Haskell confuses Int with Int -> Int

Question

I recently came across this weird issue in Haskell. The following code is supposed to return a value trimmed down to a range (if it's above high it should return high if it's under low it should return low.

inRange :: Int -> Int -> Int -> Int
inRange low high = max low $ min high

the error message is:

scratch.hs:2:20:
    Couldn't match expected type ‘Int -> Int’ with actual type ‘Int’
    In the expression: max low $ min high
    In an equation for ‘inRange’: inRange low high = max low $ min high

scratch.hs:2:30:
    Couldn't match expected type ‘Int’ with actual type ‘Int -> Int’
    Probable cause: ‘min’ is applied to too few arguments
    In the second argument of ‘($)’, namely ‘min high’
    In the expression: max low $ min high

Shouldn't it take another argument and put it into high? I already tried other possibilities like:

\x -> max low $ min high x

and

\x -> max low $ (min high x)

When trying it in GHCI I'm getting the following error:

<interactive>:7:5:
    Non type-variable argument in the constraint: Num (a -> a)
    (Use FlexibleContexts to permit this)
    When checking that ‘inRange’ has the inferred type
      inRange :: forall a.
                 (Num a, Num (a -> a), Ord a, Ord (a -> a)) =>
                 a -> a

Answer 1

($) is defined as:

f $ x = f x

So your example actually is:

max low (min high)

which is wrong, because you actually want

max low (min high x)

Using function composition, which is defined as:

f . g = \x -> f (g x)

and your working example \x -> max low (min high x) we get:

\x -> max low (min high x)
== max low . min high -- by definition of (.)