How to apply a function to a tuple?
ScalaScala Problem Overview
This should be an easy one. How do I apply a function to a tuple in Scala? Viz:
scala> def f (i : Int, j : Int) = i + j f: (Int,Int)Intscala> 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!
Many thanks in advance.
Scala Solutions
Solution 1 - Scala
In Scala 2.8 and newer:
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>
In Scala 2.7:
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>
Solution 2 - Scala
Following up on the other answer, one could write (tested with 2.11.4):
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
See def tupled: ((T1, T2)) ⇒ R:
> Creates a tupled version of this function: instead of 2 arguments, it > accepts a single scala.Tuple2 argument.
Solution 3 - Scala
Scala 2.13
def f (i : Int, j : Int) = i + j
import scala.util.chaining._
(3,4).pipe((f _).tupled) //res0: Int = 7
Solution 4 - Scala
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
Solution 5 - Scala
In Scala 3, you can use TupledFunction:
For a single use, you can do
summon[TupledFunction[(Int, Int) => Int, ((Int, Int)) => Int]].tupled(f)((2, 3))
To make it easier to use, you can use an extension (copied from Dotty's own documentation)
extension [F, T <: Tuple, R](f: F)(using tf: TupledFunction[F, T => R])
def tupled(t: T): R = tf.tupled(f)(t)
And then you can do f.tupled((2, 3)) to get 5.