CODEWITHSUNDEEP
haskellbinary-tree
Jose Miguel Alemán
Jose Miguel Alemán

Reputation: 1

BinTree haskell

I try to implement a function that returns a tree with the same structure but with all duplicate values ​​( nodes )

data BinTreeInt = Void | Node Int BinTreeInt BinTreeInt deriving Show

dupElem :: (Num a) => a -> BinTreeInt -> BinTreeInt
dupElem f Void = Void
dupElem f (Node n l r) = Node (f n) (dupElem f l)(dupElem f r)

I got this error:

P4.hs:382:32:
    Could not deduce (a ~ (Int -> Int))
    from the context (Num a)
      bound by the type signature for
                 dupElem :: Num a => a -> BinTreeInt -> BinTreeInt
      at P4.hs:380:12-51
      ‘a’ is a rigid type variable bound by
          the type signature for
            dupElem :: Num a => a -> BinTreeInt -> BinTreeInt
          at P4.hs:380:12
    Relevant bindings include
      f :: a (bound at P4.hs:382:9)
      dupElem :: a -> BinTreeInt -> BinTreeInt (bound at P4.hs:381:1)
    The function ‘f’ is applied to one argument,
    but its type ‘a’ has none
    In the first argument of ‘Node’, namely ‘(f n)’
    In the expression: Node (f n) (dupElem f l) (dupElem f r)
Failed, modules loaded: none.

Seems to be easy but... Haskell is not my love!

Upvotes: 0

Views: 271

Answers (2)

Bargros
Bargros

Reputation: 531

According to your function signature "a" is constrained as a Num type but you are calling a as a function over n.

dupElem :: (Num a) => a -> BinTreeInt -> BinTreeInt

So my suggestion would be to change:

Node (f n) (dupElem f l)(dupElem f r)

for

Node f (dupElem f l)(dupElem f r)

as chepner suggested above.

Upvotes: 0

chepner
chepner

Reputation: 531165

If I understand correctly, you simply want to replace n with f, not apply f as a function to n.

dupElem :: (Num a) => a -> BinTreeInt -> BinTreeInt
dupElem _ Void = Void
dupElem f (Node n l r) = Node f (dupElem f l)(dupElem f r)

For instance, dupElem 3 (BinTreeInt 1 (BinTreeInt 2 Void Void) Void) would return (BinTreeInt 3 (BinTreeInt 3 Void Void) Void).

Upvotes: 3

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