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.

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