Decomposing tuples in function arguments

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