octavian octavian - 1 month ago 9
Scala Question

Type mismatch for function composition

I have defined an identity function and a composition function:

def identity[T](x: T): T = x
def composition[A, B, C](f: A => B, g: B => C)(x: A) = g(f(x))


I am trying to assert that the identity function can be applied on both sides with the same result:

assert(composition((x: Int) => x + 1, identity)(3) == composition(identity, (x: Int) => x + 1)(3))


However, I am getting these errors:

Error:(7, 40) type mismatch;
found : Nothing => Nothing
required: Int => Nothing
assert(composition((x: Int) => x + 1, identity)(3) == composition(identity, (x: Int) => x + 1)(3));}


And:

Error:(7, 68) type mismatch;
found : Nothing => Nothing
required: A => Nothing
assert(composition((x: Int) => x + 1, identity)(3) == composition(identity, (x: Int) => x + 1)(3));}


Why is this the case?

Answer

This is the case because the compiler is having difficulty inferring the types properly, specifically inferring what A is. You can help him by placing A as the first argument and each function in a separate parameter list:

def composition[A, B, C](x: A)(f: A => B)(g: B => C) = g(f(x))

And now this works as expected:

scala> :pa
// Entering paste mode (ctrl-D to finish)

  def identity[T](x: T): T = x
  def composition[A, B, C](x: A)(f: A => B)(g: B => C) = g(f(x))

// Exiting paste mode, now interpreting.

identity: [T](x: T)T
composition: [A, B, C](x: A)(f: A => B)(g: B => C)C

scala> println(composition(3)((x: Int) => x + 1)(identity) == composition(3)(identity)((x: Int) => x + 1))
true

Alternatively, you can specify the type parameter explicitly to help the compiler infer the right type:

println(composition((x: Int) => x + 1, identity[Int], 3) == 
        composition(identity[Int], (x: Int) => x + 1, 3))