Scala如何将Map转换为元组的varargs？

Question

在Scala Play 2.2.x测试的上下文中,我有一个Map[String, String]我需要传递给一个接受(String, String)*即(String, String)元组varargs的函数.

例如

val data : Map[String, String] = Map("value" -> "25", "id" -> "", "columnName" -> "trades")
route(FakeRequest(POST, "/whatever/do").withFormUrlEncodedBody(data))

但这会导致类型不匹配,因为withFormUrlEncodedBody只接受一种(String, String)*类型.

Answer 1

只是:

def foo(names: (String, String)*) = names.foreach(println)
val folks = Map("john" -> "smith", "queen" -> "mary")
foo(folks.toSeq:_*)
// (john,smith)
// (queen,mary)

_* 编译器的提示在哪里？