Anonymous Scala function syntax

Jeff Storey picture Jeff Storey · Oct 11, 2009 · Viewed 9.1k times · Source

I'm learning more about Scala, and I'm having a little trouble understanding the example of anonymous functions in http://www.scala-lang.org/node/135. I've copied the entire code block below:

object CurryTest extends Application {
    def filter(xs: List[Int], p: Int => Boolean): List[Int] =
        if (xs.isEmpty) xs
        else if (p(xs.head)) xs.head :: filter(xs.tail, p)
        else filter(xs.tail, p)

    def modN(n: Int)(x: Int) = ((x % n) == 0)

    val nums = List(1, 2, 3, 4, 5, 6, 7, 8)
    println(filter(nums, modN(2)))
    println(filter(nums, modN(3)))
}

I'm confused with the application of the modN function

def modN(n: Int)(x: Int) = ((x % n) == 0)

In the example, it's called with one argument

modN(2) and modN(3)

What does the syntax of modN(n: Int)(x: Int) mean?

Since it's called with one argument, I'm assuming they're not both arguments, but I can't really figure out how the values from nums get used by the mod function.

Answer

Flaviu Cipcigan picture Flaviu Cipcigan · Oct 11, 2009

This is a fun thing in functional programming called currying. Basically Moses Schönfinkel and latter Haskell Curry (Schonfinkeling would sound weird though...) came up with the idea that calling a function of multiple arguments, say f(x,y) is the same as the chain of calls {g(x)}(y) or g(x)(y) where g is a function that produces another function as its output.

As an example, take the function f(x: Int, y: Int) = x + y. A call to f(2,3) would produce 5, as expected. But what happens when we curry this function - redefine it as f(x:Int)(y: Int)and call it as f(2)(3). The first call, f(2) produces a function taking an integer y and adding 2 to it -> therefore f(2) has type Int => Int and is equivalent to the function g(y) = 2 + y. The second call f(2)(3) calls the newly produced function g with the argument 3, therefore evaluating to 5, as expected.

Another way to view it is by stepping through the reduction (functional programmers call this beta-reduction - it's like the functional way of stepping line by line) of the f(2)(3) call (note, the following is not really valid Scala syntax).

f(2)(3)         // Same as x => {y => x + y}
 | 
{y => 2 + y}(3) // The x in f gets replaced by 2
       |
     2 + 3      // The y gets replaced by 3
       |
       5

So, after all this talk, f(x)(y) can be viewed as just the following lambda expression (x: Int) => {(y: Int) => x + y} - which is valid Scala.

I hope this all makes sense - I tried to give a bit of a background of why the modN(3) call makes sense :)