分解函数参数中的元组

Question

In python I can do this:

def f((a, b)):
    return a + b

d = (1, 2)
f(d)

Here the passed in tuple is being decomposed while its being passed to f .

Right now in scala I am doing this:

def f(ab: (Int, Int)): Int = {
    val (a, b) = ab
    a + b
}

val d = (1, 2)
f(d)

Is there something I can do here so that the decomposition happens while the arguments are passed in? Just curious.

Answer 1

You can create a function and match its input with pattern matching:

scala> val f: ((Int, Int)) => Int = { case (a,b) => a+b }
f: ((Int, Int)) => Int

scala> f(1, 2)
res0: Int = 3

Or match the input of the method with the match keyword:

scala> def f(ab: (Int, Int)): Int = ab match { case (a,b) => a+b }
f: (ab: (Int, Int))Int

scala> f(1, 2)
res1: Int = 3

Another way is to use a function with two arguments and to "tuple" it:

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

scala> val g = f.tupled // or Function.tupled(f)
g: ((Int, Int)) => Int = <function1>

scala> g(1, 2)
res10: Int = 3

// or with a method
scala> def f(a: Int, b: Int): Int = a+b
f: (a: Int, b: Int)Int

scala> val g = (f _).tupled // or Function.tupled(f _)
g: ((Int, Int)) => Int = <function1>

scala> g(1, 2)
res11: Int = 3

// or inlined
scala> val f: ((Int,Int)) => Int = Function.tupled(_+_)
f: ((Int, Int)) => Int = <function1>

scala> f(1, 2)
res12: Int = 3

Answer 2

Starting in Scala 3 , with the improved tupled function feature:

// val tuple = (1, 2)
// def f(a: Int, b: Int): Int = a + b
f.tupled(tuple)
// 3

Play with it in Scastie

Answer 3

object RandomExperiments extends App{
  def takeTuple(t:(Int,Int))=print (s"$t ${t._1}\n")
  takeTuple(1,3)
  takeTuple((1,3))
  takeTuple(((1,3)))

}

prints:

(1,3) 1
(1,3) 1
(1,3) 1