CODEWITHSUNDEEP
functionhaskellcurrying
Liamh101
Liamh101

Reputation: 49

Haskell Curry Function and Brackets

Hi this is probably a very simple problem but I'm having issue with it. I'm trying to make a roots function with the formula:

 roots a b c = ((-b + t)/a', (-b - t)/a')
 where
 t  = b ^ 2 - 4 * a * c
 a' = 2 * a

I'm now trying to make it a curried function however I can't seem to get it to work this is what I've put:

roots:: Double -> (Double -> (Double -> Double))

Could someone please help me out?

Thanks!

Upvotes: 0

Views: 200

Answers (2)

Sibi
Sibi

Reputation: 48654

In Haskell, functions are automatically curried. So you don't have to do anything special to make them curried.

Your function roots is of the type roots:: Double -> Double -> Double -> (Double, Double). Something like this will typecheck: let a = roots 3.0 because of currying.

In case your roots function was not curried, then it is likely to have a type like this: roots:: (Double , Double , Double) -> (Double, Double) which is not the proper way to write function definitons.

Upvotes: 3

markubik
markubik

Reputation: 671

As far as I know (but I'm not the expert, just had couple of lessons 'bout Haskell so far) function that gets 3 input parameters and produces one output (like in your example) should be written like:

roots:: Double -> Double -> Double -> Double

Last element in the chain (forth Double) is return type, all previous ones are input parameter types. This should do the trick

Upvotes: 0

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