Hello – in myFoldl :: (t -> a -> t) -> t -> [a] -> t the t and the a are type variables. There are no “classes” at all. Since you didn’t name the type variables, Haskell has provided default names for you “at random”, and it’s in some ways a little unfortunate that it’s named one of them t. You could (probably should) provide your own declaration, e.g. renaming t to b: myFoldl :: (b -> a -> b) -> b -> [a] -> b This represents exactly the same thing, and says that myFoldl will work for any two types a and b. It takes three params: (b -> a -> b) A function of values of type b and type a that returns a value of type b. (The accumulation function) b A value of type b (The initial accumulated value) [a] A list of values of type a (This list to accumulate over) and returns b A value of type b (The final accumulated value) The prelude function: foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b is then almost identical. This time, though, instead of being restricted to a list of values (of type a) to fold over, it can fold over any type of foldable collection of values (of type a). t represents the type of the foldable collection, and Foldable t => is a constraint that says the function will only work for a type t that is an instance of the class Foldable. [BTW: does a type declaration like this one reflect in any way the recursion going on? I didn’t really understand the question, but you can often deduce what a function does, and how it might be implemented, from its type signature.] I don’t know how you’re going about learning Haskell, but following a tutorial (such as http://learnyouahaskell.com/) might be helpful, and I think explains all of the above step by step. Regards, David. From: Lawrence Bottorffmailto:borgauf@gmail.com Sent: 02 January 2021 17:52 To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskellmailto:beginners@haskell.org Subject: [Haskell-beginners] Type explanation of foldl? I've got this myFoldl f init [] = init myFoldl f init (x:xs) = myFoldl f newInit xs where newInit = f init x which when evaluated gives this for its type
:t myFoldl myFoldl :: (t -> a -> t) -> t -> [a] -> t
. . . not totally sure what that all means, e.g., how the t as a generic class is behaving is very mysterious. Obviously, the final result is of type t . . . and that must relate to the incoming argument init, correct? (BTW, does a type declaration like this one reflect in any way the recursion going on?) But then with Prelude foldl I'm really clueless
:t foldl foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b
I'm guessing the t is again a generic type of the Foldable class, but then. . . Can someone walk me through this? LB