## 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 style^{1}. 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-associative^{2}. 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 $ bar`

Will pass

`bar`

as an argument to`foo`

, whereas:will pass

`($)`

and`bar`

as arguments to`foo`

.“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 $.