Function composition and $ operator in Haskell

Today I was trying to understand how function application, composition and $ work in Haskell. Consider two following lines of code:

take 3 (reverse (filter even [1..10]))
take 3 . reverse . filter even $ [1..10]

These are equivalent and both produce a list [10,8,6]. First version is based on grouping function calls with parentheses and this is pretty straightforward. My problem was the second version which is based on function composition and a special function application operator, denoted as . (dot) and $ respectively. This form is often used to write functions in a pointfree style1. I was trying to work out in what order function calls are executed and why do I have to use $ operator. Here’s what I came up with.

The function composition (dot) operator is defined as:

(.) :: (b -> c) -> (a -> b) -> a -> c
(f . g) x = f (g x)

This operator has priority of 9 and is right-associative2. This means that a . b . c . d is the same as (a . (b . (c . d))). The $ operator is defined as:

($) :: (a -> b) -> a -> b
f $ x = f x

This operator simply applies a function to a given parameter. In contrast to standard function application, which has highest possible priority of 10 and is left-associative, the $ operator has priority of 0 and is right-associative (that second property doesn’t matter in my example). Such a low priority means that all other operators on both sides of $ will be evaluated before applying the $. So the call

take 3 . reverse . filter even $ [1..10]

will first evaluate take 3 . reverse . filter even constructing a partially applied function that becomes fully applied when it receives one more parameter. The missing parameter is the list on the right side of $. By definition of dot operator, this call is therefore equivalent to take 3 (reverse (filter even [1..10])). That’s what we expected.

Why do we need the $ operator? If it wasn’t there then the function call filter even [1..10] would evaluate first – remember that function application has priority of 10, while function composition has a lower priority of 9. This would lead to take 3 . reverse . [2,4,6,8,10], but a list cannot be composed with a function. The dot operator expects it’s second argument to be a function of one argument, not a list, and that’s the reason we need $ – to allow function composition to evaluate first.

  1. The name “pointfree” doesn’t come from the point operator used to compose functions []
  2. You can verify this by typing :i (.) in ghci. This will display the type definition, fixity (infixr in that case) and priority of the dot operator []

7 Responses to “Function composition and $ operator in Haskell”

  1. Magnap says:

    Thank you for this. I was reading “learn you a haskell for great good” and was wondering why you couldn’t just write take 3 $ reverse $ filter even $ [1..10]

  2. Jan Stolarek says:

    Well, actually you can write that. “take 3 $ reverse $ filter even $ [1..10]” is a valid Haskell code.

  3. Rick Bradford says:

    I had trouble understanding function composition until I redefined it in terms of pipelining.

    Let’s write it :: [1..10] |> filter even |> reverse |> take 3

    I find that more readable, but it is equivalent to:

    take 3 . reverse . filter even $ [1..10]

    And, although Prelude doesn’t have pipelining built in, it can simply be defined:

    (|>) f g = g f

    Works like a charm, and makes my Haskell code much more readable

  4. Matt says:

    How does Haskell know to pass the function and parameter to $ rather than pass $ as a parameter to the function?

  5. Jan Stolarek says:

    To pass an operator as an argument to a function you have to put in parentheses, so that it becomes operator section with no arguments. So having:

    foo $ bar

    Will pass bar as an argument to foo, whereas:

    foo ($) bar

    will pass ($) and bar as arguments to foo.

  6. Fahad Aijaz says:

    “Thank you for this. I was reading “learn you a haskell for great good” and was wondering why you couldn’t just write take 3 $ reverse $ filter even $ [1..10]”.

    I would have to say exactly the same. Thanks

  7. Tom Ellis says:

    It’s not really about “evaluating”, just type checking.

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