# Passing Ord Function as a Function Parameter

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 = ...
``````

### >Solution :

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`.)