varargs难题？

Question

我确定答案很简单,但我陷入了困境:

Welcome to Scala version 2.7.1.final (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_14).
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scala> def f(x:Int*)=0
f: (Int*)Int

scala> val xs:Seq[Int]=1::2::3::4::Nil
xs: Seq[Int] = List(1, 2, 3, 4)

scala> f (xs)
<console>:7: error: type mismatch;
 found   : Seq[Int]
 required: Int
       f (xs)
          ^

我如何构建'Int*'？

Answer 1

要将序列解压缩到参数列表中,请使用 _*

scala> f(xs: _*)
res1: Int = 0