The Applicative typeclass has a somewhat infamous reputation for having opaque laws. There are a lot of great alternative rephrasing of these laws, from many different approaches. For this post, however, I want to talk about Applicative in terms of one of my favorites: Const .

Const

The Const data type from the standard libraries is relatively simple as far as newtypes go:

newtype Const w a = Const { getConst :: w } w aw }

However, let’s look at a less polymorphic version, IntConst , which is essentially Const Int :

newtype IntConst a = IntConst { getIntConst :: Int }

For a IntConst a , the a is a phantom type parameter. This means that there are not necessarily any values of type a in a value of type IntConst a . In modern GHC with PolyKinds, this means that a might not even be a type that can have values — you might have, say, a value of type IntConst Maybe , or a value of type IntConst Monad , and GHC would be perfectly happy.

IntConst admits a straightforward Functor instance that is a lot like the Functor instance for Proxy and Either e :

instance Functor IntConst where fmap _ ( IntConst w) = IntConst w _ (w)

In fact, sometimes I like to refer to Const w a as “an Either w a with only Left , no Right ”. The Functor instance reflects this pretty well:

instance Functor ( Either e) where e) fmap _ ( Left e) = Left e -- just like 'Const' _ (e) fmap f ( Right x) = Right (f x) -- who cares f (x)(f x)

However, the Applicative instance of IntConst is one of my favorite things about it. Let’s try to imagine how we’d write it.

First of all, let’s look at the types of the functions we need:

pure :: a -> IntConst a (<*>) :: IntConst (a -> b) -> IntConst a -> IntConst b (ab)

Now, remember that IntConst ’s type parameter is phantom, so we don’t have any actual values of type a -> b , a , or b involved. An IntConst a , for any a , is really just an Int . Essentially, once we strip out the newtype wrapper shenanigans (replacing IntConst a with its contents, Int ), we just get:

pure :: a -> Int (<*>) :: Int -> Int -> Int

We now have a few options on how to implement these. Let’s try one and see if it works:

instance Applicative IntConst where pure _ = IntConst 42 IntConst x <*> IntConst _ = IntConst x

A perfectly reasonable implementation, right? Our Applicative instance type-checks. And if it type-checks, ship it! Time to call it a day and go home, right?

Not quite.

Applicative

Let’s take a detour through the essense of the Applicative typeclass. Or, at least, one particular interpretation of it that makes sense for instances that represent some sort of effectful action.

The essence of Applicative, to me, is a way to combine values representing some sort of “effect” in a sane way. In Haskell, we often talk about data types as representing/describing some sort of effects. Applicative lets us combine (“sequence”) them in a way that allows us to write powerful generic combinators.

One way to look at it is as a generalization of fmap to taking two parameters:

fmap1 :: (a -> b ) -> F a -> F b (ab ) fmap2 :: (a -> b -> c) -> F a -> F b -> F c (ac)

fmap alone lets you take a single F a and transform it into an F b . fmap2 (or, liftA2 in the standard libraries) is a way of taking two F -values and squishing them into one fat F value.

It does this by letting us talk about combining the effects of an F a , independent of the a (the result). For example, sequenceA_ :

sequenceA_ :: Applicative f => [f a] -> f () [f a]f ()

Basically will take a list of f a s, and return a new f () that has all of the effects of the f a s in the list.

To do this sensibly, we need also to talk about a “no-op” value:

pure :: a -> F a

pure x is intended to be a no-op with “no effects”.

With this, we can say something about the behavior of <*> or fmap2 or liftA2 . Namely:

The effects of f <*> x must have the effects of f once and the effects of x once – no more, and no less. pure ’s results must have no “effects”, and so if used with <*> , introduces no extra effects: pure f <*> x = fmap f x f x f <*> pure x = fmap ( $ x) f x) f (Remember that fmap is not allowed to affect “effects” in any way) Combining effects must be associative.

With this guarantee, we can write sequenceA_ in a polymorphic way:

sequenceA_ :: Applicative f => [f a] -> f () [f a]f () = pure () sequenceA_ []() : xs) = x *> sequenceA_ xs sequenceA_ (xxs)sequenceA_ xs

And we can say with certainty that:

Each “effect” of each value in the [f a] will be executed exactly once: no more, and no less. sequenceA_ of an empty list has no effects.

This makes sequenceA_ a useful combinator. The fact that we can talk about how sequenceA_ behaves for all Applicative instances makes it something that is worth defining. If you use sequenceA_ for your type, you can do it knowing that it will behave in a well-defined way: it must execute every action once (no more, and no less), and sequencing an empty list must have no effects.

If it weren’t for those Applicative laws and expectations, sequenceA_ would be a pretty useless function. It might completely ignore all effects, or it might perform some of the effects multiple times…who knows! The fact that we have Applicative laws and expectations means we can look at the implementation of sequenceA_ and know with certainty (and make bold claims about) how sequenceA_ combines effects.

For example, we can always make the program substitution:

*> sequenceA_ ys sequenceA_ xssequenceA_ ys -- can be substituted with ++ ys) sequenceA_ (xsys)

If sequenceA_ combines all the effects in xs once, and sequenceA_ combines all the effects in ys once, and *> combines the effects of either side once, then this is a legal substitution that doesn’t change what your program does.

However, if the Applicative instance doesn’t have any rules, we can’t do this. For example, the original form uses pure () twice (once for each list’s [] end), and the second form uses pure () once (since there’s only one list). If pure () was allowed to have effects…then the first version would have more effects than the second.

Back to Const

With this new information in mind, let’s revisit our instance for IntConst .

In order for IntConst ’s Applicative instance to behave meaningfully, and in order to be able to match with user expectations of Applicative instance, you need to make sure it follows the basic principles I mentioned earlier (the effects of f <*> x has the effects of f and x exactly once, and the properties about pure ).

We haven’t defined what the “effects” of IntConst are yet, but let’s at least look at if our pure behaves sensibly with <*> . Namely, let’s check pure f <*> x = fmap f x .

Note that this is a meaningful starting point because fmap ’s definition is fixed. For any type, there is at most one possible fmap that is legal and lawful — and in Haskell, we only have to check that fmap id leaves all inputs unchanged.

With that out of the way, let’s check our pure f <*> x = fmap f x law with a simple example for x …say, IntConst 5 . On the left hand side, we have:

-- pure _ = IntConst 42 -- IntConst x <*> InstConst _ = IntConst x pure f <*> IntConst 5 = IntConst 42 <*> IntConst 5 = IntConst 42

On the right hand side, we have:

-- fmap f (IntConst x) = IntConst x fmap f ( IntConst 5 ) = IntConst 5 f (

It is clear that this definition does not work, since IntConst 42 is not the same as IntConst 5 .

What if we defined:

instance Applicative IntConst where pure _ = IntConst 42 IntConst _ <*> IntConst y = IntConst y

Is that any better? Well, pure f <*> IntConst 5 is now equal to IntConst 5 , so that works out. But what about f <*> pure x = fmap ($ x) f ? Let’s use IntConst 3 as our f . On the left hand side:

-- pure _ = IntConst 42 -- IntConst _ <*> IntConst y = IntConst y IntConst 3 <*> pure x = IntConst 3 <*> IntConst 42 = IntConst 42

And on the right hand side:

-- fmap f (IntConst x) = IntConst x fmap ( $ x) ( IntConst 3 ) = IntConst 3 x) (

Ah, that’s wrong too, then.

At this point it might seem like I am facetiously moving very slowly to an answer that has to use both inputs. After all, my earlier statement claimed that f <*> x has to use both the effects of f and the effects of x , each exactly once. Because we didn’t really know what the “effects” of IntConst are, we don’t know exactly “how” to combine them…but we can probably guess it has to use both Int s. So, with that in mind, let’s try another definition:

instance Applicative IntConst where pure _ = IntConst 42 IntConst x <*> IntConst y = IntConst (x + y) (xy)

Alright, now we use both x and y in the result. Let’s see again if this follows our expectations about pure – if pure f <*> x is the same as fmap f x . Using IntConst 5 again as x :

-- pure _ = IntConst 42 -- IntConst x <*> InstConst y = IntConst (x + y) pure f <*> IntConst 5 = IntConst 42 <*> IntConst 5 = IntConst 47

On the right hand side, we have:

-- fmap f (IntConst x) = IntConst x fmap f ( IntConst 5 ) = IntConst 5 f (

Another dead end. It looks like it isn’t just enough that we use both Int s…we have to use them in a way such that the Int we use as the result of pure f as to be an identity to our operation. Whatever Int is returned by pure f has to leave any other Int unchanged when used with <*> .

Thinking back, we remember that if our operation is + , we can use 0 , since 0 + x = x and x + 0 = x , for all x . Luckily, our operation x + y is one that even has an identity. If we had chosen another operation (like x + 2 * y ), we wouldn’t be so lucky. Finally:

instance Applicative IntConst where pure _ = IntConst 0 IntConst x <*> IntConst y = IntConst (x + y) (xy)

At last this feels like something that should make sense. And, does it? Testing out, again, pure f <*> x = fmap f x :

-- pure _ = IntConst 0 -- IntConst x <*> InstConst y = IntConst (x + y) pure f <*> IntConst 5 = IntConst 0 <*> IntConst 5 = IntConst 5

-- fmap f (IntConst x) = IntConst x fmap f ( IntConst 5 ) = IntConst 5 f (

Perfect! And, checking now f <*> pure x = fmap ($ x) f :

-- pure _ = IntConst 0 -- IntConst x <*> IntConst y = IntConst (x + y) IntConst 3 <*> pure x = IntConst 3 <*> IntConst 0 = IntConst 3

-- fmap f (IntConst x) = IntConst x fmap ( $ x) ( IntConst 3 ) = IntConst 3 x) (

This definition works for both !

The Effect of Const

With our definition picked out, what do we think sequenceA_ does for IntConst ?

sequenceA_ :: [ IntConst a] -> IntConst () a]()

Well, if each application of <*> adds together the Int in the IntConst a , and sequenceA_ uses <*> once per every IntConst a …we can guess that sequenceA_ for IntConst is just sum !

This might be more clear if we strip away the newtype wrappers (replacing IntConst a with its contents, Int ):

sequenceA_ :: [ Int ] -> Int

From this definition of <*> , we can form an idea of what the effects of the IntConst Applicative are: they add to some accumulator environment! And pure _ = IntConst 0 means that pure _ adds zero to our accumulator – it leaves our accumulator unchanged, and so effectively has no effect.

That’s why sequenceA_ is sum – it sequences every effect of the IntConst , which means that it sequences all of those “add to the accumulator” effects one after the other.

Alternative Pictures

Note that the requirements we gave for the Applicative instance doesn’t necessarily imply that the one we have is the only instance. For example, the following instance is also valid:

instance Applicative IntConst where pure _ = IntConst 1 IntConst x <*> IntConst y = IntConst (x * y) (xy)

If our “combining” action is * , then pure has to be an identity. So, pure _ = IntConst 1 works fine as an identity here, since 1 * x = x and x * 1 = x , for all x .

-- pure _ = IntConst 1 -- IntConst x <*> InstConst y = IntConst (x * y) pure f <*> IntConst 5 = IntConst 1 <*> IntConst 5 = IntConst 5

-- fmap f (IntConst x) = IntConst x fmap f ( IntConst 5 ) = IntConst 5 f (

A General Alternative

It looks like the Applicative for IntConst has to follow some pattern:

<*> has to combine the Int s inside somehow using some operation f . ( f also has to be associative, which is a point we didn’t touch on specifically)

has to combine the s inside somehow using some operation . ( also has to be associative, which is a point we didn’t touch on specifically) The Int that pure returns has to be an identity to f .

Sound familiar?

This is all satisfied if and only if f and the result of pure form a monoid on the integers!

There is a very fundamental link here: the Applicative laws for IntConst are satisfied if and only if <*> acts monoidally on the contents, with pure ’s result as the identity of that monoid.

(For those unfamiliar with monoids, a Monoid in Haskell is a type w that has an associative operation (<>) :: w -> w -> w along with an identity mempty :: w that leaves values unchanged when used with <> .)

So, any f and pure works, as long as they form a monoid. And any monoid in the Integers is a valid Applicative instance for IntConst !

General Const

Let’s revisit our original Const type:

newtype Const w a = Const { getConst :: w } w aw }

The Functor instance is unique, so there isn’t any leeway we have ( fmap is always fixed for every type):

instance Functor ( Const w) where w) fmap _ ( Const w) = Const w _ (w)

This is the only definition that preserves fmap id = id .

Now we can actually write an Applicative instance for Const w …as long as we provide a Monoid to use with w !

instance Monoid w => Applicative ( Const w) where w) pure _ = Const mempty Const x <*> Const y = Const (x <> y) (xy)

Like how we said, as long as the “combining” function for x and y have the identity that is given by the result of pure , this is a valid Applicative.

The “effects” of this Applicative instance are “accumulate to some accumulator”. If this sounds familiar, this is because this is exactly the effect of the Writer w Applicative instance. Const w and Writer w have the same effects (“accumulate to some accumulator”), and this can be seen clearly by comparing the two types:

data Const w a = Const w w a data Writer w a = Writer w a w aw a

( Const is just Writer without the a value)

What is Applicative Really?

If you think about this, this seems like a bit of a crazy coincidence. Applicatives are an interesting concept, and Monoids are a different one.

But it looks like in order to make an Applicative instance for Const w , the behavior of <*> and pure have to follow some certain properties in order to fit the Applicative laws…and those properties are exactly monoidal properties and the Monoid laws.

To illustrate this link, we can see the type of pure and <*> for Const w , without the newtype wrappers (and ignored arguments)

instance Monoid w => Applicative ( Const w) where w) pure :: w (<*>) :: w -> w -> w

And let’s look at the Monoid typeclass:

instance Monoid w where mempty :: w (<>) :: w -> w -> w

It seems like Const is nothing more than a (type-level) function on a Monoid. As an * -> (k -> *) , it takes a * -kinded Monoid and turns it into a k -> * -kinded Monoid:

instance Monoid ( w :: * ) => Applicative ( Const w :: k -> * )

“Give me a Monoid and I’ll give you something k -> * that is also a monoid!”

Const is a monoid homomorphism: it takes a monoid w with mempty and (<>) , and turns it into a monoid Const w with pure _ and <*> :

Const (x <> y) = Const x <*> Const y (xy) Const mempty = pure () ()

Meaning “ <> then Const ” is the same as “ Const then <*> ”, and “ Const the mempty ” is the same as pure () . Both things essentially convey the exact same monoid – one with <> and mempty , and the other with <*> and pure () . In fact, it’s a bit more than a monoid homomorphism – it’s a monoid isomorphism:

<> getConst y = getConst (x <*> y) getConst xgetConst ygetConst (xy) mempty = getConst ( pure ()) getConst (())

Which means “ getConst then <> ” is the same as “ <*> then getConst ”, and mempty is the same as getConst (pure ()) . getConst takes you from one monoid (with <*> and pure () ) to another (with <> and mempty ).

One incidental observation – sequenceA_ for Const w might look familiar:

sequenceA_ :: Monoid w => [ Const w a] -> Const w () w a]w () -- strip out newtype wrappers sequenceA_ :: Monoid w => [w] -> w [w]

It’s just mconcat !

As an exercise, see if you can understand this definition of mconcat in terms of Const and traverse :

mconcat :: Monoid w => [w] -> w [w] mconcat = getConst . traverse_ Const getConsttraverse_

traverse_ , if you aren’t familiar with it, an “effectful” function (in our case, Const :: w -> Const w w ) over all values in a container, and sequences all of their effects.

Monoid is the Key

All of this actually witnesses the core of Applicative. A lot of people describe Applicative as a “lax monoidal functor”.

In this post, I was really handwavey with how I talked about “effects” (“ f <*> x must use the effects of f and x each once and only once”, I claimed, without defining what an effect was). The notion of what an “effect” is really comes from each individual Applicative, and each type really has its own conceptual picture of what counts as an effect. The rigorous test of what is a meaningful way to have an effect that can be combined comes from those laws ( pure f <*> x = fmap f x , etc.) and the overall sentiment that the combination of effects is monoidal.

At its heart, Applicative enforces that liftA2 and <*> are supposed to be “monoidal” in some way. This fact is hidden by the normal form of the Applicative laws, but I feel like seeing this play out in the Applicative instance for Const — how Monoid is exactly the constraint necessary to implement the instance, and how Const forms a monoid isomorphism — really helps hammer in the monoidal nature of all Applicative instances.

Applicative instances must be monoidal in how they sequence their effects. Because Const ’s effects are so simple (“accumulate a value”), this makes it an especially obvious demonstration of this.