# 99 questions/Solutions/3

### From HaskellWiki

(*) Find the K'th element of a list. The first element in the list is number 1.

This is (almost) the infix operator !! in Prelude, which is defined as:

(!!) :: [a] -> Int -> a (x:_) !! 0 = x (_:xs) !! n = xs !! (n-1)

Except this doesn't quite work, because !! is zero-indexed, and element-at should be one-indexed. So:

elementAt :: [a] -> Int -> a elementAt list i = list !! (i-1)

Or without using the infix operator:

elementAt' :: [a] -> Int -> a elementAt' (x:_) 1 = x elementAt' [] _ = error "Index out of bounds" elementAt' (_:xs) k | k < 1 = error "Index out of bounds" | otherwise = elementAt' xs (k - 1)

Alternative version:

elementAt'' :: [a] -> Int -> a elementAt'' (x:_) 1 = x elementAt'' (_:xs) i = elementAt'' xs (i - 1) elementAt'' _ _ = error "Index out of bounds"

**This does not work correctly on invalid indexes and infinite lists, e.g.:**

elementAt'' [1..] 0

A few more solutions using prelude functions:

elementAt'' xs n | length xs < n = error "Index out of bounds" | otherwise = fst . last $ zip xs [1..n] elementAt''' xs n = head $ foldr ($) xs $ replicate (n - 1) tail -- Negative indices not handled correctly: -- Main> elementAt''' "haskell" (-1) -- 'h' elementAt'''' xs n | length xs < n = error "Index out of bounds" | otherwise = last $ take n xs elementAt''''' xs n | length xs < n = error "Index out of bounds" | otherwise = head . reverse $ take n xs elementAt'''''' xs n | length xs < n = error "Index out of bounds" | otherwise = head $ drop (n - 1) xs

elementAt_w'

elementAt_w'pf = (last .) . take . (+ 1)

elementAt_w'pf

elementAt_w'pf' = flip $ (last .) . take . (+ 1)