Question

I got this function:

foo [] = []
foo (x:xs) = foo us ++ foo ys
where us = filter (<=x) xs
      ys = filter (>=x) xs

type of this function is Ord a => [a] -> [b] .

I don’t understand why the output type is [b] and not [a]. I think it should be [a] since the elements of the output list will be part of the elements of the input list.

I am using Hugs, but I don’t think it changes anything.

Answer 1

The type Ord a => [a] -> [b] is internally consistent, though!

The problem is that you never actually add any elements from the input list to the output list. You need a base case; something like foo [x] = [x]. As it stands, you never actually say that any elements from the input list get added to the output list; the result of this function will always be [], which can have type [b] regardless of input.

If you’re trying to implement something like Quicksort here, though, there are two logical problems in your implementation:

1. x, the pivot, doesn’t get added to the output list.
2. Any values in the list that are equal to x other than x itself will be added twice, once from us and once from ys.