Chaining max with . operator in Haskell

Question

I'm trying to use the . operator to write a function that takes three values and returns their maximum.

Obviously, the following work

max3 a b c = max a (max b c)
max3 a b c = max a $ max b c

but I would like to use .. I tried

max3 a b c = max a . max b c

but get the error

Couldn't match expected type `a0 -> Float' with actual type `Float'
In the first argument of `max', namely `b'

I know that the example is silly, but a nice explanation of the correct way to do this and why would be greatly appreciated.

One must remember that (.) isn't some built-in magic operator; it's a normal function defined as any other function would be. — AJF
– AJF, Commented Dec 23, 2016 at 9:25
@AJFarmar The issue (although I did not realize at the time of asking) was primarily that the error message was not helpful. I still don't understand why max should expect a0 -> Float. — Kittsil
– Kittsil, Commented Dec 23, 2016 at 13:11

Daniel Wagner · Accepted Answer · 2016-12-23 06:45:22Z

6

Recall the definition of (.):

(f . g) x = f (g x)

You have the expression

max a (max b c)

which matches the right hand side of (f . g) x if we set f = max a, g = max b, and x = c. Using these substitutions on the left-hand side of the definition of (.), we get:

(max a . max b) c = max a (max b c)

answered Dec 23, 2016 at 6:45

Daniel Wagner

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Comments

Zeta · Accepted Answer · 2016-12-23 07:41:13Z

Let's introduce some more parentheses in your first example:

max3 a b c = (max a) ((max b) c)

Now compare that to the ones in your last one:

max3 a b c = (max a) . ((max b) c)

Or, if we write (.) in prefix notation:

max3 a b c = (.) (max a) ((max b) c)

And now we see why you get that error. In order to type-check, (max b) c needs to be a function:

(.)       :: (b     -> c    )  -> (a -> b   ) -> a -> c
max a     ::  Float -> Float
max b c   ::                           Float
                                  ^^^^^^^^^^^

We get a better error message if we use a constrained version of max instead:

maxFloat :: Float -> Float -> Float
maxFloat = max

max3 a b c = max a . max b c

Now the error message is a lot better:

Couldn't match expected type ‘a0 -> Float’ with actual type ‘Float’
Possible cause: ‘maxFloat’ is applied to too many arguments
In the second argument of ‘(.)’, namely ‘maxFloat b c’
In the expression: maxFloat a . maxFloat b c

That being said, let's actually tackle the problem:

max3 a b c = max a ((max b) c)
           = (max a . max b) c

Note that you can also write

max3 :: Ord a => a -> a -> a -> a   
max3 a b = max a . max b

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