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.
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]
Well, actually you can write that. “take 3 $ reverse $ filter even $ [1..10]” is a valid Haskell code.
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
How does Haskell know to pass the function and parameter to $ rather than pass $ as a parameter to the function?
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 $ barWill pass
baras an argument tofoo, whereas:will pass
($)andbaras arguments tofoo.“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
It’s not really about “evaluating”, just type checking.
Thanks for this post, very nice example with $.