如何将函数应用于元组？

Question

这应该是一件容易的事。 如何将函数应用于 Scala 中的元组？ 可视化：

scala> def f (i : Int, j : Int) = i + j
f: (Int,Int)Int

scala> val p = (3,4)
p: (Int, Int) = (3,4)

scala> f p
:6: error: missing arguments for method f in object $iw;
follow this method with `_' if you want to treat it as a partially applied function
       f p
       ^

scala> f _ p
:6: error: value p is not a member of (Int, Int) => Int
       f _ p
           ^

scala> (f _) p
:6: error: value p is not a member of (Int, Int) => Int
       (f _) p
             ^

scala> f(p)
:7: error: wrong number of arguments for method f: (Int,Int)Int
       f(p)
       ^

scala> grr!

非常感谢提前。

Answer 1

在 Scala 2.8 及更新版本中：

scala> def f (i : Int, j : Int) = i + j
f: (i: Int,j: Int)Int

// Note the underscore after the f
scala> val ff = f _
ff: (Int, Int) => Int = <function2>

scala> val fft = ff.tupled
fft: ((Int, Int)) => Int = <function1>

在 Scala 2.7 中：

scala> def f (i : Int, j : Int) = i + j
f: (Int,Int)Int

// Note the underscore after the f
scala> val ff = f _
ff: (Int, Int) => Int = <function>

scala> val fft = Function.tupled(ff)
fft: ((Int, Int)) => Int = <function>

Answer 2

跟进另一个答案，可以写（用 2.11.4 测试）：

scala> def f (i: Int, j: Int) = i + j
f: (i: Int, j: Int)Int

scala> val ff = f _
ff: (Int, Int) => Int = <function2>

scala> val p = (3,4)
p: (Int, Int) = (3,4)

scala> ff.tupled(p)
res0: Int = 7

参见def tupled: ((T1, T2)) ⇒ R ：

创建此函数的元组版本：它接受单个scala.Tuple2参数而不是 2 个参数。

Answer 3

斯卡拉 2.13

def f (i : Int, j : Int) = i + j

import scala.util.chaining._
(3,4).pipe((f _).tupled)   //res0: Int = 7

Answer 4

scala> def f (i: Int, j: Int) = i + j
f: (i: Int, j: Int)Int
scala> val p = (3,4)
p: (Int, Int) = (3,4)
scala> val ft = (f _).tupled
ft: ((Int, Int)) => Int = <function1>
scala> ft apply(p)
res0: Int = 7

Answer 5

在 Scala 3 中，您可以使用TupledFunction ：

一次性使用，你可以做

summon[TupledFunction[(Int, Int) => Int, ((Int, Int)) => Int]].tupled(f)((2, 3))

为了更容易使用，您可以使用扩展名（复制自 Dotty 自己的文档）

extension [F, T <: Tuple, R](f: F)(using tf: TupledFunction[F, T => R])
  def tupled(t: T): R = tf.tupled(f)(t)

然后你可以做f.tupled((2, 3))得到5 。