Understanding how Either is an instance of Functor

Question

In my free time I'm learning Haskell, so this is a beginner question.

In my readings I came across an example illustrating how Either a is made an instance of Functor:

instance Functor (Either a) where
    fmap f (Right x) = Right (f x)
    fmap f (Left x) = Left x

Now, I'm trying to understand why the implementation maps in the case of a Right value constructor, but doesn't in the case of a Left?

Here is my understanding:

First let me rewrite the above instance as

instance Functor (Either a) where
    fmap g (Right x) = Right (g x)
    fmap g (Left x) = Left x

Now:

1. I know that fmap :: (c -> d) -> f c -> f d

2. if we substitute f with Either a we get fmap :: (c -> d) -> Either a c -> Either a d

3. the type of Right (g x) is Either a (g x), and the type of g x is d, so we have that the type of Right (g x) is Either a d, which is what we expect from fmap (see 2. above)

4. now, if we look at Left (g x) we can use the same reasoning to say that its type is Either (g x) b, that is Either d b, which is not what we expect from fmap (see 2. above): the d should be the second parameter, not the first! So we can't map over Left.

Is my reasoning correct?

Answer 1

This is right. There is also another quite important reason for this behavior: You can think of Either a b as a computation, that may succeed and return b or fail with an error message a. (This is also, how the monad instance works). So it's only natural, that the functor instance won't touch the Left values, since you want to map over the computation, if it fails, there's nothing to manipulate.

Answer 2

Your account is right of course. Maybe the reason why we have a difficulty with instances like this is that we are really defining infinitely many functor instances at once -- one for each possible Left type. But a Functor instance is a systematic way of operating on the infinitely many types in the system. So we are defining infinitely many ways of systematically operating on the infinitely many types in the system. The instance involves generality in two ways.

If you take it by stages, though, maybe it's not so strange. The first of these types is a longwinded version of Maybe using the unit type () and its only legitimate value ():

data MightBe b     = Nope ()    | Yep b
data UnlessError b = Bad String | Good b
data ElseInt b     = Else Int   | Value b

Here we might get tired and make an abstraction:

data Unless a b    = Mere a     | Genuine b

Now we make our Functor instances, unproblematically, the first looking a lot like the instance for Maybe:

instance Functor MightBe where
  fmap f (Nope ()) = Nope ()   -- compare with Nothing
  fmap f (Yep x)   = Yep (f x) -- compare with Just (f x)

instance Functor UnlessError where
  fmap f (Bad str) = Bad str   -- a more informative Nothing
  fmap f (Good x)  = Good (f x)

instance Functor ElseInt where
  fmap f (Else n) = Else n 
  fmap f (Value b) = Value (f b)

But, again, why bother, let's make the abstraction:

instance Functor (Unless a) where
  fmap f (Mere a) = Mere a
  fmap f (Genuine x) = Genuine (f x)

The Mere a terms aren't touched, as the (), String and Int values weren't touched.

Answer 3

As others mentioned, Either type is a functor in its both arguments. But in Haskell we are able to (directly) define only functors in a type's last arguments. In cases like this, we can get around the limitation by using newtypes:

newtype FlipEither b a = FlipEither { unFlipEither :: Either a b }

So we have constructor FlipEither :: Either a b -> FlipEither b a that wraps an Either into our newtype with swapped type arguments. And we have dectructor unFlipEither :: FlipEither b a -> Either a b that unwraps it back. Now we can define a functor instance in FlipEither's last argument, which is actually Either's first argument:

instance Functor (FlipEither b) where
    fmap f (FlipEither (Left x))  = FlipEither (Left (f x))
    fmap f (FlipEither (Right x)) = FlipEither (Right x)

Notice that if we forget FlipEither for a while we get just the definition of Functor for Either, just with Left/Right swapped. And now, whenever we need a Functor instance in Either's first type argument, we can wrap the value into FlipEither and unwrap it afterward. For example:

fmapE2 :: (a -> b) -> Either a c -> Either b c
fmapE2 f = unFlipEither . fmap f . FlipEither

---

Update: Have a look at Data.Bifunctor, of which Either and (,) are instances of. Each bifunctor has two arguments and is a functor in each of them. This is reflected in Bifunctor's methods first and second.

The definition of Bifunctor of Either is very symetric:

instance Bifunctor Either where
    bimap f _ (Left a)  = Left (f a)
    bimap _ g (Right b) = Right (g b)

    first  f = bimap f id

    second f = bimap id f

Answer 4

Now, I'm trying to understand why the implementation maps in the case of a Right value constructor, but doesn't in the case of a Left?

Plug in here and it might make sense.

Assume a = String (an error message) You apply Either a to an Float.

So you have an f: Float -> Integer say for example roundoff.

(Either String) (Float) = Either String Float.

now (fmap f):: Either String Float -> Either String Int So what are you going to do with f? f doesn't have a clue what to do with strings so you can't do anything there. That is obviously the only thing you can act on are the right values while leaving the left values unchanged.

In other words Either a is a functor because there is such an obvious fmap given by:

• for Right values apply f
• for Left values do nothing