Scala：將元組列表擴展為元組的變長參數列表

Question

我對如何將List / Seq / Array擴展為變長參數列表感到困惑。

鑒於我有接受元組的test_func函數：

scala> def test_func(t:Tuple2[String,String]*) = println("works!")
test_func: (t: (String, String)*)Unit

當我通過元組時哪個有效：

scala> test_func(("1","2"),("3","4"))
works!

通過閱讀Scala參考資料，我對以下人員也有深刻印象：

scala> test_func(List(("1","2"),("3","4")))
<console>:9: error: type mismatch;
 found   : List[(java.lang.String, java.lang.String)]
 required: (String, String)
              test_func(List(("1","2"),("3","4")))
                        ^

還有另一種絕望的嘗試：

scala> test_func(List(("1","2"),("3","4")).toSeq)
<console>:9: error: type mismatch;
 found   : scala.collection.immutable.Seq[(java.lang.String, java.lang.String)]
 required: (String, String)
              test_func(List(("1","2"),("3","4")).toSeq)

如何將列表/序列/數組擴展為參數列表？

先感謝您！

Answer 1

您需要添加:_* 。

scala> test_func(List(("1","2"),("3","4")):_*)
works!
scala> test_func(Seq(("1","2"),("3","4")):_*)
works!
scala> test_func(Array(("1","2"),("3","4")):_*)
works!