从应用（<*>）混淆中导出monad绑定

Question

Working through the Haskell Book and my brain is breaking on the following example. I really don't know what the flip function is doing on line 21

1 class Functor f where
2   fmap :: (a -> b) -> f a -> f b
3
4 class Functor f => Applicative f where
5   pure :: a -> f a
6   (<*>) :: f (a -> b) -> f a -> f b
7
8 class Applicative f => Monad f where
9   return :: a -> f a
10   (>>=) :: f a -> (a -> f b) -> f b
11
12 instance Functor ((->) r) where
13   fmap = (.)
14
15 instance Applicative ((->) r) where
16   pure = const
17   f <*> a = \r -> f r (a r)
18
19 instance Monad ((->) r ) where
20   return = pure
21   ra >>= arb = flip arb <*> ra


-- flip :: (a -> b -> c) -> b -> a -> c

-- ra >>= arb = flip arb <*> ra

As I understand it, flip takes a function that takes two arguments, then two individual arguments, and returns a value. In the bind example, are we passing arb as the (a -> b -> c) , then <*> as the b in flip's signature, and finally ra as the a ? I can't see it.

I've tried making the types more specific to my actual example so you could rewrite <*> as

(<*>) :: (r -> a -> b) -> (r -> a) -> (r -> b)

and I can do the same for bind

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

so even a dummy like myself can see that if we could swap <*> around we could line then up like

(<???>) :: (r -> a) -> (r -> a -> b) -> (r -> b)
(>>=) ::   (r -> a) -> (a -> r -> b) -> (r -> b)

but looking at the second arguments there, the first one wants an r as its first argument and bind wants an a

So somehow flip is the book's example is doing that for us, but I really don't understand how. Any help would be greatly appreciated.

Thank you!

Answer 1

Top-level point of confusion, I think: flip is modifying arb , not modifying <*> as you seem to believe. We have "modified" <*> to have the "right" argument order just by giving <*> its arguments in the opposite order we got them!

Now for the details. We have, as you noted:

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

So, since we have on the left hand side written

ra >>= arb = ...

then what we have in scope is:

ra :: r -> a
arb :: a -> r -> b

(Note how the names were chosen to reflect the types!) Rewriting the type you gave for flip , we have

flip :: (a -> b -> c) -> b -> a -> c -- original
flip :: (a -> r -> b) -> r -> a -> b -- rename variables

hence:

flip arb :: r -> a -> b

Now recall the type of (<*>) that you wrote:

(<*>) :: (r -> a -> b) -> (r -> a) -> (r -> b)

So for the first argument to (<*>) , we want something of type r -> a -> b . Hey! flip arb has that type! For the second argument, we want something of type r -> a . And we're in luck again, because ra has that type, so...

flip arb <*> ra :: r -> b

(As usual with infix operators, this is the application of the operator (<*>) the arguments flip arb and ra .) What type were we hoping to have? Well, we go back to the type of (>>=) now:

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

After taking two arguments, this is supposed to return something of type r -> b . Cool, that's what we have built.

ra >>= arb = flip arb <*> ra