Il 16 gennaio 2021 alle 23:32 Lawrence Bottorff ha scritto:
This is the definition of list foldr
foldr :: (a -> b -> b) -> b -> [a] -> b foldr _ z [] = z foldr f z (x:xs) = f x (foldr f z xs)
In both foldl and foldr in the OP the n variable in lambda functions would seem to be for the accumulator, hence, I assume the n is considered the free variable? And then the wildcard in each lambda function refers to the bound variable, i.e., the list argument's elements to be folded.
The ‘_’ means «do not bind this parameter». Hence λ> (\x _ -> x + 1) 10 20 11
but the second example I'm not sure how the (\x n -> x + n) is being applied in the form . . . f x (foldr f z xs) It obviously must be doing the same (+) 1 ((+) 2 ((+) 3 ((+) 4 5))) but how the function is being applied I don't understand.
foldr (+) 0 ([1,2,3]) (+) 1 (foldr (+) 0 [2,3]) (+) 1 ((+) 2 (foldr (+) 0 [3])) (+) 1 ((+) 2 ((+) 3 (foldr (+) 0 []))) (+) 1 ((+) 2 ((+) 3 0)) (+) 1 ((+) 2 3) (+) 1 5 6
BTW, is the t a format in
:type foldr foldr :: Foldable t => (a -> b -> b) -> b -> t a -> b
something from category theory, i.e., for the list instance, t a is [a] What is the algebraic syntax where t a becomes [a] in the case of lists? It would be nice to understand some day exactly what :i Foldable is saying
What book are you studying on that does not talk about Typeclasses? I feel you are making your life harder by not following a good study program.