"fold" in Haskell?

Question

Write a function of higher order atEach f xs applying the default function f to each element of the list xs.

atEach succ [1 to 5] = [2,3,4,5,6]
atEach length ["Haskell", "go", "forward"] = [7,5,8]

Why not `atEach length ["Haskell", "go", "forward"] = [7, 2, 7]`? If so, you are being asked to rewrite the standard `map` function. **What have you tried?** — dave4420, May 06 '13 at 09:11
@AntonGuryanov Please don't do the OP's homework for him, he will not learn. (Help him: yes; do it for him: no.) — dave4420, May 06 '13 at 10:01
yes, it's for my homework and I would like concrete assistance — , May 06 '13 at 12:56
https://docs.google.com/document/d/1ky6tSD5Z63UI2xJx5beZvdLgcva2iWZeDP5bnwYJ-wc/edit?usp=sharing How to I am writing the homework, I was wondering if you could help me — , May 06 '13 at 13:38
possible duplicate of [How would you define map and filter using foldr in Haskell?](http://stackoverflow.com/questions/5726445/how-would-you-define-map-and-filter-using-foldr-in-haskell) — Daniel Wagner, May 06 '13 at 15:09

score 5 · Answer 1 · answered May 06 '13 at 09:36

As dave4420 already pointed out, your atEach seems to be the standard map function (please clarify if not). If this is the case, you have different ways to implement it, e.g.:

-- direct recursion
atEach _ [] = []
atEach f (x:xs) = ??? 

-- list comprehension
atEach f xs = [??? | x <- xs]

--using a fold
atEach f = foldr ??? []

I don't want to spoil the fun, so you can try to fill out the ???.

"fold" in Haskell?

1 Answers1