How would you define map and filter using foldr in Haskell?

Question

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

Answer 1

For your first question, foldr already has a case for the empty list, so you need not and should not provide a case for it in your own map.

map' f = foldr (\x xs -> f x : xs) []

The same holds for filter'

filter' p = foldr (\x xs -> if p x then x : xs else xs) []

Nothing is wrong with your lambda expressions, but there is something wrong with your definitions of filter' and map'. In the cons case (x:xs) you eat the head (x) away and then pass the tail to foldr. The foldr function can never see the first element you already ate. :)

Alse note that:

filter' p = foldr (\x xs -> if p x then x : xs else xs) []

is equivalent (η-equivalent) to:

filter' p xs = foldr (\x xs -> if p x then x : xs else xs) [] xs