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Home/ Questions/Q 4604378
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T00:17:14+00:00 2026-05-22T00:17:14+00:00

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

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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?

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  1. Editorial Team
    Editorial Team
    2026-05-22T00:17:15+00:00Added an answer on May 22, 2026 at 12:17 am

    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.

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