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. So I can recreate foldr (+) 5 [1,2,3,4] with foldr (\x n -> x + n) 5 [1,2,3,4] They both return 15. The first one results in (+) 1 ((+) 2 ((+) 3 ((+) 4 5))) 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. Beta reduction doesn't get me very far \x n -> x + n (5)([1,2,3,4]) \x 5 -> x + 5 ([1,2,3,4]) but obviously the enclosing lambda calc for foldr is doing something to create the (+) 1 ((+) 2 ((+) 3 ((+) 4 5))) form. 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 On Sat, Jan 16, 2021 at 4:36 PM Francesco Ariis <fa-ml@ariis.it> wrote:
Il 16 gennaio 2021 alle 16:10 Lawrence Bottorff ha scritto:
I have this
myLength1 = foldl (\n _ -> n + 1) 0
and this
myLength2 = foldr (\_ n -> n + 1) 0
I am guessing that foldl knows to assign the accumulator-seed argument to the dependent variable and the list argument's elements recursively to the independent variable; and with foldr to do the opposite. Is this a fair assumption? BTW, where can I get a look at the code for fold functions; or does the type definition answer my original question? Not really able to decipher it so well
:t foldl foldl :: Foldable t => (b -> a -> b) -> b -> t a -> b
foldl and foldr have slightly different signatures,
λ> :t +d foldl foldl :: (b -> a -> b) -> b -> [a] -> b λ> :t +d foldr foldr :: (a -> b -> b) -> b -> [a] -> b
(Notice `b -> a -> b` vs. `a -> b -> b`), hence the lambdas have a different non-matched parameter. Does this answer your question? —F _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners