This code works: let d = [1: 1, 2: 2, 3: 3] let intToInt = d.map { $0 + 1 } var ...

afthonos · on June 27, 2014

The code works, but the type of the generic is not guaranteed at compile time. FunctorResult could be anything at all. There is no compile-time obligation for the code to return a Dictionary<KeyType, P>.

The more canonical example of a Functor is really the Optional. The mapping method for an optional looks like this:

  func fmap(f: A -> B) -> Optional<B> {
    switch self {
    case Some(let a):
      return Some(f(A))
    case None:
      return None
    }
  }

However, with your protocol, I can define the mapping function for the Optional as:

  func fmap(f: A -> B) -> B[] {
    switch self {
    case Some(let a):
      return [f(A)]
    case None:
      return B[]()
    }
  }

This would pass the type-checker, but is not what a Functor does.