为什么以下用于创建Tree的Foldable实例的haskell代码导致无限递归？

Question

I am trying to write a Foldable instance of the following Tree class I created:

data Tree a b = Tree b a [Tree a b]

But I wanted to operate on type 'a' in the definition therefore I made a wrapper data type the flips the types in Tree:

data FlipTree a b = FlipTree (Tree b a)

Now to write the actual definition I came up with the following:

instance Foldable (FlipTree a) where
  foldr f x (FlipTree (Tree _ s [])) = (f s x)
  foldr f x (FlipTree (Tree _ s children)) =
              let updated_x = (f s x)
                  tqs = (map (\n -> FlipTree n) children) in
              foldl (foldr f) updated_x tqs

But this results in infinite recursion when foldr is called. I have spent two days and have not been able to figure out the thing that I am doing wrong?

this is lpaste link to the code file . I isolated the code as much as I could. This is another file1 which is read to run the code using the command:

runghc code.hs 30000000.0 20000.0 < file1.txt

Answer 1

Not an answer (the questions premise of infinite looping doesn't seem to be correct) but stylistic advice – too long to comment.

First, you probably want to make this a newtype (there's no reason to have an extra layer of laziness).

newtype FlipTree a b = FlipTree (Tree b a)

Second, for an easier understandable solution it tends IMO to be preferrable to implement foldMap instead of foldr ; it makes the semantics clearer. There are multiple ways to fold over this structure, the most reasonable seems

  foldMap f (FlipTree (Tree _ s [])) = f s
  foldMap f (FlipTree (Tree _ s children))
             = f s <> foldMap (foldMap f . FlipTree) children

where the outer foldMap folds over the list of children, using again foldMap for FlipTree s as the folding-mapping. Now, if you want to manually translate this to the foldr equivalent, you basically just need to thread through the argument.

  foldr f x (FlipTree (Tree _ s children))
             = f s . flip (foldr $ flip (foldr f) . FlipTree) children $ x

Now, possibly that instance doesn't actually give the desired behaviour, so please tell us how you want it to behave!

Answer 2

Turned out the definition is 100% correct. The unwanted behaviour was being caused by the rest of the code. But thanks anyway to the stackoverflow community for always being there.