Hi, I think your interpretation of (replicate (n-1) tail) is wrong. First note that tail is not applied to anything, is just the function tail. So replicate (n-1) tail is [tail, tail, tail, .... , tail]. For example replicate 3 tail = [tail, tail, tail] Note that if you try to compute this in Haskell it won't be able to show that result, because there is no "show" defined for something that has the type of a list of functions. Having said this, your example can now be written elementAt [1,2] 2 = = head (foldr ($) [1,2] (replicate 1 tail)) = head (foldr ($) [1,2] [tail]) = head (tail $ (foldr ($) [1,2] [])) = head (tail $ [1,2]) = head (tail ([1,2])) = head [2] = 2 For a more general example you can try elementAt [1,2,3,4] 3 = = head (foldr ($) [1,2,3,4] (replicate 2 tail)) = head (foldr ($) [1,2,3,4] [tail,tail]) = head (tail $ (foldr ($) [1,2,3,4] [tail])) = head (tail $ tail $ [1,2,3,4]) = head (tail (tail [1,2,3,4))) = head (tail [2,3,4]) = head [3,4] = 3 Cheers, Ut http://www.avg.com/email-signature?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail Mail priva di virus. www.avg.com http://www.avg.com/email-signature?utm_medium=email&utm_source=link&utm_campaign=sig-email&utm_content=webmail <#DAB4FAD8-2DD7-40BB-A1B8-4E2AA1F9FDF2> Il giorno gio 14 gen 2021 alle ore 07:25 Lawrence Bottorff < borgauf@gmail.com> ha scritto:
With 99 questions https://wiki.haskell.org/99_questions/Solutions/3 Problem 3 wants a function elementAt that will take a list and an index and return the element for that index. One very odd version in the solutions is
elementAt xs n = head $ foldr ($) xs $ replicate (n - 1) tail
So the function "passed" is ($) and the accumulator "seed" is the incoming list xs and the list to be worked on is (replicate (n-1) tail) which . . . and I can't fathom what's happening -- other than perhaps (replicate (n-1) (tail xs))
elementAt [1,2] 2 would be
foldr ($) [1,2] (replicate 1 (tail [1,2]) foldr ($) [1,2] ([2])
. . . now I'm lost. Can someone walk me through this?
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