是的,有可能。使用 .tupled 可以将lambda转换为接受元组作为参数。

鳞皮

scala> def foo(i: Int, s: String): Int = i
foo: (i: Int, s: String)Int

scala> (foo _).tupled
res3: ((Int, String)) => Int = scala.Function2$$Lambda$226/234698513@45984654

scala> val tuple: (Int, String) = (1, "s")
tuple: (Int, String) = (1,s)

scala> (foo _).tupled(tuple)
res5: Int = 1

foo(tuple._1, tuple._2) 应该工作。

如果您想要更易于维护的内容,我建议您执行以下操作:

type MyTuple = (Int, String) 

def foo(t:MyTuple) = ??? // some code 

val tuple = (1, "s")

foo(tuple) // works

也在里面 foo ,打开元组的最佳方法是

val (int, string) = t