CODEWITHSUNDEEP
haskelltypeclass
Deniz Dogan
Deniz Dogan

Reputation: 26227

Defining a seemingly simple Foldable instance

What would the Foldable instance for this datatype look like?

data X t = X t [X t]

I tried this:

instance Foldable X where
    foldMap f (X x xs) = f x `mappend` foldMap f xs

But got this error:

Occurs check: cannot construct the infinite type: a = X a
When generalising the type(s) for `foldMap'
In the instance declaration for `Foldable X'

Upvotes: 1

Views: 356

Answers (1)

sth
sth

Reputation: 229593

xs is a list of items and foldMap needs to be applied to the individual items, not the list itself. Doing this with map gives a list of results that can be combined with mconcat:

instance Foldable X where
  foldMap f (X x xs) = f x `mappend` mconcat (map (foldMap f) xs)

Upvotes: 6

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