What's the purpose of Function.const?

Question

It is in ScalaDoc but without much documentation. It seems that it always returns the first parameter.

Function.const(1)(2) for instance returns 1.

Why does it exist and why is it useful?

Smalltime

Answer 1

To give a more theoretical answer: const is the K combinator of the SKI calculus. It pops sometimes up when you work with quite abstract concepts where you don't have much "to work with". Consider a (Haskell style) Functor trait:

trait Functor[F[_]] {
   def fmap[A,B](f:A=>B, fa: F[A]):F[B]
   //(<$) in Haskell
   def left[A,B](a:A, fb:F[B]):F[A] 
}

Now fmap needs to be abstract, as it is the very essence of a functor. But we can write a general implementation of left, and here we need const:

trait Functor[F[_]] {
   def fmap[A,B](f:A=>B, fa: F[A]):F[B]
   //(<$) in Haskell
   def left[A,B](a:A, fb:F[B]):F[A] = 
     fmap(Function.const(a), fb)
}

Test with Option:

case object OptionFunctor extends Functor[Option] {
   def fmap[A,B] (f:A=>B, fa:Option[A]):Option[B] = fa match {
      case Some(a) => Some(f(a))
      case None => None
   }
}

//left works:
OptionFunctor.left("test",Some(42))
//--> Option[java.lang.String] = Some(test)
OptionFunctor.left("test",None:Option[Int])
//--> Option[java.lang.String] = None

As you can see left does what it should (wrapping a value in some functor when we have already a "role model" or "pattern" for this functor in the second argument). Defining it very abstract without knowing anything about the kind of functor was only possible by using const.

Lemay

Answer 2

It's useful for passing as an argument to a higher-order function. For example, to replace all elements of a list with the same element:

scala> List(1, 2, 3, 4, 5).map(Function.const(7))
res1: List[Int] = List(7, 7, 7, 7, 7)

You could of course also write

scala> List(1, 2, 3, 4, 5).map(_ => 7)
res2: List[Int] = List(7, 7, 7, 7, 7)

Depending on the context, one might be more readable than the other.

Wakerobin