CODEWITHSUNDEEP
haskellfunctional-programming
Luis Fernando Pineda
Luis Fernando Pineda

Reputation: 91

Haskell High order functions. passing two functions as arguments.

I have recently been teaching myself Haskell, and one of my exercises was to implement a function that takes two functions as arguments and passes number 3 to the first function and the result to the second function. I have more experience with racket. The function that i created for racket is 

(define (nestfun3 function function2)
  (function2 (function 3)))

I'm trying to replicate this function in racket. i know that high order functions with only one function as argument can be like

twice function = function . function

How can I pass two functions? I tried 

twice function = function . function
three function = function . twice

Upvotes: 2

Views: 540

Answers (1)

Random Dev
Random Dev

Reputation: 52300

the direct translation would be:

apply3 :: Num a => (a -> b) -> (b -> c) -> c
apply3 f g = g (f 3)

remember: you don't want to just compose functions - you want to first apply 3 and then apply the result to the other function

sorry for the mixing ... this should work now:

λ> let f = (+1)
λ> let g = (*2)
λ> apply3 f g
8
λ> apply3 g f
7

in case you wondered - you don't have to actually start with the signature (although it's good practice to add it) - if you do this in ghci:

λ> let apply3 f g = g (f 3)
λ> :t apply3
apply3 :: Num a => (a -> t1) -> (t1 -> t) -> t

it will give it do you (or you can use ghc-mod or whatever) - I actually just renamed the type-parameters (those t are ugly)

This way it should look very similar to what you did in Scheme

of course you might want to make sure not to swap functions while renaming :|

Upvotes: 5

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