# 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
```

or `elementAt_w'`

correctly in point-free style:

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

Pedantic note: the above definition of `elementAt_w'pf`

does not conform to the order of arguments specified by the question, but the following does:

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