I need to write the Functor instances for the Flip datatype:
data K a b = K a
newtype Flip f a b = Flip (f b a) deriving (Eq, Show)
instance Functor (Flip K a) where
fmap=undefined
The solution I was given in class is:
instance Functor (Flip K a) where
fmap f (Flip (K b)) = Flip (K (f b))
I really don't understand what's going on here and I'm beginning to doubt my whole understanding of data types and functors. What I do understand is this (please correct me if any of this is wrong):
K is a data type that turns 2 parameters into a structure K a ( that only keeps the first parameter)
Flip is a datatype that turns 3 arguments into a structure with one
Because in fmap :: (a-> b) -> f a -> f b, f a has kind *, to write the Functor instance of Flip, we write it on the last type in Flip. Aka f and a are "constants" in a way, and we write the functor for the type b. I would write something like:
instance Functor (Flip f a) where
fmap f (Flip x y z) = fmap Flip x y (f z)
I know that that is completely wrong but I'm not sure why.
Also, why would we bring K into the Functor instance of Flip? Can someone explain thoroughly the process of coming up with this solution and why it is correct?
K is a data type that turns 2 parameters into a structure K a ( that only keeps the first parameter)
This isn't quite right. K a b is a data type formed using two parameters, but it's not really right to say that it "turns them into" anything. Instead, it's simply just stating to the world that there now exists a new type: K a b. "So what?" you might ask. Well, the second half of the data type defines how to make new values of this type. That part says, "You can make a new value of type K a b with this function I'll call K which has type a -> K a b." It's really important to recognize that there is a distinction between the type K and the constructor K.
So, it's not that K "only keeps the first parameter"—it's that the constructor K (which is a function) happens to not take any arguments of type b.
Flip is a datatype that turns 3 arguments into a structure with one
Just as above, this is not quite right. The Flip declaration states that there can be values of type Flip f a b, and the only way to make them is by using the constructor Flip that has type f b a -> Flip f a b.
In case you're wondering how I'm coming up with the type signatures for the constructors K and Flip, it's not actually mysterious, and you can double check by typing :t K or :t Flip into GHCi. These types are assigned based entirely on the right hand side of the data type declaration. Also, note that the type name and constructor don't have to be the same. For instance, consider this data type:
data Foo a = Bar Int a | Foo String | Baz a a
This declares a type Foo a with three constructors:
Bar :: Int -> a -> Foo a
Foo :: String -> Foo a
Baz :: a -> a -> Foo a
Basically, each of the types after the constructor name are the arguments, in order, to the constructor.
Because in fmap :: (a-> b) -> f a -> f b, f a has kind *, to write the Functor instance of Flip, we write it on the last type in Flip. Aka f and a are "constants" in a way, and we write the functor for the type b.
This is basically right! You could also say that f has kind * -> *. Since Flip has kind (* -> *) -> * -> * -> *, you need to provide it two type arguments (the first of kind * -> * and the second of kind *) to get it to the right kind. Those first two arguments become fixed ("constants" in a way) in the instance.
I would write something like: ... I know that that is completely wrong but I'm not sure why.
The reason your instance is completely wrong is that you've mixed up the type with the constructor. It doesn't make sense to put (Flip x y z) in the pattern position where you did because the constructor Flip only takes one argument—remember, it's type is Flip :: f b a -> Flip f a b! So you'd want to write something like:
instance Functor (Flip f a) where
fmap f (Flip fxa) = ...
Now, what do you fill in for the ...? You have a value fxa :: f x a, and you have a function f :: x -> y, and you need to produce a value of type f y a. Honestly, I don't know how to do that. After all, what is a value of typ f x a? We don't know what f is?!
Also, why would we bring K into the Functor instance of Flip? Can someone explain thoroughly the process of coming up with this solution and why it is correct?
We saw just above that we can't write the Functor instance for an arbitrary f, but what we can do is write it for a particular f. It turns out that K is just such a particular f that works. Let's try to make it work:
instance Functor (Flip K a) where
fmap f (Flip kxa) = ...
When f was arbitrary, we got stuck here, but now we know that kxa :: K x a. Remember that the only way to make a value of type K x a is using the constructor K. Therefore, this value kxa must have been made using that constructor, so we can break it apart as in: kxa ⩳ K x' where x' :: x. Let's go ahead and put that into our pattern:
fmap f (Flip (K x')) = ...
Now we can make progress! We need to produce a value of type Flip K a y. Hmm. The only way to produce a value of type Flip is using the Flip constructor, so let's start with that:
fmap f (Flip (K x')) = Flip ...
The Flip constructor at type Flip K a y takes a value of type K y a. The only way to produce one of those is with the K constructor, so let's add that:
fmap f (Flip (K x')) = Flip (K ...)
The K constructor at type K y a takes a value of type y, so we need to provide a value of type y here. We have a value x' :: x and a function f :: x -> y. Plugging the first into the second gives us the value we need:
fmap f (Flip (K x')) = Flip (K (f x'))
Just rename x' to b, and you have exactly the code your teacher provided.
DDub wrote in their answer:
You have a value fxa :: f x a, and you have a function f :: x -> y, and you need to produce a value of type f y a. Honestly, I don't know how to do that. After all, what is a value of type f x a? We don't know what f is?!
And I agree, but I woulld like to add a bit. Your teacher's idea as to how to deal with this is pretty cool (things like this K come in quite handy when you are trying to write down some counterexample, like here), and yet, I reckon we can make this code way broader. I use Data.Bifunctor.
So, what are Bifunctors? They are just what their name says: a * -> * -> * type (which we call bifunctors as well sometimes, yet they are not the same thing) which allows mapping over its both arguments (snippet from the source):
class Bifunctor p where
-- | Map over both arguments at the same time.
--
-- #'bimap' f g ≡ 'first' f '.' 'second' g#
bimap :: (a -> b) -> (c -> d) -> p a c -> p b d
bimap f g = first f . second g
{-# INLINE bimap #-}
-- | Map covariantly over the first argument.
--
-- #'first' f ≡ 'bimap' f 'id'#
first :: (a -> b) -> p a c -> p b c
first f = bimap f id
{-# INLINE first #-}
-- | Map covariantly over the second argument.
--
-- #'second' ≡ 'bimap' 'id'#
second :: (b -> c) -> p a b -> p a c
second = bimap id
{-# INLINE second #-}
So, here is how I would go about that:
instance Bifunctor f => Functor (Flip f a) where
fmap x2y (Flip fxa) = Flip (first x2y fxa)
Speaking of your teacher's code, it's a very nice idea, yet a more narrow one as K is a Bifunctor:
instance Bifunctor K where
bimap f _g (K a) = K (f a)
A lawful one:
bimap id id (K a) = K (id a) = id (K a)
As it says in the link above, having bimap only written down, that's the only law we need to worry about.
We just need to use sane and helpful naming, and suddenly it all becomes simple and clear (as opposed to torturous and contorted):
data K b a = MkK b -- the type (K b a) "is" just (b)
newtype Flip f a b = MkFlip (f b a) -- the type (Flip f a b) "is" (f b a)
deriving (Eq, Show)
instance Functor (Flip K a) where
-- fmap :: (b -> c) -> Flip K a b -> Flip K a c
fmap g (MkFlip (MkK b)) = MkFlip (MkK (g b))
-- MkK b :: K b a
-- MkFlip (_ :: K b a) :: Flip K a b
There's not even one question arising in our minds now looking at this, not one doubt we aren't able to immediately resolve.
Using same names for types and for data constructors while teaching, as well as using f both for "f"unction and "f"unctor, is pure abuse of the students.
Only when you've become fed up with all the Mks and don't feel they are helpful to you in any way, you can safely and easily throw them away, as experts usually do.
Related
import Control.Applicative
main = print $ fmap (*2) (1,2)
produces (1,4). I would expect it it to produce (2,4) but instead the function is applied only to the second element of the tuple.
Update I've basically figured this out almost straight away. I'll post my own answer in a minute..
Let me answer this with a question: Which output do you expect for:
main = print $ fmap (*2) ("funny",2)
You can have something as you want (using data Pair a = Pair a a or so), but as (,) may have different types in their first and second argument, you are out of luck.
Pairs are, essentially, defined like this:
data (,) a b = (,) a b
The Functor class looks like this:
class Functor f where
fmap :: (a -> b) -> f a -> f b
Since the types of function arguments and results must have kind * (i.e. they represent values rather than type functions that can be applied further or more exotic things), we must have a :: *, b :: *, and, most importantly for our purposes, f :: * -> *. Since (,) has kind * -> * -> *, it must be applied to a type of kind * to obtain a type suitable to be a Functor. Thus
instance Functor ((,) x) where
-- fmap :: (a -> b) -> (x,a) -> (x,b)
So there's actually no way to write a Functor instance doing anything else.
One useful class that offers more ways to work with pairs is Bifunctor, from Data.Bifunctor.
class Bifunctor f where
bimap :: (a -> b) -> (c -> d) -> f a c -> f b d
bimap f g = first f . second g
first :: (a -> b) -> f a y -> f b y
first f = bimap f id
second :: (c -> d) -> f x c -> f x d
second g = bimap id g
This lets you write things like the following (from Data.Bifunctor.Join):
newtype Join p a =
Join { runJoin :: p a a }
instance Bifunctor p => Functor (Join p) where
fmap f = Join . bimap f f . runJoin
Join (,) is then essentially the same as Pair, where
data Pair a = Pair a a
Of course, you can also just use the Bifunctor instance to work with pairs directly.
The Functor instance is actually from the GHC.Base module which is imported by Control.Applicative.
Trying to write the instance I want, I can see that it won't work, given the definition of tuples; the instance requires just one type parameter, while the 2-tuple has two.
A valid Functor instance would at least have to be on tuples, (a,a) that have the same type for each element, but you cannot do anything sneaky, like define the instance on:
type T2 a = (a,a)
because instance types aren't permitted to be synonyms.
The above restricted 2-tuple synonym is logically the same as the type:
data T2 a = T2 a a
which can have a Functor instance:
instance Functor T2 where
fmap f (T2 x y) = T2 (f x) (f y)
As Gabriel remarked in the comments, this can be useful for branching structures or concurrency.
The type of fmap in Functor is:
fmap :: Functor f => (a -> b) -> f a -> f b
it looks like ,first apply function (a -> b) to the parameter of f a to create a result of type b, then apply f to it, and result is f b
using Maybe a for example:
fmap show (Just 1)
result is : Just "1"
same as saying:
Just (show 1)
but when (->) is used as a Functor (in Control.Monad.Instances)
import Control.Monad.Instances
(fmap show Just) 1
result is : "Just 1"
that is, Just is applied first, then show is applied. In another example ,result is same:
fmap (*3) (+100) 1
result is 303
why not *3 first, then +100?
The fmap instance for (->) r (i.e. functions) is literally just composition. From the source itself:
instance Functor ((->) r) where
fmap = (.)
So, in your example, we can just replace fmap with (.), and do some transformations
fmap (*3) (+100) 1 =>
(.) (*3) (+100) 1 =>
(*3) . (+100) $ 1 => -- put (.) infix
(*3) (1 + 100) => -- apply (+100)
(1 + 100) * 3 -- apply (*3)
That is, fmap for functions composes them right to left (exactly the same as (.), which is sensible because it is (.)).
To look at it another way (for (double) confirmation!), we can use the type signature:
-- general fmap
fmap :: Functor f => (a -> b) -> f a -> f b
-- specialised to the function functor (I've removed the last pair of brackets)
fmap :: (a -> b) -> (r -> a) -> r -> b
So first the value of type r (the third argument) needs to be transformed into a value of type a (by the r -> a function), so that the a -> b function can transform it into a value of type b (the result).
the type of fmap in Functor is:
fmap :: Functor f => (a -> b) -> f a -> f b
it looks like ,first apply function (a -> b) to the parameter of f a
to create a result of type b, then apply f to it, and result is f b
That is the type of fmap, but your interpretation of what that type means is wrong.
You seem to assume that f a has one parameter, and that that parameter has type a.
Consider xs :: [a]:
Perhaps xs = [].
Perhaps xs = [x1].
Perhaps xs = [x1, x2].
...
The type f a is a functor f with a single type parameter a. But values of type f a do not necessarily take the form F x, as you can see from the first and third cases above.
Now consider fmap f xs:
Perhaps fmap f xs = [].
Perhaps fmap f xs = [f x1].
Perhaps fmap f xs = [f x1, f x2].
...
We don't necessarily apply f at all (first case)! Or we might apply it more than once (third case).
What we do is replace the things of type a, with things of type b. But we leave the larger structure intact --- no new elements added, no elements removed, their order is left unchanged.
Now let's think about the functor (c ->). (Remember, a functor takes one type parameter only, so the input to (->) is fixed.)
Does a c -> a even contain an a? It might not contain any as at all, but it can somehow magic one out of thin air when we give it a c. But the result from fmap has type c -> b: we only have to provide a b out of that when we're presented with a c.
So we can say fmap f x = \y -> f (x y).
In this case, we're applying f on demand --- every time the function we return gets applied, f gets applied as well.
It needs to be defined that way to make the types work out. As you pointed out, the type of fmap is:
fmap :: Functor f => (a -> b) -> f a -> f b
Let's consider the case when the functor f is ((->) c)
(Note: we'd actually like to write this as (c ->), i.e. functions from c, but Haskell doesn't allow us to do this.)
Then f a is actually ((->) c a), which is equivalent to (c -> a), and similarly for f b, so we have:
fmap :: (a -> b) -> (c -> a) -> (c -> b)
i.e. we need to take two functions:
f :: a -> b
g :: c -> a
and construct a new function
h :: c -> b
But there's only one way to do that: you have to apply g first to get something of type a, and then apply f to get something of type b, which means that you have to define
instance Functor ((->) c) where
fmap f g = \x -> f (g x)
or, more succinctly,
instance Functor ((->) c) where
fmap = (.)
fmap for (->) is defined like fmap = (.).
So, (fmap f g) x is (f . g) x is f (g x). In your case (*3) ((+100) 1), which equals 3 * (100 + 1) which results in 303.
In order to form a function type, you need 2 kind parameters for (->), that is the single input argument type and the return type.
A Functor can only take 1 type parameter, so you have to nail down the input argument type(since it's the first one from left to right), which makes the return type of the function to be the type parameter of the Functor.
So for function (the Functor) a->b, you need to give fmap a function ff of type b->xxx other than a->xxx to work, and that means the function ff can only be applied after a->b is apply.
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:
I know that fmap :: (c -> d) -> f c -> f d
if we substitute f with Either a we get fmap :: (c -> d) -> Either a c -> Either a d
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)
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?
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.
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.
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
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
Take a quick peek at the following interactive session in GHCi:
Prelude> import Control.Applicative
Prelude Control.Applicative> (+1) <$> [1,2]
[2,3]
Prelude Control.Applicative> (+1) <$> (1,2)
(1,3)
I guess there is a good reason for the behavior of <$> regarding pairs, but I wasn't able to find one so far, so:
Why is <$> (or fmap) defined to act only on the second member of a pair and not on both values?
<$> (aka fmap) is a member of the Functor class like so:
class Functor f where
fmap :: (a -> b) -> f a -> f b
So whatever f is must be a parameterised type with one type argument. Lists are one such type, when written in their prefix form [] ([] a is the same as [a]). So the instance for lists is:
instance Functor [] where
-- fmap :: (a -> b) -> [] a -> [] b
fmap = map
Pairs can also be written in prefix form: (,) a b is the same as (a, b). So let's consider what we do if we want a Functor instance involving pairs. We can't declare an instance Functor (,) because the pair constructor (,) takes two types -- and they can be different types! What we can do is declare an instance for (,) a -- that's a type that only needs one more type:
instance Functor ( (,) a ) where
-- fmap :: (b -> c) -> (,) a b -> (,) a c
fmap f (x, y) = (x, f y)
Hopefully you can see that the definition of fmap is the only sensible one we can give. The answer as to why the functor instance operates on the second item in a pair is that the type for the second item comes last in the list! We can't easily declare a functor instance that operates on the first item in a pair. Incidentally, this generalises to larger tuples, e.g. the quadruple (,,,) a b c d (aka (a, b, c, d)) can also have a Functor instance on the last item:
instance Functor ( (,,,) a b c) where
-- fmap :: (d -> e) -> (,,,) a b c d -> (,,,) a b c e
fmap f (p, q, r, s) = (p, q, r, f s)
Hope that helps explain it all!
Consider the definition of the Functor typeclass:
class Functor f where
fmap :: (a -> b) -> f a -> f b
Obviously, f has kind * -> *. So you can only declare instances for a datatype, which has kind * -> *. What you can do, is doing some stuff like this:
instance Functor (,) where
fmap :: (a -> b) -> (,) a -> (,) b
This would work on partitial applicated tuples and is really unhandy. So one defined the instance like this:
instance Functor ((,) a) where
fmap :: (b -> c) -> (,) a b -> (,) a c
fmap f (x,y) = (x,f y)
In a nutshell: It is not possible in plain Haskell 98 (although I believe tht there's a syntax extension for this) to define an instance as
What you can do, is defining your own tuple:
data T a = T a a
instance Functor T where
fmap f (T a b) = T (f a) (f b)
Then you can do whatever you like. You see, because the kind is * -> * instead of * -> * -> *, anything's okay.
I guess, a tuple doesn't need to be homogeneous, I mean both types can be different. If you want a homogeneous tuple you can use a list and then fmap will work.
How would you expect (+1) ("Hello", 2) to work ?
Prelude> import Control.Applicative
Prelude Control.Applicative> (+1) <$> ("hello",2)
("hello",3)
That's just work, but there is not special behavior when both type are same.
By the way I don't know why the second value is not used rather than the first one, but anyway you can only use one value.
Learn You a Haskell has an example about functors. I can read LYAH, and text, and figure out what is supposed to happen -- but I don't know enough to write something like this. I'm finding this problem often in Haskell.
instance Functor (Either a) where
fmap f (Right x) = Right (f x)
fmap f (Left x) = Left x
However, I'm confused.. Why doesn't this comple
instance Functor (Either a) where
fmap f (Right x) = Right (x)
fmap f (Left x) = Left (f x)
If f isn't being used in the top definition, then what else constrains x such that it can't satisfy Left
Here's the functor class:
class Functor f where
fmap :: (a -> b) -> f a -> f b
Note that "f" by itself is a type constructor because it's applied to a type variable in the fmap line. Here are some examples to make this clear:
Type constructors:
IO
Maybe
Either String
Types:
IO Char
Maybe a
Either String String
"Maybe a" is a type with one type constructor (the "Maybe") and one type variable (the "a"). It's not something concrete yet, but it is usable in type signatures for polymorphic functions.
"Either" is a type constructor that takes two type arguments, so even after you apply one (e.g. Either String it's still a type constructor because it can take another type argument.
The point of this is: when you define a Functor instance, the type constructor f cannot change. This is because it's represented by the same variable, f, as both the argument and result of fmap. The only type that's allowed to change is the type that's applied to the f constructor.
When you write instance Functor (Either c), Either c is filled in for f everywhere in the declaration of fmap. This gives fmap the following type for this instance:
fmap :: (a -> b) -> (Either c) a -> (Either c) b
With the definition of Either, the only useful way to get this type is by applying the Right value to the function. Remember that "Either" has two possible values with possibly different types. Here the Left value has type 'c', so you can't apply it to the function (which expects an 'a')[1], and the result also wouldn't be correct because you'd be left with Either b a, which doesn't match the class definition.
After replacing "f" with "Either c" to get the above type signature for fmap with the "Either c" instance, writing the implementation is next. There are two cases to consider, the Left and the Right. The type signature tells us that the type of the Left side, "c", can't change. We also don't have any way to change the value because we don't know what type it actually is. All we can do is leave it alone:
fmap f (Left rval) = Left rval
For the Right side, the type signature says that we have to change from a value with type "a" to a value with type "b". The first argument is a function to do exactly that, so we use the function with the input value to get the new output. Putting the two together gives the full definition
instance Functor (Either c) where
fmap f (Right rval) = Right (f rval)
fmap f (Left lval) = Left lval
There's a more general principle at work here which is why writing a Functor instance that adjusts the Left side is impossible, at least with the Prelude definitions. Copying some code from above:
class Functor f where
fmap :: (a -> b) -> f a -> f b
instance Functor (Either c) where ...
Even though we have a type variable 'c' in the instance definition, we can't use it in any of the class methods because it's not mentioned in the class definition. So you can't write
leftMap :: (c -> d) -> Either c a -> Either d a
leftMap mapfunc (Left x) = Left (mapfunc x)
leftMap mapfunc (Right x) = Right x
instance Functor (Either c) where
--fmap :: (c -> d) -> Either c a -> Either d a
fmap = leftMap
The result of leftMap, and thus fmap, is now (Either d) a. The (Either c) has changed to an (Either d), but this isn't allowed because there's no way to express it in the Functor class. To express this, you'd need a class with two type variables, e.g.
class BiFunctor f where
lMap :: (a -> b) -> f a c -> f b c
rMap :: (c -> d) -> f a c -> f a d
biMap :: (a -> b) -> (c -> d) -> f a c -> f b d
In this class, since both the left and right type variables are in scope, it's possible to write methods that operate on either (or both) sides.
instance BiFunctor Either where
lMap = leftMap
rMap = rightMap --the same as the standard fmap definition
biMap fl fr e = rMap fr (lMap fl e)
Although in practice people usually just write "biMap" for the BiFunctor class and use "id" for the other function if a left or right mapping is necessary.
[1] More accurately, the Left value has type 'c', the function expects an 'a', but the type checker can't unify those types because the 'c' type isn't in scope in the class definition.
Left and Right aren't types, and Left x and Right y are of the same type. They are just constructors of Either. You may consider
Left :: c -> Either c d
Right :: d -> Either c d
You can have 2 fmap declarations because we know the Left's and the Right's are different values. It's just like
g :: Int -> Int
g 1 = 2
g 2 = 4
g n = n
Here we can't say 1 and 2 and n are different "types" just because pattern matching works.
The Functor class is defined such that
class Functor f where
fmap :: (a -> b) -> f a -> f b
Note that a and b are arbitrary types. For clarity, let's rename the a in your instance to c, and the function f to func.
instance Functor (Either c) where
fmap func (Right x) = Right (x)
fmap func (Left x) = Left (func x)
Assume your Either follows the default definition
data Either c d = Left c | Right d
then by your definition,
fmap func (Right x) = Right x
-- # (a -> b) -> f a f b
-- # f = Either c
this forces a = b, and
fmap func (Left x) = Left (func x)
-- # (a -> b) -> f a f b
-- # f = Either c
forces c = a = b. Both are not valid considering a, b and c are independent arbitrary types.
Ok so here's another very simple try at this.
You ask why this doesn't compile:
instance Functor (Either a) where
fmap f (Right x) = Right (x)
fmap f (Left x) = Left (f x)
So let's try to simplify the problem by trying to define the same function without putting it as part of a class instance declaration:
That gives us
foo f (Right x) = Right (x)
foo f (Left x) = Left (f x)
Which indeed does compile. ghci tells us the type signature:
*Main> :t foo
foo :: (t1 -> a) -> Either t1 t -> Either a t
We'll rename some of the variables to get something more uniform looking:
foo :: (a -> b) -> Either a c -> Either b c
That makes perfect sense. It takes a function and applies it to the Left of an Either.
But what's the signature for fmap?
*Main> :t fmap
fmap :: (Functor f) => (a -> b) -> f a -> f b
So let's substitute Either c for f in the fmap signature (I renamed Either a to Either c to keep our two different as from getting mixed up):
fmap :: (a -> b) -> Either c a -> Either c b
Do you see the problem? Your function is perfectly valid -- it just has a different type than what fmap for Either a must necessarily have.
This is a sort of beautiful thing about types. Given the signature for fmap, there is really only one meaningful implementation for fmap on Either a.
Sometimes, when we're lucky and careful, we can end up in similar situations -- given a type signature, the function almost writes itself.
Edit: trying to answer the questions below.
1) There's no "composition of two functions" going on. To get the type signature for fmap over Either a just go through the fmap function signature, and every place you see f, replace it with Either a. We would call that a "specialization" of the type signature of fmap. Which is to say, it is strictly less general than the normal type of fmap -- anyplace that requires a function of the more specialized type, you can pass in something of the general type with no problems.
2) Your function for mapping over the left side (which I named "foo" in the above examples) is just fine. It works fine, it does what you want. You just can't name it fmap and use it in a Functor instance. Personally, I'd name it something like onLeft or mapLeft.
All the following can be ignored/is for information, but not a suggestion for future reading in the near future/actual use:
If one wants to get very technical, because you can map over both the left and the right side (although you can only declare Functor for the latter), Either is not only a Functor, but a Bifunctor. This is provided in, e.g., ekmett's Category-Extras library ( see http://hackage.haskell.org/packages/archive/category-extras/0.44.4/doc/html/Control-Bifunctor.html).
There's lots of cool stuff involving calculating with programs, and "origami programming" that uses bifunctors more rigorously. You can read about it here: http://lambda-the-ultimate.org/node/1360. But, you probably don't want to, at least until you're much more familiar with Haskell. It is computer-sciency, mathy, researchy, and very cool, but not necessary at all to understand idiomatic Haskell programming.
While I will eventually cleave to your format, I'm going to start with something in a slightly different format, as I think it will make my explanation clearer.
Let's consider a different datatype
data Choice a = Default Integer | Chosen a
-- This corresponds to your top, working, instance.
instance Functor Choice where
fmap f (Default i) = Default i
fmap f (Chosen a) = Chosen (f a)
It should be clear why this instance works. However, what about the following:
-- Broken!
instance Functor Choice where
fmap f (Default i) = Default (f i)
fmap f (Chosen a) = Chosen a
You should be able to see why this doesn't work. The type of fmap is Functor f => (a -> b) -> f a -> f b; in this context, it's (a -> b) -> Choice a -> Choice b. Thus, the f argument has the type a -> b. However, in the second (failed) instance declaration, you write f i. We know, because of the datatype declaration, that i must be an Integer, so we can't apply f to it. Similarly, since a has type a, Chosen a will have type Chosen a, not type Chosen b. Thus, the Functor instance on the bottom can't work.
Well, your top instance for Either works because, like in the Choice example, it obeys the types. Let's look at it, with a few renamings:
instance Functor (Either c) where
fmap f (Left c) = Left c
fmap f (Right a) = Right (f a)
This instance declaration doesn't declare an instance of Functor for Either—it can't. Something which is an instance of Functor must take one type parameter. Thus, Int can't be a functor, since Int takes no type parameters, but [] and Maybe can be, since [a] and Maybe a are complete types. Either, however, takes two type parameters: Either a b. Thus, what this instance does is declare that Either c is a functor for any possible c. That c is fixed for the duration of the instance declaration. So let's go through and add types (this is not legal syntax!):
instance Functor (Either c) where
fmap :: forall a b. (a -> b) -> (Either c) a -> (Either c) b
fmap f (Left (c :: c)) = Left c
fmap f (Right (a :: a)) = Right (f a :: b)
Since f has type a -> b, but c's type is fixed at c, we can't possibly write Left (f c); and even if we could, we want the c to be left alone, so that we can return an (Either c) b. Similarly, we must apply f to a in order to get something of type b.
This is also why your bottom instance doesn't work: you have a function which needs to work for any type being applied only to the fixed type c, and you leave the type you need to transform alone. Let's look at it, again with type signatures added:
instance Functor (Either c) where
fmap :: forall a b. (a -> b) -> (Either c) a -> (Either c) b
fmap f (Left (c :: c)) = Left (f c)
fmap f (Right (a :: a)) = Right a
Here, your first part of the function definition attempts to apply a function f :: a -> b to something of the fixed type c, which cannot work, so this already fails. But let's look at what type this generates. In this case, we'd expect that (somehow) f c would have the type b, and a would have the type a. In that case, we're returning a value of type Either b a, which is still not allowed.
Basically, the problem stems from this. First, note that f is the same in between the two function definition clauses, so it can't change between lines. Second, note that we are fixing c, and declaring an instance for that c. This is true for any c, but we only look at one at a time. Finally, because of this, Left's argument is not parametrized by the type that f expects; it's guaranteed to have some fixed type c. This means that (a) you can't apply f to it, and (b) you must apply it to Right's argument, since otherwise you won't change the type you're expected to change.
(Edit to try to answer the question better)
The definition of Either is:
data Either a b = Left a | Right b
So "Either" takes two type arguments. By the way, technically "Either" is not actually a type but a type constructor; it takes type arguments to create a type.
The definition of Functor is:
class Functor f where
fmap :: (p -> q) -> f p -> f q
So in this class definition any type "f" that is an instance of Functor must take a type argument. This isn't declared; it is inferred from the "f p" and "f q"; since "f" is being given a type parameter here it must be a type that takes one.
(Note: the original definition used "a" and "b" instead of "p" and "q". I'm using different letters to keep things distinct from "Either a b" when I get to that later)
In most cases "f" is a container type like a list or a tree. So for instance you have
data Tree a = ...
instance Functor Tree where
fmap func_a2b tree_of_a = ... -- tree of b.
However "Either" takes two type parameters, so how can we fit it into this scheme? The answer is that types can have partial application just like functions. In the same way as
I can declare a function
foo x y = ...
and then say "foo 2" in order to get a new function that expects the second argument, so I can say "Either a" to get a new type that expects the second type argument.
Now look at the original instance:
instance Functor (Either a) where ....
So here "Either a" is a type constructor that expects one more argument, just like Functor expects of its instances. So the type of "fmap" for "Either a" will be
fmap :: (p -> q) -> Either a p -> Either a q
So now in the "where" clause you have to give a definition of "fmap" that has this type. The first one you quote has this type because the second type parameter is used for the "Right" constructor, and that is the one that the function is applied to. The second one won't work, because it would have the type
fmap :: (p -> q) -> Either p a -> Either q a
And that is not what the Functor class says its going to be.
Belive it or not, this isn't magic. It all has to do with the type Either a b not being the same type as Either b a. Here is the definition of Either from Prelude
data Either a b = Left a | Right b
... Notice How the type variable a comes first, then b, and also notice that we only include a in the declaration of the Either Functor:
instance Functor (Either a) where
fmap f (Right x) = Right (f x)
fmap f (Left x) = Left x
Now lets look at the definition of the Maybe Functor
instance Functor Maybe where
fmap = map
Here there is no type variable, Although Maybe takes one type parameter (like Maybe Int). What I am trying to get to is that types aren't functors, type constructors are functors (functors have kind *->*).
So let f :: b -> c, in the version of the Either Functor that works, the x from (Left x) is of type a, which is fine since it's (Either a) that is a functor, the x in (Right x) is of Type b so (Right x) is of type ((Either a) b), and (Right (f x)) is of type ((Either a) c), therefore fmap is of type (b->c) -> ((Either a) b) -> ((Either a) c), as required.
In your version that failed, we have that x in (Right (x)) is not of type a, but of type b, So (Right (x)) is not of type ((Either a) c) which doesn't fit with the type of fmap.
so to sum it up: the fmap that works comes out : (b -> c) -> (Either a) b -> (Either a) c,
but the one that doesn't work comes out: (b -> c) -> (Either b) a -> (Either c) a and thats not the right type for fmap.
Hopefully, this will help...
First, though, some background:
1) Functor needs a "type constructor", a (well, not a type per se,...) type that "needs" another regular type given to it to form a "full type", just like a generic in Java/C++.
So, for example, List is a Functor (it is, by the way), or Array, because (among other things) the full type isn't just List, but List<A>. So, :
A Functor takes a "type constructor", an incomplete type.
2) Either is a constructor type that Haskell folks (read: Edward Kmett, and other well-math-endowed all-stars) call a bifunctor. It needs two types given to it to be complete. For example, a full use of Either is: Either Integer String which means (yeah, yeah, "duh!") it's either a (Left) Integer, or a (Right) String. So, this means Either Integer is an incomplete type that is either a Left Integer or a Right...b when you
decide what that "b" is supposed to be.
Now, for the fun part!
The top stuff works because, fmap uses some type constructor, and uses it with an a -> b function to make a similar function from f a to f b - the hands-down favorite example for this in Haskell is the one for lists, AKA map :: (a -> b) -> ([a] -> [b]), where the Functor is the [ ] part. Now, using something like Either a (let's go ahead and use Either Integer from earlier), fmap's type signature looks like this:
fmap :: (a -> b) -> (Either Integer a -> Either Integer b)
and the two examples (from the Top part) show what fmap does with representative values of that Either Integer a type, in order to get an Either Integer b-typed value.
Now, yours -bottom- doesn't work, because:
You have a function, f, that takes
as to bs.
Your Left type has to be type
Integer, and stay an Integer (or
type Float, and stay a Float, what
ever type is the left one of the
two types of the Either type
you're using).
Your Right type has to be of
whatever type that the function
takes ("a"), and go to the type
that the function makes ("b").
It has to do this (but your stuff doesn't - that's why it doesn't work), because that's the type that fmap needs to have. Specifically, you have these equations:
fmap f (Right x) = Right (x)
fmap f (Left x) = Left (f x)
Your equations give fmap the types:
fmap :: (a -> b) -> Either c d -> Either c d
fmap :: (a -> b) -> Either a d -> Either b d
which not only doesn't fit what fmap wants, but it isn't even consistent with each other!
I'm sorry I wrote half a book to wade through, but I hope that gives some insight to you.
Top works because fmap::(b -> c) -> Either a b -> Either a c and yours -bottom- doesn't work because that would require fmap::(a -> c) -> Either a b -> Either a c. But, it would work if you changed Either to
data Either' a b = Left' b | Right' a deriving (Eq, Show)
instance Functor (Either' a) where
fmap f (Right' x) = Right' (x)
fmap f (Left' x) = Left' (f x)
The instance you're trying to write, let's call it fmap2 for now, has the following type:
fmap2 :: (a -> b) -> Either a c -> Either b c
If you set the LANGUAGE pragma TypeOperators, GHC also accepts the type
fmap2 :: (a -> b) -> (a `Either` c) -> (b `Either` c)
In an ideal world this also would work:
fmap2 :: (a -> b) -> (`Either` c) a -> (`Either` c) b
which would give a Functor instance for (`Either` c) but the similarity between normal operators (and their sections) and type operators breaks down at this point (unless there's a GHC option I'm missing!)
In short: your understanding of functors is okay, but you're bitten by the lack of type-level lambdas.
Ehm... How about a few words about "kinds" ?..
That may simplify understanding, I guess.
Remember what is currying. I.e. in ghci:
Prelude> let f x y z = x + y * z
f :: (Num a) => a -> a -> a -> a
Prelude> :t f 1
f 1 :: (Num t) => t -> t -> t
Prelude> :t f 1 2
f 1 2 :: (Num t) => t -> t
Prelude> :t f 1 2 3
f 1 2 3 :: (Num t) => t
The same things you have with types. When you say Either kind of that type is * -> * -> * (i.e. it takes two types and produces type) and when you say Either a kind is * -> * and for Either a b it's * (btw Monad a and Functor a requires a to be of kind * -> *, as I remember).
So when you say type Either a that means type that is still incomplete (requires some "argument" to be bound), so fmap :: (a -> b) -> f a -> f b becomes fmap :: (a -> b) -> (Either c) a -> (Either c) b when f substituted by Either c.