CODEWITHSUNDEEP
haskellmonoidsfoldable
OneEyeQuestion
OneEyeQuestion

Reputation: 742

How does Foldable knows the implementation of mappend

I am learning Haskell with the help of "http://learnyouahaskell.com". I am following the example of a BST (Binary Search Tree) that is an instance of Foldable:

data Tree a = Nil | Node a (Tree a) (Tree a) deriving (Show, Read, Eq)

instance F.Foldable Tree where
    foldMap f Nil = mempty
    foldMap f (Node x l r) = (F.foldMap f l) `mappend` (f x) `mappend` (F.foldMap f r)

When I run F.foldMap (\x -> [x]) testTree I get a list that represents the folded tree. I implemented my own data type:

newtype OnlySum a = OnlySum {value :: a} deriving (Eq, Ord, Read, Show, Bounded)
instance Num a => Monoid (OnlySum a) where
    mempty = OnlySum 0
    OnlySum x `mappend` OnlySum y = OnlySum (x + y)

And ran this command: F.foldMap (\x -> OnlySum x) testTree to obtain the tree folded as this OnlySum {value = 34}.

The question is: How does Foldable know the definition of mempty and mappend depending on the return type of the function f passed to foldMap? Does it infer it or is there a way for Haskell to automatically know which is the definition?

Upvotes: 1

Views: 177

Answers (1)

lmm
lmm

Reputation: 17431

In the definition of foldMap we see that it requires a typeclass instance Monoid m. So Haskell knows that mappend is coming from Monoid. The specific type will be inferred, then a Monoid typeclass instance resolved based on that type, and that typeclass is what will be used.

Upvotes: 3

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