I'm trying to write an instance of Applicative for my datatype
data Multinode k v = Multinode { key :: k
, values :: [v]
} deriving Show
data Multimap k v = Multimap [Multinode k v] deriving Show
I'm still learning Haskell. So I may make funny mistakes. If I understood properly, Applicative is a subset of Functor. The operators that have to be defined are pure and <*>.
pure :: a -> f a
(<*>) :: f (a -> b) -> f a -> f b
My attempt is the following
instance Applicative (Multimap k) where
pure x = Multimap [(Multinode x [x])]
(Multimap mf#((Multinode kf vsf):nsf)) <*> (Multimap mx#((Multinode kx vsx):nsx)) =
Multimap [Multinode kf (vsf <*> vsx)]
Maps and multimaps are not naturally Applicative. Here's why: you need
pure
:: (...) -- Possible constraint involving k, but not v
=> v -> MultiMap k v
Since there are no constraints on v, pure cannot use its argument to generate any keys. So the set of keys in the result must always be the same, which doesn't work out for any meaningful notion of <*> that I know of.
The semigroupoids package offers a class called Apply, which is much more promising:
class Functor f => Apply f where
(<.>) :: f (a -> b) -> f a -> f b
liftF2 :: (a -> b -> c) -> f a -> f b -> f c
(.>) :: f a -> f b -> f b
(<.) :: f a -> f b -> f a
This is just like Applicative but without pure. <.> is analogous to <*>, liftF2 is analogous to liftA2, etc. The semigroupoids package includes an instance Ord k => Apply (Map k). You should be able to think of a couple different instances for your type based on the general idea of that one.
Related
In Haskell Applicatives are considered stronger than Functor that means we can define Functor using Applicative like
-- Functor
fmap :: (a -> b) -> f a -> f b
fmap f fa = pure f <*> fa
and Monads are considered stronger than Applicatives & Functors that means.
-- Functor
fmap :: (a -> b) -> f a -> f b
fmap f fa = fa >>= return . f
-- Applicative
pure :: a -> f a
pure = return
(<*>) :: f (a -> b) -> f a -> f b
(<*>) = ??? -- Can we define this in terms return & bind? without using "ap"
I have read that Monads are for sequencing actions. But I feel like the only thing a Monad can do is Join or Flatten and the rest of its capabilities comes from Applicatives.
join :: m (m a) -> m a
-- & where is the sequencing in this part? I don't get it.
If Monad is really for sequencing actions then How come we can define Applicatives (which are not considered to strictly operate in sequence, some kind of parallel computing)?
As monads are Monoids in the Category of endofunctors. There are Commutative monoids as well, which necessarily need not work in order. That means the Monad instances for Commutative Monoids also need an ordering?
Edit:
I found an excellent page
http://wiki.haskell.org/What_a_Monad_is_not
If Monad is really for sequencing actions then How come we can define Applicatives (which are not considered to strictly operate in sequence, some kind of parallel computing)?
Not quite. All monads are applicatives, but only some applicatives are monads. So given a monad you can always define an applicative instance in terms of bind and return, but if all you have is the applicative instance then you cannot define a monad without more information.
The applicative instance for a monad would look like this:
instance (Monad m) => Applicative m where
pure = return
f <*> v = do
f' <- f
v' <- v
return $ f' v'
Of course this evaluates f and v in sequence, because its a monad and that is what monads do. If this applicative does not do things in a sequence then it isn't a monad.
Modern Haskell, of course, defines this the other way around: the Applicative typeclass is a subset of Functor so if you have a Functor and you can define (<*>) then you can create an Applicative instance. Monad is in turn defined as a subset of Applicative, so if you have an Applicative instance and you can define (>>=) then you can create a Monad instance. But you can't define (>>=) in terms of (<*>).
See the Typeclassopedia for more details.
We can copy the definition of ap and desugar it:
ap f a = do
xf <- f
xa <- a
return (xf xa)
Hence,
f <*> a = f >>= (\xf -> a >>= (\xa -> return (xf xa)))
(A few redundant parentheses added for clarity.)
(<*>) :: f (a -> b) -> f a -> f b
(<*>) = ??? -- Can we define this in terms return & bind? without using "ap"
Recall that <*> has the type signature of f (a -> b) -> f a -> f b, and >>= has m a -> (a -> m b) -> m b. So how can we infer m (a -> b) -> m a -> m b from m a -> (a -> m b) -> m b?
To define f <*> x with >>=, the first parameter of >>= should be f obviously, so we can write the first transformation:
f <*> x = f >>= k -- k to be defined
where the function k takes as a parameter a function with the type of a -> b, and returns a result of m b such that the whole definition aligns with the type signature of bind >>=. For k, we can write:
k :: (a -> b) -> m b
k = \xf -> h x
Note that the function h should use x from f <*> x since x is related to the result of m b in some way like the function xf of a -> b.
For h x, it's easy to get:
h :: m a -> m b
h x = x >>= return . xf
Put the above three definations together, and we get:
f <*> x = f >>= \xf -> x >>= return . xf
So even though you don't know the defination of ap, you can still get the final result as shown by #chi according to the type signature.
Some Haskell source code (see ref):
-- | Sequential application.
--
-- A few functors support an implementation of '<*>' that is more
-- efficient than the default one.
(<*>) :: f (a -> b) -> f a -> f b
(<*>) = liftA2 id
-- | Lift a binary function to actions.
--
-- Some functors support an implementation of 'liftA2' that is more
-- efficient than the default one. In particular, if 'fmap' is an
-- expensive operation, it is likely better to use 'liftA2' than to
-- 'fmap' over the structure and then use '<*>'.
liftA2 :: (a -> b -> c) -> f a -> f b -> f c
liftA2 f x = (<*>) (fmap f x)
Three things seem to be quite confusing to me:
1) (<*>) is defined in terms of liftA2, where liftA2 is defined in terms of (<*>). How does it work? I see no obvious "recursion-break" case...
2) id is an a -> a function. Why is it passed into liftA2 as an (a -> b -> c) function?
3) fmap id x always equals x, since functors must preserve appropriate identities. Thus (<*>) (fmap id x) = (<*>) (x) where x = f a - an a-typed functor itself (by the way, how can a-typifying of the functor can be explained from the pure category theory's point of view? functor is just a mapping between categories, it has no further "typification"... seems like it is better to say - "a container of type a with an (endo)functor defined for each instance of asummed category Hask of well-defined Haskell types). So (<*>) (f a) while by definition (<*>) expects f(a' -> b'): thus, the only way to make it work is to deliberately bound a to be an (a' -> b'). However when I run :t \x -> (<*>) (fmap id x) in the gchi, it spits out something mind-blowing: f (a -> b) -> f a -> f b - which I fail to explain.
Can someone step by step explain how does that work and why it even compiles?
P.S. Category theory terms, if needed, are welcome.
For question 1, you left out a very important piece of context.
class Functor f => Applicative f where
{-# MINIMAL pure, ((<*>) | liftA2) #-}
Those definitions you quoted belong to a class. That means instances can override them. Furthermore, the MINIMAL pragma says that in order to work, at least one of them must be overridden in the instance. So the breaking of the recursion happens whenever one is overridden in a particular instance. This is just like how the Eq class defines (==) and (/=) in terms of each other so that you only need to provide a definition for one in a hand-written instance.
For question two, a -> b -> c is shorthand for a -> (b -> c). So it unifies with (let's rename variables to avoid collision) d -> d as (b -> c) -> (b ->c). (tangentially, that's also the type of ($).)
For three - you're absolutely right. Keep simplifying!
\x -> (<*>) (fmap id x)
\x -> (<*>) x
(<*>)
So it shouldn't really be a surprise ghci gave you the type of (<*>) back, should it?
1) (<*>) is defined in terms of liftA2, where liftA2 is defined in terms of (<*>). How does it work? I see no obvious "recursion-break" case...
It's not recursion. In your instance of Applicative you can either define both of them or just one. If you define just (<*>) then liftA2 is defined from (<*>), and vice versa.
2) id is an a -> a function. Why is it passed into liftA2 as an (a -> b -> c) function?
Unification works as follows,
(<*>) :: f (a -> b) -> f a -> f b
(<*>) = liftA2 id
liftA2 :: (a -> b -> c) -> f a -> f b -> f c
id : u -> u
liftA2 : (a -> (b -> c) -> f a -> f b -> f c
------------------------------------------------------
u = a
u = b->c
id : (b->c) -> (b->c)
liftA2 : ((b->c) -> (b->c)) -> f (b->c) -> f b -> f c
------------------------------------------------------
liftA2 id : f (b->c) -> f b -> f c
3.
liftA2 :: (a -> b -> c) -> f a -> f b -> f c
liftA2 h x = (<*>) (fmap h x)
Renamed the first argument from f to h, to prevent confusion since f also shows in the type
h :: a -> (b -> c)
x :: f a
fmap :: (a -> d) -> f a -> f d
------------------------------
d = b -> c
h :: a -> (b->c)
x :: f a
fmap :: (a -> (b->c)) -> f a -> f (b->c)
----------------------------------------
fmap h x :: f (b -> c)
fmap h x :: f (b -> c)
(<*>) :: f (b -> c) -> f b -> f c
-------------------------------------
(<*>) fmap h x :: f b -> f c
Edit:
Consistency
To show the consistency of both formulas, first lets first rewrite liftA2 into something simpler. We can use the formula below to get rid of fmap and use only pure and <*>
fmap h x = pure h <*> x
and it's best to put all points in the definition. So we get,
liftA2 h u v
= (<*>) (fmap h u) v
= fmap h u <*> v
= pure h <*> u <*> v
So we want to check the consistency of,
u <*> v = liftA2 id u v
liftA2 h u v = pure h <*> u <*> v
For the first we need the property that pure id <*> u = u
u <*> v
= liftA2 id u v
= pure id <*> u <*> v
= u <*> v
For the second we need a property of liftA2. Properties of applicative are usually given in terms of pure and <*> so we need to derive it first. The required formula is derived from pure h <*> pure x = pure (h x).
liftA2 h (pure x) v
= pure h <*> pure x <*> v
= pure (h x) <*> v
= liftA2 (h x) v
This requires h : t -> a -> b -> c. The proof of consistency becomes,
liftA2 h u v
= pure h <*> u <*> v
= pure h `liftA2 id` u `liftA2 id` v
= liftA2 id (liftA2 id (pure h) u) v
= liftA2 id (liftA2 h u) v
= liftA2 h u v
1) (<*>) is defined in terms of liftA2, where liftA2 is defined in terms of (<*>). How does it work? I see no obvious "recursion-break" case...
Each instance is responsible for overriding at least one of the two. This is documented in a machine-readable way in the pragma at the top of the class:
{-# MINIMAL pure, ((<*>) | liftA2) #-}
This pragma announces that instance writers must define at least the pure function and at least one of the other two.
id is an a -> a function. Why is it passed into liftA2 as an (a -> b -> c) function?
If id :: a -> a, we can choose a ~ d -> e to get id :: (d -> e) -> d -> e. Traditionally, this particular specialization of id is spelled ($) -- maybe you've seen that one before!
3) ...
I don't... actually see any contradiction set up in the facts you state. So I'm not sure how to explain away the contradiction for you. However, you have a few infelicities in your notation that might be related to mistakes in your thinking, so let's talk about them briefly.
You write
Thus (<*>) (fmap id x) = (<*>) (x) where x = f a.
This is not quite right; the type of x is f a for some Functor f, but it is not necessarily equal to f a.
by the way, how can a-typifying of the functor can be explained from the pure category theory's point of view? functor is just a mapping between categories, it has no further "typification"... seems like it is better to say - "a container of type a with an (endo)functor defined for each instance of assumed category Hask of well-defined Haskell types
A functor constitutes two things: a mapping from objects to objects, and a mapping from arrows to arrows that is compatible with the object mapping. In a Haskell Functor instance declaration like
instance Functor F where fmap = fmapForF
the F is the mapping from objects to objects (objects in both the source and target categories are types, and F is a thing which takes a type and produces a type) and the fmapForF is the mapping from arrows to arrows.
I run :t \x -> (<*>) (fmap id x) in the gchi, it spits out something mind-blowing: f (a -> b) -> f a -> f b - which I fail to explain.
Well, you already observed that fmap id x = x, which means \x -> (<*>) (fmap id x) = \x -> (<*>) x. And for any function f, f = \x -> f x (up to some niggles that aren't important right now), so in particular \x -> (<*>) (fmap id x) = (<*>). So ghci gives you the type of (<*>), as it should.
Here I have to disagree with the GHC devs on their coding style :)
I would like to argue that one should never write
ap = liftA2 id
but, instead, use the equivalent
ap = liftA2 ($)
since the latter makes it clear that we are lifting the application operation.
(Actually, for very technical reasons GHC devs can not use $ here in this internal module, as pointed out below in the comments. So, at least they have a very good reason for their choice.)
Now, you might wonder why id can be used instead of $. Formally, we have
($) f x
= f x
= (id f) x
= id f x
hence, eta-contracting x then f, we get ($) = id.
Indeed, ($) is a "special case" of id.
id :: a -> a
-- choose a = (b -> c) as a special case
id :: (b -> c) -> (b -> c)
id :: (b -> c) -> b -> c
($):: (b -> c) -> b -> c
Hence, the main difference is: id is the identity on any type a, while ($) is the "identity" on any functional type b -> c. The latter is best visualized as a binary function (application), but it can equivalently be considered a unary function (identity) on a function type.
What is the general term for a functor with a structure resembling QuickCheck's promote function, i.e., a function of the form:
promote :: (a -> f b) -> f (a -> b)
(this is the inverse of flip $ fmap (flip ($)) :: f (a -> b) -> (a -> f b)). Are there even any functors with such an operation, other than (->) r and Id? (I'm sure there must be). Googling 'quickcheck promote' only turned up the QuickCheck documentation, which doesn't give promote in any more general context AFAICS; searching SO for 'quickcheck promote' produces no results.
(<*>) :: Applicative f => f (a -> b) -> f a -> f b
(=<<) :: Monad m => (a -> m b) -> m a -> m b
Given that Monad is more powerful an interface than Applicative, this tell us that a -> f b can do more things than f (a -> b). This tells us that a function of type (a -> f b) -> f (a -> b) can't be injective. The domain is bigger than the codomain, in a handwavey manner. This means there's no way you can possibly preserve behavior of the function. It just doesn't work out across generic functors.
You can, of course, characterize functors in which that operation is injective. Identity and (->) a are certainly examples. I'm willing to bet there are more examples, but nothing jumps out at me immediately.
So far I found these ways of constructing an f with the promote morphism:
f = Identity
if f and g both have promote then the pair functor h t = (f t, g t) also does
if f and g both have promote then the composition h t = f (g t) also does
if f has the promote property and g is any contrafunctor then the functor h t = g t -> f t has the promote property
The last property can be generalized to profunctors g, but then f will be merely a profunctor, so it's probably not very useful, unless you only require profunctors.
Now, using these four constructions, we can find many examples of functors f for which promote exists:
f t = (t,t)
f t = (t, b -> t)
f t = (t -> a) -> t
f t = ((t,t) -> b) -> (t,t,t)
f t = ((t, t, c -> t, (t -> b) -> t) -> a) -> t
Also note that the promote property implies that f is pointed.
point :: t -> f t
point x = fmap (const x) (promote id)
Essentially the same question: Is this property of a functor stronger than a monad?
Data.Distributive has
class Functor g => Distributive g where
distribute :: Functor f => f (g a) -> g (f a)
-- other non-critical methods
Renaming your variables, you get
promote :: (c -> g a) -> g (c -> a)
Using slightly invalid syntax for clarity,
promote :: ((c ->) (g a)) -> g ((c ->) a)
(c ->) is a Functor, so the type of promote is a special case of the type of distribute. Thus every Distributive functor supports your promote. I don't know if any support promote but not Distributive.
I want to map over Applicative form.
The type of map-like function would be like below:
mapX :: (Applicative f) => (f a -> f b) -> f [a] -> f [b]
used as:
result :: (Applicative f) => f [b]
result = mapX f xs
where f :: f a -> f b
f = ...
xs :: f[a]
xs = ...
As the background of this post, I try to write fluid simulation program using Applicative style referring to Paul Haduk's "The Haskell School of Expression", and I want to express the simulation with Applicative style as below:
x, v, a :: Sim VArray
x = x0 +: integral (v * dt)
v = v0 +: integral (a * dt)
a = (...calculate acceleration with x v...)
instance Applicative Sim where
...
where Sim type means the process of simulation computation and VArray means Array of Vector (x,y,z). X, v a are the arrays of position, velocity and acceleration, respectively.
Mapping over Applicative form comes when definining a.
I've found one answer to my question.
After all, my question is "How to lift high-order functions (like map
:: (a -> b) -> [a] -> [b]) to the Applicative world?" and the answer
I've found is "To build them using lifted first-order functions."
For example, the "mapX" is defined with lifted first-order functions
(headA, tailA, consA, nullA, condA) as below:
mapX :: (f a -> f b) -> f [a] -> f [b]
mapX f xs0 = condA (nullA xs0) (pure []) (consA (f x) (mapA f xs))
where
x = headA xs0
xs = tailA xs0
headA = liftA head
tailA = liftA tail
consA = liftA2 (:)
nullA = liftA null
condA b t e = liftA3 aux b t e
where aux b t e = if b then t else e
First, I don't think your proposed type signature makes much sense. Given an applicative list f [a] there's no general way to turn that into [f a] -- so there's no need for a function of type f a -> f b. For the sake of sanity, we'll reduce that function to a -> f b (to transform that into the other is trivial, but only if f is a monad).
So now we want:
mapX :: (Applicative f) => (a -> f b) -> f [a] -> f [b]
What immediately comes to mind now is traverse which is a generalization of mapM. Traverse, specialized to lists:
traverse :: (Applicative f) => (a -> f b) -> [a] -> f [b]
Close, but no cigar. Again, we can lift traverse to the required type signature, but this requires a monad constraint: mapX f xs = xs >>= traverse f.
If you don't mind the monad constraint, this is fine (and in fact you can do it more straightforwardly just with mapM). If you need to restrict yourself to applicative, then this should be enough to illustrate why you proposed signature isn't really possible.
Edit: based on further information, here's how I'd start to tackle the underlying problem.
-- your sketch
a = liftA sum $ mapX aux $ liftA2 neighbors (x!i) nbr
where aux :: f Int -> f Vector3
-- the type of "liftA2 neighbors (x!i) nbr" is "f [Int]
-- my interpretation
a = liftA2 aux x v
where
aux :: VArray -> VArray -> VArray
aux xi vi = ...
If you can't write aux like that -- as a pure function from the positions and velocities at one point in time to the accelerations, then you have bigger problems...
Here's an intuitive sketch as to why. The stream applicative functor takes a value and lifts it into a value over time -- a sequence or stream of values. If you have access to a value over time, you can derive properties of it. So velocity can be defined in terms of acceleration, position can be defined in terms of velocity, and soforth. Great! But now you want to define acceleration in terms of position and velocity. Also great! But you should not need, in this instance, to define acceleration in terms of velocity over time. Why, you may ask? Because velocity over time is all acceleration is to begin with. So if you define a in terms of dv, and v in terms of integral(a) then you've got a closed loop, and your equations are not propertly determined -- either there are, even given initial conditions, infinitely many solutions, or there are no solutions at all.
If I'm thinking about this right, you can't do this just with an applicative functor; you'll need a monad. If you have an Applicative—call it f—you have the following three functions available to you:
fmap :: (a -> b) -> f a -> f b
pure :: a -> f a
(<*>) :: f (a -> b) -> f a -> f b
So, given some f :: f a -> f b, what can you do with it? Well, if you have some xs :: [a], then you can map it across: map (f . pure) xs :: [f b]. And if you instead have fxs :: f [a], then you could instead do fmap (map (f . pure)) fxs :: f [f b].1 However, you're stuck at this point. You want some function of type [f b] -> f [b], and possibly a function of type f (f b) -> f b; however, you can't define these on applicative functors (edit: actually, you can define the former; see the edit). Why? Well, if you look at fmap, pure, and <*>, you'll see that you have no way to get rid of (or rearrange) the f type constructor, so once you have [f a], you're stuck in that form.
Luckily, this is what monads are for: computations which can "change shape", so to speak. If you have a monad m, then in addition to the above, you get two extra methods (and return as a synonym for pure):
(>>=) :: m a -> (a -> m b) -> m b
join :: m (m a) -> m a
While join is only defined in Control.Monad, it's just as fundamental as >>=, and can sometimes be clearer to think about. Now we have the ability to define your [m b] -> m [b] function, or your m (m b) -> m b. The latter one is just join; and the former is sequence, from the Prelude. So, with monad m, you can define your mapX as
mapX :: Monad m => (m a -> m b) -> m [a] -> m [b]
mapX f mxs = mxs >>= sequence . map (f . return)
However, this would be an odd way to define it. There are a couple of other useful functions on monads in the prelude: mapM :: Monad m => (a -> m b) -> [a] -> m [b], which is equivalent to mapM f = sequence . map f; and (=<<) :: (a -> m b) -> m a -> m b, which is equivalent to flip (>>=). Using those, I'd probably define mapX as
mapX :: Monad m => (m a -> m b) -> m [a] -> m [b]
mapX f mxs = mapM (f . return) =<< mxs
Edit: Actually, my mistake: as John L kindly pointed out in a comment, Data.Traversable (which is a base package) supplies the function sequenceA :: (Applicative f, Traversable t) => t (f a) => f (t a); and since [] is an instance of Traversable, you can sequence an applicative functor. Nevertheless, your type signature still requires join or =<<, so you're still stuck. I would probably suggest rethinking your design; I think sclv probably has the right idea.
1: Or map (f . pure) <$> fxs, using the <$> synonym for fmap from Control.Applicative.
Here is a session in ghci where I define mapX the way you wanted it.
Prelude>
Prelude> import Control.Applicative
Prelude Control.Applicative> :t pure
pure :: Applicative f => a -> f a
Prelude Control.Applicative> :t (<*>)
(<*>) :: Applicative f => f (a -> b) -> f a -> f b
Prelude Control.Applicative> let mapX fun ma = pure fun <*> ma
Prelude Control.Applicative> :t mapX
mapX :: Applicative f => (a -> b) -> f a -> f b
I must however add that fmap is better to use, since Functor is less expressive than Applicative (that means that using fmap will work more often).
Prelude> :t fmap
fmap :: Functor f => (a -> b) -> f a -> f b
edit:
Oh, you have some other signature for mapX, anyway, you maybe meant the one I suggested (fmap)?
At different questions I've found hints in comments concerning using the (->) instance of Monads e.g. for realizing point-free style.
As for me, this is a little too abstract. Ok, I've seen Arrow instances on (->) and it seems to me, that (->) can be used in instance notations but not in type declarations (that would alone be stuff for another question).
Has anyone examples using (->) as instance of Monad? Or a good link?
Sorry if this question may already have been discussed here, but searching for "(->) Monad instance" gives you many many hits as you can imagine ... since nearly every question about Haskell somewhere involves (->) or "Monad".
For a given type r, the function of type r -> a can be thought of as a computation delivering an a using an environment typed r. Given two functions r -> a and a -> (r -> b), it's easy to imagine that one can compose these when given an environment (again, of type r).
But wait! That's exactly what monads are about!
So we can create an instance of Monad for (->) r that implements f >>= g by passing the r to both f and g. This is what the Monad instance for (->) r does.
To actually access the environment, you can use id :: r -> r, which you can now think of as a computation running in an environment r and delivering an r. To create local sub-environments, you can use the following:
inLocalEnvironment :: (r -> r) -> (r -> a) -> (r -> a)
inLocalEnvironment xform f = \env -> f (xform env)
This pattern of having an environment passed to computations that can then query it and modify it locally is useful for not just the (->) r monad, which is why it is abstracted into the MonadReader class, using much more sensible names than what I've used here:
http://hackage.haskell.org/packages/archive/mtl/2.0.1.0/doc/html/Control-Monad-Reader-Class.html
Basically, it has two instances: (->) r that we've seen here, and ReaderT r m, which is just a newtype wrapper around r -> m a, so it's the same thing as the (->) r monad I've described here, except it delivers computations in some other, transformed monad.
To define a monad for (->) r, we need two operations, return and (>>=), subject to three laws:
instance Monad ((->) r) where
If we look at the signature of return for (->) r
return :: a -> r -> a
we can see its just the constant function, which ignores its second argument.
return a r = a
Or alternately,
return = const
To build (>>=), if we specialize its type signature with the monad (->) r,
(>>=) :: (r -> a) -> (a -> r -> b) -> r -> b
there is really only one possible definition.
(>>=) x y z = y (x z) z
Using this monad is like passing along an extra argument r to every function. You might use this for configuration, or to pass options way down deep into the bowels of your program.
We can check that it is a monad, by verifying the three monad laws:
1. return a >>= f = f a
return a >>= f
= (\b -> a) >>= f -- by definition of return
= (\x y z -> y (x z) z) (\b -> a) f -- by definition of (>>=)
= (\y z -> y ((\b -> a) z) z) f -- beta reduction
= (\z -> f ((\b -> a) z) z) -- beta reduction
= (\z -> f a z) -- beta reduction
= f a -- eta reduction
2. m >>= return = m
m >>= return
= (\x y z -> y (x z) z) m return -- definition of (>>=)
= (\y z -> y (m z) z) return -- beta reduction
= (\z -> return (m z) z) -- beta reduction
= (\z -> const (m z) z) -- definition of return
= (\z -> m z) -- definition of const
= m -- eta reduction
The final monad law:
3. (m >>= f) >>= g ≡ m >>= (\x -> f x >>= g)
follows by similar, easy equational reasoning.
We can define a number of other classes for ((->) r) as well, such as Functor,
instance Functor ((->) r) where
and if we look at the signature of
-- fmap :: (a -> b) -> (r -> a) -> r -> b
we can see that its just composition!
fmap = (.)
Similarly we can make an instance of Applicative
instance Applicative ((->) r) where
-- pure :: a -> r -> a
pure = const
-- (<*>) :: (r -> a -> b) -> (r -> a) -> r -> b
(<*>) g f r = g r (f r)
What is nice about having these instances is they let you employ all of the Monad and Applicative combinators when manipulating functions.
There are plenty of instances of classes involving (->), for instance, you could hand-write the instance of Monoid for (b -> a), given a Monoid on a as:
enter code here
instance Monoid a => Monoid (b -> a) where
-- mempty :: Monoid a => b -> a
mempty _ = mempty
-- mappend :: Monoid a => (b -> a) -> (b -> a) -> b -> a
mappend f g b = f b `mappend` g b
but given the Monad/Applicative instance, you can also define this instance with
instance Monoid a => Monoid (r -> a) where
mempty = pure mempty
mappend = liftA2 mappend
using the Applicative instance for (->) r or with
instance Monoid a => Monoid (r -> a) where
mempty = return mempty
mappend = liftM2 mappend
using the Monad instance for (->) r.
Here the savings are minimal, but, for instance the #pl tool for generating point-free code, which is provided by lambdabot on the #haskell IRC channel abuses these instances quite a bit.