CODEWITHSUNDEEP
haskell
D idsea J
D idsea J

Reputation: 113

Haskell using map to call function with more than one parameter

Let's say we have the following code representing binary trees where decodeInt searches the tree for an integer:

import Text.Show.Functions
data BinTree a b = Leaf b | Node a (BinTree a b) (BinTree a b) deriving Show

example :: BinTree (Int -> Bool) Char
example = Node (\x -> x > 4) (Node (\x -> x * x == x) (Leaf 'g') (Node (\x -> x == 0)
          (Leaf 'u') (Leaf 'l'))) (Node (\x -> x >= 7) (Leaf 'f') (Leaf 'i'))

countInnerNodes :: BinTree a b -> Int
countInnerNodes (Node a b c) = 1 + countInnerNodes b + countInnerNodes c
countInnerNodes (Leaf x) = 0

decodeInt :: BinTree (Int -> Bool) b -> Int -> b
decodeInt (Leaf b) p = b
decodeInt (Node x y z) p = if (x(p) == True) then decodeInt z p else decodeInt y p

decode :: BinTree (Int -> Bool) b -> [Int] -> String
decode x [] = "empty list"
decode x xs = ??????

How can I use map to get a result like this when calling decode?

decode Tree [1,2,3,4] 
  = [decodeInt Tree (1), decodeInt Tree (2), 
     decodeInt Tree (3), decodeInt Tree (4)]

/edit: Followup Let's also say we would like to create a function like the following

mapTree (\x -> ’e’) example

mapTree should return a BinTree just like example with the only difference being that the Char of every Leaf has been replaced with an 'e'. How do I accomplish that? I've started Haskell yesterday so I'm quite new to functional programming.

Upvotes: 1

Views: 364

Answers (2)

Coding Newbie
Coding Newbie

Reputation: 21

A bit too late but maybe this helps :

decode :: BinTree (Int -> Bool) b -> [Int] -> [b]
decode _ [] = []
decode x (y:ys) = decodeInt x y .... maybe you have an idea how to continue here ;)

Upvotes: 2

chepner
chepner

Reputation: 532153

decodeInt :: BinTree (Int -> Bool) b -> Int -> b, so assuming t :: BinTree (Int -> Bool) b, then decodeInt t :: Int -> b. You map that function over your list of Ints.

decode t xs = let y = map (decodeInt t) xs
              in ...

You'll still have to figure out how to convert y to the String value that decode is expected to return.

Upvotes: 2

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