Function composition in Haskell
(This post is intended for 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
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, two-argument 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
addone, we might want to try writing a function like
both that composes
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
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.
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
(.), 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
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:
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
Int -> Int. From the type signature of
(.)we read that the first argument to
(.)should have type
b -> cfor certain
c; here we get that
- The second argument to
addone, again of type
Int -> Int. This means that
a -> b(which, from the previous point, we know to be
a -> Int) is really
Int -> Int, so
We don't apply any more arguments to
(.), so the return type is
a -> c; since we know that both
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
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 is
timestwo :: Int -> Int, which should match
b -> c. So:
Second argument to
(.): given is
plus :: Int -> Int -> Int, which should match
a -> b; since we already know that
Int, this is
a -> Int. So
Int -> Int -> Int, which is the same as
Int -> (Int -> Int), should match
a -> Intfor some
Int -> (Int -> Int) a -> Int
But this can never be true! We can make
Int, sure, but
Int -> Intwill never match
Int. 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 right-associative: its parentheses collect to the right. This is exactly like you may have learned that subtraction is left-associative (
1 - 2 - 3 = (1 - 2) - 3) and exponentiation is right-associative (
1 ^ 2 ^ 3 = 1 ^ (2 ^ 3)= 123).
- Multi-argument functions in Haskell aren't multi-argument 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.
g to the result, and returns the result of that.
(.) 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
timestwo . plus.
plus really takes only a single argument (an
Integer), 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
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.
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
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
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.