Bind operator for non-monadic functions

Question

I more or less wrapped my head around monads, but i can't deduct how expression

(>>=) id (+) 3

evaluates to 6. It just seems the expression somehow got simplified to

(+) 3 3

but how? How is 3 applied twice? Could someone explain what is happening behind the scenes ?

• chi: The key is to realize that (+) :: a -> m a where m ~ (->) a which is a monad. (The fact that we need Num a for (+) does not change this.) Also, id :: m a for the same monad. So, we can have id >>= (+) :: m a, which is a function a -> a and can be applied to 3.

Answer 1

This follows from how >>= is defined for the ((->) r) types:

(f =<< g) x = f (g x) x

Thus

(>>=) id (+) 3
=
(id >>= (+)) 3
=
((+) =<< id) 3
=
(+) (id 3) 3
=
3 + 3

see the types:

> :t let (f =<< g) x = f (g x) x in (=<<)
let (f =<< g) x = f (g x) x in (=<<)
 :: (t1 -> (t2 -> t)) -> (t2 -> t1) -> (t2 -> t)
> :t (=<<)
(=<<) :: Monad m => (a -> m b) -> m a -> m b

The types match with

t1 ~ a
(t2 ->) ~ m -- this is actually ((->) t2)`
t ~ b

Thus the constraint Monad m here means Monad ((->) t2), and that defines the definition of =<< and >>= which get used.

If you want to deduce the definition from the type,

(>>=) :: Monad m => m a -> (a -> m b) -> m b
m ~ ((->) r)
(>>=) :: (r -> a) -> (a -> r -> b) -> (r -> b)
(>>=) f g r = b
 where
 a = f r
 rb = g a
 b = rb r

which after the simplification becomes the one we used above.

And if you want to understand it "with words",

(=<<) :: (Monad m, m ~ ((->) r)) => (a -> m b) -> m a -> m b
(f =<< g) x = f (g x) x

• g is a "monadic value" that "can compute" an "a", represented as r -> a
• f a calculates a "monadic value" that "can compute" a "b", represented as r -> b,
• thus \x -> f (g x) x is a monadic value that "can compute" a "b", given an "r".

So these "non-monadic functions" are, in fact, monadic values, which happen to be functions.

Thus in your example, g = id, f = (+), and

• id is a "monadic value" that "can compute" an "a", an a -> a
• (+) a calculates a "monadic value" that "can compute" a "b", an a -> b, which b is actually also an a,
• thus \x -> (+) (id x) x is a monadic value that "can compute" an "a", given an "a":

(>>=) id (+)
=
((+) =<< id) 
=
\x -> (+) (id x) x
=
\x -> (+) x x

• shree.pat18: Not sure you need the intermediate translation to =<<, but perhaps I am missing something?
• Will Ness: @shree.pat18 that's the definition that I remember. :) its type (=<<) :: Monad m => (a -> m b) -> m a -> m b also aligns nicely with the other types, of (<$>) and (<*>), all ending with the m a -> m b.
• pharmaz0ne: Thanks, great exhaustive answer, everything makes sense now. Also thanks to @shree.pat18 and chi. I would like to use the chance to ask about type abbreviations, e.g. what r in (->) r stands for and why isn't it t1 or a . What is the difference between t1, a, r etc... I cannot find anything on the web about it.
• shree.pat18: You might find this useful, although I too cant find an authoritative document: kowainik.github.io/posts/naming-conventions
• Will Ness: they are just mnemonics of course. r on the right of -> is usually for "result", here on the left of -> I use it as a reminder of "reader" although of course the whole type is a "reader" from an "environment" so it probably should have been e instead (but e is also often an "error" as in e.g. Either e a). a, t, etc are for the more generic types.

Answer 2

Adding some colour to Will's excellent answer.

If we look at the source, we have this:

instance Monad ((->) r) where
 f >>= k = \ r -> k (f r) r

If we rearrange the input expression slightly, we get (id >>= (+)) 3. This now resembles the form shown above. Now fitting the input into the above 'template', we can rewrite the input as \ r -> (+) (id r) r

This is the same expression we arrived at for the final evaluation i.e. (+) (id 3) 3