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 i

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 = 

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

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 = 

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

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

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