: (args: Seq[String])Unit

scala> def bar(args: String*) = fooV(args: _*)
bar: (args: String*)Unit

scala> def barS(args: Seq[String]) = args foreach println
barS: (args: Seq[String])Unit

scala> def barV(args: String*) = barS(args)
barV: (args: String*)Unit

scala> def barV(args: String*) = barS(args.toSeq)
barV: (args: String*)Unit