Haskell functional programming, function as parameter

Question

I'm am training funtional Haskell and cannot usderstand what is wrong with this:

maxBy :: (a -> Int) -> a -> a -> a
maxBy measure a b = max (measure a) (measure b)

Here are examples how this maxBy function should work:

maxBy (*2)   3       5      ==>  5
maxBy length [1,2,3] [4,5]  ==>  [1,2,3]
maxBy head   [1,2,3] [4,5]  ==>  [4,5]

Error message is:

Couldn't match expected type ‘a’ with actual type ‘Int’ ‘a’ is a rigid type variable bound by the type signature for: maxBy :: forall a. (a -> Int) -> a -> a -> a

Welcome to Haskell! This subtle type disagreement is exactly the sort of thing that happens all the time in more dynamic languages and everyone thinks it's fine and no-one notices... until they get burned by it, and then they have to examine their entire program to find the contradiction. It's much easier to give the burden of spotting these contradictions to a typechecking compiler like GHC. — James Brock
– James Brock, Commented Nov 30, 2020 at 12:53
This is one of the cases where it really helps to comment out the type signature and get surprised by the type the compiler infers. — Ingo
– Ingo, Commented Nov 30, 2020 at 23:17

chepner · Accepted Answer · 2020-11-30 14:26:14Z

5

Your function is returning measure a or measure b when you want it to return a or b.

edited Nov 30, 2020 at 14:26

chepner

538k77 gold badges594 silver badges746 bronze badges

answered Nov 30, 2020 at 12:04

Alistair Wall

3321 gold badge2 silver badges3 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Blue Over a year ago

Yeah I just realised that :D

Collectives™ on Stack Overflow

Haskell functional programming, function as parameter

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related