Difference between Monad and Applicative in Haskell

Question

I just read the following from typeclassopedia about the difference between Monad and Applicative. I can understand that there is no join

Answer 1

My favorite example is the "purely applicative Either". We'll start by analyzing the base Monad instance for Either

instance Monad (Either e) where
  return = Right
  Left e  >>= _ = Left e
  Right a >>= f = f a

This instance embeds a very natural short-circuiting notion: we proceed from left to right and once a single computation "fails" into the Left then all the rest do as well. There's also the natural Applicative instance that any Monad has

instance Applicative (Either e) where
  pure  = return
  (<*>) = ap

where ap is nothing more than left-to-right sequencing before a return:

ap :: Monad m => m (a -> b) -> m a -> m b
ap mf ma = do 
  f <- mf
  a <- ma
  return (f a)

---

Now the trouble with this Either instance comes to light when you'd like to collect error messages which occur anywhere in a computation and somehow produce a summary of errors. This flies in the face of short-circuiting. It also flies in the face of the type of (>>=)

(>>=) :: m a -> (a -> m b) -> m b

If we think of m a as "the past" and m b as "the future" then (>>=) produces the future from the past so long as it can run the "stepper" (a -> m b). This "stepper" demands that the value of a really exists in the future... and this is impossible for Either. Therefore (>>=) demands short-circuiting.

So instead we'll implement an Applicative instance which cannot have a corresponding Monad.

instance Monoid e => Applicative (Either e) where
  pure = Right

Now the implementation of (<*>) is the special part worth considering carefully. It performs some amount of "short-circuiting" in its first 3 cases, but does something interesting in the fourth.

  Right f <*> Right a = Right (f a)     -- neutral
  Left  e <*> Right _ = Left e          -- short-circuit
  Right _ <*> Left  e = Left e          -- short-circuit
  Left e1 <*> Left e2 = Left (e1 <> e2) -- combine!

Notice again that if we think of the left argument as "the past" and the right argument as "the future" then (<*>) is special compared to (>>=) as it's allowed to "open up" the future and the past in parallel instead of necessarily needing results from "the past" in order to compute "the future".

This means, directly, that we can use our purely Applicative Either to collect errors, ignoring Rights if any Lefts exist in the chain

> Right (+1) <*> Left [1] <*> Left [2]
> Left [1,2]

---

So let's flip this intuition on its head. What can we not do with a purely applicative Either? Well, since its operation depends upon examining the future prior to running the past, we must be able to determine the structure of the future without depending upon values in the past. In other words, we cannot write

ifA :: Applicative f => f Bool -> f a -> f a -> f a

which satisfies the following equations

ifA (pure True)  t e == t
ifA (pure False) t e == e

while we can write ifM

ifM :: Monad m => m Bool -> m a -> m a -> m a
ifM mbool th el = do
  bool <- mbool
  if bool then th else el

such that

ifM (return True)  t e == t
ifM (return False) t e == e

This impossibility arises because ifA embodies exactly the idea of the result computation depending upon the values embedded in the argument computations.

Answer 2

Here is my take on @J. Abrahamson's example as to why ifA cannot use the value inside e.g. (pure True). In essence, it still boils down to the absence of the join function from Monad in Applicative, which unifies the two different perspectives given in typeclassopedia to explain the difference between Monad and Applicative.

So using @J. Abrahamson's example of purely applicative Either:

instance Monoid e => Applicative (Either e) where
  pure = Right

  Right f <*> Right a = Right (f a)     -- neutral
  Left  e <*> Right _ = Left e          -- short-circuit
  Right _ <*> Left  e = Left e          -- short-circuit
  Left e1 <*> Left e2 = Left (e1 <> e2) -- combine!

(which has similar short-circuiting effect to the Either Monad), and the ifA function

ifA :: Applicative f => f Bool -> f a -> f a -> f a

What if we try to achieve the mentioned equations:

ifA (pure True)  t e == t
ifA (pure False) t e == e

?

Well, as already pointed out, ultimately, the content of (pure True), cannot be used by a later computation. But technically speaking, this isn't right. We can use the content of (pure True) since a Monad is also a Functor with fmap. We can do:

ifA' b t e = fmap (\x -> if x then t else e) b

The problem is with the return type of ifA', which is f (f a). In Applicative, there is no way of collapsing two nested ApplicativeS into one. But this collapsing function is precisely what join in Monad performs. So,

ifA = join . ifA' 

will satisfy the equations for ifA, if we can implement join appropriately. What Applicative is missing here is exactly the join function. In other words, we can somehow use the result from the previous result in Applicative. But doing so in an Applicative framework will involve augmenting the type of the return value to a nested applicative value, which we have no means to bring back to a single-level applicative value. This will be a serious problem because, e.g., we cannot compose functions using ApplicativeS appropriately. Using join fixes the issue, but the very introduction of join promotes the Applicative to a Monad.

Answer 3

I would like to share my view on this "iffy miffy" thing, as I understand this everything inside the context get applied, so for example:

iffy :: Applicative f => f Bool -> f a -> f a -> f a
iffy fb ft fe = cond <$> fb <*> ft <*> fe   where
            cond b t e = if b then t else e

case 1>> iffy (Just True) (Just “True”) Nothing ->> Nothing

upps should be Just "True" ... but

 case 2>> iffy (Just False) (Just “True”) (Just "False") ->> Just "False" 

(the "good" choice is made inside the context) I explained this to myself this way, just before the end of the computation in case >>1 we get something like that in the "chain" :

Just (Cond True "True") <*> something [something being "accidentaly" Nothing]

which according by definition of Applicative is evaluated as:

fmap (Cond True "True") something 

which when "something" is Nothing becomes a Nothing according to Functor constraint (fmap over Nothing gives Nothing). And it is not possible to define a Functor with "fmap f Nothing = something" end of story.

Answer 4

But the following description looks vague to me and I couldn't figure out what exactly is meant by "the result" of a monadic computation/action.

Well, that vagueness is somewhat deliberate, because what "the result" is of a monadic computation is something that depends on each type. The best answer is a bit tautological: the "result" (or results, since there can be more than one) is whatever value(s) the instance's implementation of (>>=) :: Monad m => m a -> (a -> m b) -> m b invokes the function argument with.

So, if I put a value into Maybe, which makes a monad, what is the result of this "computation"?

The Maybe monad looks like this:

instance Monad Maybe where
    return = Just
    Nothing >>= _ = Nothing
    Just a >>= k = k a

The only thing in here that qualifies as a "result" is the a in the second equation for >>=, because it's the only thing that ever gets "fed" to the second argument of >>=.

Other answers have gone into depth about the ifA vs. ifM difference, so I thought I'd highlight another significant difference: applicatives compose, monads don't. With Monads, if you want to make a Monad that combines the effects of two existing ones, you have to rewrite one of them as a monad transformer. In contrast, if you have two Applicatives you can easily make a more complex one out of them, as shown below. (Code is copypasted from transformers.)

-- | The composition of two functors.
newtype Compose f g a = Compose { getCompose :: f (g a) }

-- | The composition of two functors is also a functor.
instance (Functor f, Functor g) => Functor (Compose f g) where
    fmap f (Compose x) = Compose (fmap (fmap f) x)

-- | The composition of two applicatives is also an applicative.
instance (Applicative f, Applicative g) => Applicative (Compose f g) where
    pure x = Compose (pure (pure x))
    Compose f <*> Compose x = Compose ((<*>) <$> f <*> x)


-- | The product of two functors.
data Product f g a = Pair (f a) (g a)

-- | The product of two functors is also a functor.
instance (Functor f, Functor g) => Functor (Product f g) where
    fmap f (Pair x y) = Pair (fmap f x) (fmap f y)

-- | The product of two applicatives is also an applicative.
instance (Applicative f, Applicative g) => Applicative (Product f g) where
    pure x = Pair (pure x) (pure x)
    Pair f g <*> Pair x y = Pair (f <*> x) (g <*> y)


-- | The sum of a functor @f@ with the 'Identity' functor
data Lift f a = Pure a | Other (f a)

-- | The sum of two functors is always a functor.
instance (Functor f) => Functor (Lift f) where
    fmap f (Pure x) = Pure (f x)
    fmap f (Other y) = Other (fmap f y)

-- | The sum of any applicative with 'Identity' is also an applicative 
instance (Applicative f) => Applicative (Lift f) where
    pure = Pure
    Pure f <*> Pure x = Pure (f x)
    Pure f <*> Other y = Other (f <$> y)
    Other f <*> Pure x = Other (($ x) <$> f)
    Other f <*> Other y = Other (f <*> y)

Now, if we add in the Constant functor/applicative:

newtype Constant a b = Constant { getConstant :: a }

instance Functor (Constant a) where
    fmap f (Constant x) = Constant x

instance (Monoid a) => Applicative (Constant a) where
    pure _ = Constant mempty
    Constant x <*> Constant y = Constant (x `mappend` y)

...we can assemble the "applicative Either" from the other responses out of Lift and Constant:

type Error e a = Lift (Constant e) a

Answer 5

Just 1 describes a "computation", whose "result" is 1. Nothing describes a computation which produces no results.

The difference between a Monad and an Applicative is that in the Monad there's a choice. The key distinction of Monads is the ability to choose between different paths in computation (not just break out early). Depending on a value produced by a previous step in computation, the rest of computation structure can change.

Here's what this means. In the monadic chain

return 42            >>= (\x ->
if x == 1
   then
        return (x+1) 
   else 
        return (x-1) >>= (\y -> 
        return (1/y)     ))

the if chooses what computation to construct.

In case of Applicative, in

pure (1/) <*> ( pure (+(-1)) <*> pure 1 )

all the functions work "inside" computations, there's no chance to break up a chain. Each function just transforms a value it's fed. The "shape" of the computation structure is entirely "on the outside" from the functions' point of view.

A function could return a special value to indicate failure, but it can't cause next steps in the computation to be skipped. They all will have to process the special value in a special way too. The shape of the computation can not be changed according to received value.

With monads, the functions themselves construct computations to their choosing.

Answer 6

The key of the difference can be observed in the type of ap vs type of =<<.

ap :: m (a->b) -> (m a->m b)
=<< :: (a->m b) -> (m a->m b)

In both cases there is m a, but only in the second case m a can decide whether the function (a->m b) gets applied. In its turn, the function (a->m b) can "decide" whether the function bound next gets applied - by producing such m b that does not "contain" b (like [], Nothing or Left).

In Applicative there is no way for functions "inside" m (a->b) to make such "decisions" - they always produce a value of type b.

f 1 = Nothing -- here f "decides" to produce Nothing
f x = Just x

Just 1 >>= f >>= g -- g doesn't get applied, because f decided so.

In Applicative this is not possible, so can't show a example. The closest is:

f 1 = 0
f x = x

g <$> f <$> Just 1 -- oh well, this will produce Just 0, but can't stop g
                   -- from getting applied