What advantage does Monad give us over an Applicative?

Question

I\'ve read this article, but didn\'t understand last section.

The author says that Monad gives us context sensitivity, but it\'s possible to achieve the same result

Answer 1

If you try to convert the type signature of Monad's bind and Applicative <*> to natural language, you will find that:

bind : I will give you the contained value and you will return me a new packaged value

<*>: You give me a packaged function that accepts a contained value and return a value and I will use it to create new packaged value based on my rules.

Now as you can see from the above description, bind gives you more control as compared to <*>

Answer 2

With monads, subsequent effects can depend on previous values. For example, you can have:

main = do
    b <- readLn :: IO Bool
    if b
      then fireMissiles
      else return ()

You can't do that with Applicatives - the result value of one effectfull computation can't determine what effect will follow.

Somewhat related:

• Why can applicative functors have side effects, but functors can't?
• Good examples of Not a Functor/Functor/Applicative/Monad?

Answer 3

Here's a couple of functions that use the Monad interface.

ifM :: Monad m => m Bool -> m a -> m a -> m a
ifM c x y = c >>= \z -> if z then x else y

whileM :: Monad m => (a -> m Bool) -> (a -> m a) -> a -> m a
whileM p step x = ifM (p x) (step x >>= whileM p step) (return x)

You can't implement them with the Applicative interface. But for the sake of enlightenment, let's try and see where things go wrong. How about..

import Control.Applicative

ifA :: Applicative f => f Bool -> f a -> f a -> f a
ifA c x y = (\c' x' y' -> if c' then x' else y') <$> c <*> x <*> y

Looks good! It has the right type, it must be the same thing! Let's just check to make sure..

*Main> ifM (Just True) (Just 1) (Just 2)
Just 1
*Main> ifM (Just True) (Just 1) (Nothing)
Just 1
*Main> ifA (Just True) (Just 1) (Just 2)
Just 1
*Main> ifA (Just True) (Just 1) (Nothing)
Nothing

And there's your first hint at the difference. You can't write a function using just the Applicative interface that replicates ifM.

If you divide this up into thinking about values of the form f a as being about "effects" and "results" (both of which are very fuzzy approximate terms that are the best terms available, but not very good), you can improve your understanding here. In the case of values of type Maybe a, the "effect" is success or failure, as a computation. The "result" is a value of type a that might be present when the computation completes. (The meanings of these terms depends heavily on the concrete type, so don't think this is a valid description of anything other than Maybe as a type.)

Given that setting, we can look at the difference in a bit more depth. The Applicative interface allows the "result" control flow to be dynamic, but it requires the "effect" control flow to be static. If your expression involves 3 computations that can fail, the failure of any one of them causes the failure of the whole computation. The Monad interface is more flexible. It allows the "effect" control flow to depend on the "result" values. ifM chooses which argument's "effects" to include in its own "effects" based on its first argument. This is the huge fundamental difference between ifA and ifM.

There's something even more serious going on with whileM. Let's try to make whileA and see what happens.

whileA :: Applicative f => (a -> f Bool) -> (a -> f a) -> a -> f a
whileA p step x = ifA (p x) (whileA p step <*> step x) (pure x)

Well.. What happens is a compile error. (<*>) doesn't have the right type there. whileA p step has the type a -> f a and step x has the type f a. (<*>) isn't the right shape to fit them together. For it to work, the function type would need to be f (a -> a).

You can try lots more things - but you'll eventually find that whileA has no implementation that works anything even close to the way whileM does. I mean, you can implement the type, but there's just no way to make it both loop and terminate.

Making it work requires either join or (>>=). (Well, or one of the many equivalents of one of those) And those the extra things you get out of the Monad interface.

Answer 4

If you work with Applicatives, the "shape" of the result is already determined by the "shape" of the input, e.g. if you call [f,g,h] <*> [a,b,c,d,e], your result will be a list of 15 elements, regardless which values the variables have. You don't have this guarantee/limitation with monads. Consider [x,y,z] >>= join replicate: For [0,0,0] you'll get the result [], for [1,2,3] the result [1,2,2,3,3,3].

Answer 5

With Applicative, the sequence of effectful actions to be performed is fixed at compile-time. With Monad, it can be varied at run-time based on the results of effects.

For example, with an Applicative parser, the sequence of parsing actions is fixed for all time. That means that you can potentially perform "optimisations" on it. On the other hand, I can write a Monadic parser which parses some a BNF grammar description, dynamically constructs a parser for that grammar, and then runs that parser over the rest of the input. Every time you run this parser, it potentially constructs a brand new parser to parse the second portion of the input. Applicative has no hope of doing such a thing - and there is no chance of performing compile-time optimisations on a parser that doesn't exist yet...

As you can see, sometimes the "limitation" of Applicative is actually beneficial - and sometimes the extra power offered by Monad is required to get the job done. This is why we have both.

Answer 6

As Stephen Tetley said in a comment, that example doesn't actually use context-sensitivity. One way to think about context-sensitivity is that it lets use choose which actions to take depending on monadic values. Applicative computations must always have the same "shape", in a certain sense, regardless of the values involved; monadic computations need not. I personally think this is easier to understand with a concrete example, so let's look at one. Here's two versions of a simple program which ask you to enter a password, check that you entered the right one, and print out a response depending on whether or not you did.

import Control.Applicative

checkPasswordM :: IO ()
checkPasswordM = do putStrLn "What's the password?"
                    pass <- getLine
                    if pass == "swordfish"
                      then putStrLn "Correct.  The secret answer is 42."
                      else putStrLn "INTRUDER ALERT!  INTRUDER ALERT!"

checkPasswordA :: IO ()
checkPasswordA =   if' . (== "swordfish")
               <$> (putStrLn "What's the password?" *> getLine)
               <*> putStrLn "Correct.  The secret answer is 42."
               <*> putStrLn "INTRUDER ALERT!  INTRUDER ALERT!"

if' :: Bool -> a -> a -> a
if' True  t _ = t
if' False _ f = f

Let's load this into GHCi and check what happens with the monadic version:

*Main> checkPasswordM
What's the password?
swordfish
Correct.  The secret answer is 42.
*Main> checkPasswordM
What's the password?
zvbxrpl
INTRUDER ALERT!  INTRUDER ALERT!

So far, so good. But if we use the applicative version:

*Main> checkPasswordA
What's the password?
hunter2
Correct.  The secret answer is 42.
INTRUDER ALERT!  INTRUDER ALERT!

We entered the wrong password, but we still got the secret! And an intruder alert! This is because <$> and <*>, or equivalently liftAn/liftMn, always execute the effects of all their arguments. The applicative version translates, in do notation, to

do pass  <- putStrLn "What's the password?" *> getLine)
   unit1 <- putStrLn "Correct.  The secret answer is 42."
   unit2 <- putStrLn "INTRUDER ALERT!  INTRUDER ALERT!"
   pure $ if' (pass == "swordfish") unit1 unit2

And it should be clear why this has the wrong behavior. In fact, every use of applicative functors is equivalent to monadic code of the form

do val1 <- app1
   val2 <- app2
   ...
   valN <- appN
   pure $ f val1 val2 ... valN

(where some of the appI are allowed to be of the form pure xI). And equivalently, any monadic code in that form can be rewritten as

f <$> app1 <*> app2 <*> ... <*> appN

or equivalently as

liftAN f app1 app2 ... appN

To think about this, consider Applicative's methods:

pure  :: a -> f a
(<$>) :: (a -> b) -> f a -> f b
(<*>) :: f (a -> b) -> f a -> f b

And then consider what Monad adds:

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

(Remember that you only need one of those.)

Handwaving a lot, if you think about it, the only way we can put together the applicative functions is to construct chains of the form f <$> app1 <*> ... <*> appN, and possibly nest those chains (e.g., f <$> (g <$> x <*> y) <*> z). However, (=<<) (or (>>=)) allows us to take a value and produce different monadic computations depending on that value, that could be constructed on the fly. This is what we use to decide whether to compute "print out the secret", or compute "print out an intruder alert", and why we can't make that decision with applicative functors alone; none of the types for applicative functions allow you to consume a plain value.

You can think about join in concert with fmap in a similar way: as I mentioned in a comment, you can do something like

checkPasswordFn :: String -> IO ()
checkPasswordFn pass = if pass == "swordfish"
                         then putStrLn "Correct.  The secret answer is 42."
                         else putStrLn "INTRUDER ALERT!  INTRUDER ALERT!"

checkPasswordA' :: IO (IO ())
checkPasswordA' = checkPasswordFn <$> (putStrLn "What's the password?" *> getLine)

This is what happens when we want to pick a different computation depending on the value, but only have applicative functionality available us. We can pick two different computations to return, but they're wrapped inside the outer layer of the applicative functor. To actually use the computation we've picked, we need join:

checkPasswordM' :: IO ()
checkPasswordM' = join checkPasswordA'

And this does the same thing as the previous monadic version (as long as we import Control.Monad first, to get join):

*Main> checkPasswordM'
What's the password?
12345
INTRUDER ALERT!  INTRUDER ALERT!