How would you define map and filter using foldr in Haskell?
dictionary, filter, fold, functional-programming, haskell
Solution
I wish I could just comment, but alas, I don't have enough karma.
The other answers are all good ones, but I think the biggest confusion seems to be stemming from your use of x and xs.
If you rewrote it as
map' :: (a -> b) -> [a] -> [b]
map' f [] = []
map' f (x:xs) = foldr (\y ys -> (f y):ys) [] xs
you would clearly see that `x` is not even mentioned on the right-hand side, so there's no way that it could be in the solution.
Cheers
Problem
I'm doing a bit of self study on functional languages (currently using Haskell). I came across a Haskell based assignment which requires defining map and filter in terms of foldr. For the life of me I'm not fully understanding how to go about this. For example when I define a map function like: ``` map' :: (a -> b) -> [a] -> [b] map' f [] = [] map' f (x:xs) = foldr (\x xs -> (f x):xs) [] xs ``` I don't know why the first element of the list is always ignored. Meaning that: ``` map' (*2) [1,2,3,4] ``` results in [4,6,8] instead of [2,4,6,8] Similarly, my filter' function: ``` filter' :: (a -> Bool) -> [a] -> [a] filter' p [] = [] filter' p (x:xs) = foldr (\x xs -> if p x then x:xs else xs ) [] xs ``` when run as: ``` filter' even [2,3,4,5,6] ``` results in [4,6] instead of [2,4,6] Why would this be the case? And how SHOULD I have defined these functions to get the expected results? I'm assuming something is wrong with my lambda expressions...