Haskell: Function application with $

Question

In the following snippet, you can see my two collatz functions I wrote in Haskell. For the recursive application I used parentheses in the first example (collatz) to get the right precedence.

As I have just learnt function application with $, I tried to rewrite the function (collatz') using that thing. However, I encounter the following error:

Couldn't match expected type `[a]' against inferred type `a1 -> [a1]' In the second argument of `(:)', namely `collatz'' In the first argument of `($)', namely `n : collatz'' In the expression: n : collatz' $ n `div` 2

collatz :: (Integral a) => a -> [a]

collatz 1 = [1]

collatz n | even n    = n : collatz (n `div` 2)
          | otherwise = n : collatz (n * 3 + 1)

collatz' :: (Integral a) => a -> [a]

collatz' 1 = [1]

collatz' n | even n    = n : collatz' $ n `div` 2
           | otherwise = n : collatz' $ n * 3 + 1

It seamed weird to me that this didn't work. So I tried a similar example that worked:

True : [even $ 3 `div` 3]

I'd appreciate it, if somebody could take a look at it and tell me what I'm doing wrong.

Answer 1

$ has lower precedence then : (and also anything else) so your function is parsing as

(n : collatz') $ (n `div` 2)

This leads to your type error. The second argument of : expects a list but you are passing the collatz function instead.

If you still want to avoid the parenthesis around the 3n+1 part you can do something like the following

(n:) . collatz' $ n `div` 2
n : (collatz' $ n `div` 2)

although these are not necessarily cleaner then the original. In case you are wondering, the (n:) in the first example is a syntactic sugar for \x -> n : x