Passing Ord Function as a Function Parameter

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Hello I am learning Haskell at the moment and I was wondering if and how I can pass a function of Eq as a parameter to another function since it works with (+), (-) and (*) (and I know passing functions in general works in haskell). I want to work with the Eq operator in my bar function depending on if the head of my list is smaller, equal or bigger than 0. But the way I did it in my code snippet below I get a compiler error. Sorry if I am missing something obvious here I am pretty new to haskell, your help and explanations are appreciated.

foo :: [x] -> [x]
foo x
  | head x < 0  = bar (<) x
  | head x == 0 = bar (==) x
  | head x > 0  = bar (>) x

bar :: (a -> a -> Bool) -> [x] -> [x]
bar func x = ...
haskell
1 Answers

You need to add an Ord constraint to bar, since you can't use values of any type with (<) et al, only those with an Ord instance. You also need to use a throughout the type; using x implies that the type of list passed is independent of the Ord constraint, which would prevent you from using func on an element of the argument x.

 bar :: Ord a => (a -> a -> Bool) -> [a] -> [a]
 bar func x = ...

Since Eq is a superclass of Ord, you don't need to specify Eq explicitly; an instance of Ord implies an instance of Eq.

You need the constraint on foo as well, since you use <, ==, and > in the guards, not just as arguments to bar. To emphasize exhaustivity, try using compare instead of three separate guards

foo :: Ord a => [a] -> [a]
foo x = let f = case compare (head x) 0 of
                  LT -> (<)
                  EQ -> (==)
                  GT -> (>)
        in bar f x

(And consider what, if anything, needs to be done about an empty list argument to foo.)

user_11268410
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