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 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...

Answer 1

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