Question

我很困惑如何将List / Seq / Array扩展为可变长度的参数列表。

鉴于我有test_func函数接受元组：

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

当我传递元组时，哪个有效：

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

通过阅读Scala参考资料，我得到了强烈的印象，即以下内容也可以起作用：

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")))
                        ^

还有一次绝望的尝试：

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)

如何将List / Seq / Array扩展为参数列表？

提前谢谢！

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!

Scala：将元组列表展开为元组的可变长度参数列表

1 个答案: