Function composition in Haskell
(This post is intended for Haskell beginners.)
In Haskell, the dot operator (.)
, written infix like f . g
, is function composition.
For example, suppose you have two functions:
addone :: Int > Int
addone n = n + 1
timestwo :: Int > Int
timestwo n = n * 2
then you can compose those two functions like timestwo . addone
:
 computes 2 * (n + 1), which is 2 * n + 2
both :: Int > Int
both = timestwo . addone
And we can try it out in ghci
:
Prelude> both 10
22
Prelude> timestwo (addone 10)
22
Prelude> 2 * (10 + 1)
22
This works well, like we expect. These functions are intentionally very simple; the point of this post is to explain how function composition works, not how to write complex Haskell functions.
Now suppose we have an additional, twoargument function to play with:
plus :: Int > Int > Int
plus a b = a + b
If we give plus
some arguments, we can apply timestwo
to the result:
Prelude> timestwo (plus 3 4)  2 * (3 + 4) = 14
14
Getting inspiration from the first example with timestwo
and addone
, we might want to try writing a function like both
that composes timestwo
and plus
using (.)
.
both2 = timestwo . plus
But the compiler won't accept this:
• Couldn't match type ‘Int > Int’ with ‘Int’
Expected type: Int > Int
Actual type: Int > Int > Int
• Probable cause: ‘plus’ is applied to too few arguments
In the second argument of ‘(.)’, namely ‘plus’
In the expression: timestwo . plus
In an equation for ‘both2’: both2 = timestwo . plus

11  both2 = timestwo . plus
 ^^^^
Why is this case different than the first example with both
?
The difference is that plus
takes two arguments, not one.
As always in Haskell, let's take a look at the types to see what's going on here.
Types
What are the types of the functions we're using?
addone :: Int > Int
timestwo :: Int > Int
plus :: Int > Int > Int
(.) :: (b > c) > (a > b) > (a > c)
Remember that you can look up the type of standard library functions online: e.g. here for (.)
.
Function composition, (.)
, takes a function from type b
to type c
and a function from type a
to type b
, and returns a function from type a
to type c
.
Note that we can also write the type of (.)
slightly differently, without the final pair of parentheses:
(.) :: (b > c) > (a > b) > (a > c)
(.) :: (b > c) > (a > b) > a > c
These both mean the exact same thing.
When at first, we wrote timestwo . addone
as the definition of both
, we were really applying (.)
to two arguments: timestwo
and addone
.
Indeed, we could also have written this:
both :: Int > Int
both = (.) timestwo addone
which means exactly the same thing as both = timestwo . addone
.
Let's do some of the compiler's work here and figure out how the types match up. (If you want to learn more about how you can do type inference by hand, one guide is this one.)
 The first argument to
(.)
istimestwo
of typeInt > Int
. From the type signature of(.)
we read that the first argument to(.)
should have typeb > c
for certainb
andc
; here we get thatb
andc
are bothInt
.  The second argument to
(.)
isaddone
, again of typeInt > Int
. This means thata > b
(which, from the previous point, we know to bea > Int
) is reallyInt > Int
, soa
is alsoInt
.
We don't apply any more arguments to (.)
, so the return type is a > c
; since we know that both a
and c
are Int
, this is Int > Int
, so both
indeed has the right type signature.
Now let's try to do the same thing with our second attempt in both2
, namely timestwo . plus
, which is the same as (.) timestwo plus
.
Remember the types:
timestwo :: Int > Int
plus :: Int > Int > Int
(.) :: (b > c) > (a > b) > (a > c)

First argument to
(.)
: given istimestwo :: Int > Int
, which should matchb > c
. So:b
andc
are bothInt
. 
Second argument to
(.)
: given isplus :: Int > Int > Int
, which should matcha > b
; since we already know thatb
isInt
, this isa > Int
. SoInt > Int > Int
, which is the same asInt > (Int > Int)
, should matcha > Int
for somea
:Int > (Int > Int) a > Int
But this can never be true! We can make
a
equal toInt
, sure, butInt > Int
will never matchInt
. Remember the compiler error we got earlier:• Couldn't match type ‘Int > Int’ with ‘Int’ Expected type: ...
Suspiciously similar, isn't it?
Seen it coming
Could we have seen this coming? Of course. Remember that the following Haskell types mean exactly the same:
a > b > c > d > e
a > (b > (c > (d > e)))
This can be explained (or remembered) in two ways:
>
is rightassociative: its parentheses collect to the right. This is exactly like you may have learned that subtraction is leftassociative (1  2  3 = (1  2)  3
) and exponentiation is rightassociative (1 ^ 2 ^ 3 = 1 ^ (2 ^ 3)
= 1^{23}). Multiargument functions in Haskell aren't multiargument functions; instead, they're functions that take one argument and return a function taking the rest.
The second interpretation is the one that you should use, because it is by far the most useful one to really understand what is going on in Haskell with functions  and since Haskell is a functional programming language, (almost) everything is functions, so understanding them is important!
Now look again at the type of (.)
(all three mean the same):
(.) :: (b > c) > (a > b) > a > c
(.) :: (b > c) > (a > b) > (a > c)
(.) :: (b > c) > ((a > b) > (a > c))
(Why can't we remove the parentheses around the b > c
and a > b
in there? That is because it the type would then suddenly mean something different. Try to work that out for yourself!)
We can read this as: (.)
takes a function g :: b > c
, a function f :: a > b
and a value x :: a
.
It applies f
to x
, applies g
to the result, and returns the result of that.
Indeed, (.)
can be implemented like this:
(.) :: (b > c) > (a > b) > a > c
(.) g f x = g (f x)
Finally, now look again at our failed attempt to write both2
as timestwo . plus
.
plus
really takes only a single argument (an Int
eger), and it returns a function that takes a second integer and returns the sum of those two integers.
So the value f x
in the definition of (.)
is a function in the case of both2
!
And timestwo
takes an Int
as input, not a function.
So this is not going to work, as we've seen in a different way above.
Conclusion
We figured out why timestwo . plus
doesn't work in two different ways: by following what the compiler does, and by reasoning on a somewhat more abstract level about what (.)
does.
Both arrive at the same conclusion, which is always good: if you have two different ways of deriving something and they give the same result, then the chances are already smaller that both are wrong.
Is there a way to fix the error, and write the composition of timestwo
and plus
somehow?
Yes; there are multiple ways, in fact.
In this case, the best one is probably to just write it out:
both2 a b = timestwo (plus a b)
Arguably, this makes it the clearest what's going on.
If you really want a pointfree (also jokingly called "pointless") version that doesn't mention any variable names, try one of the following:
both2 = (timestwo .) . plus  same as: (\f > timestwo . f) . plus
both2 = curry (timestwo . uncurry plus)
But that's only for fun. Don't actually do that.