# 99 questions/Solutions/2

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m (Add description to foldl solution.) |
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myButLast'''' = head . tail . reverse | myButLast'''' = head . tail . reverse | ||

+ | </haskell> | ||

− | myButLast''''' = | + | We can also use a pair to keep the last two elements processed from the list. We take advantage of lazy evaluation, such that if the list is too small, the initial elements in the pair get evaluated, resulting in an error: |

+ | <haskell> | ||

+ | myButLast''''' = snd.(foldl (\(a,b) c -> (c,a)) (e1, e2)) | ||

+ | where e1 = error "List too small!" | ||

+ | e2 = error "List is null!" | ||

</haskell> | </haskell> | ||

+ | |||

+ | |||

+ | [[Category:Programming exercise spoilers]] |

## Revision as of 15:27, 15 May 2014

(*) Find the last but one element of a list.

myButLast :: [a] -> a myButLast = last . init myButLast' x = reverse x !! 1 myButLast'' [x,_] = x myButLast'' (_:xs) = myButLast'' xs myButLast''' (x:(_:[])) = x myButLast''' (_:xs) = myButLast''' xs myButLast'''' = head . tail . reverse

We can also use a pair to keep the last two elements processed from the list. We take advantage of lazy evaluation, such that if the list is too small, the initial elements in the pair get evaluated, resulting in an error:

myButLast''''' = snd.(foldl (\(a,b) c -> (c,a)) (e1, e2)) where e1 = error "List too small!" e2 = error "List is null!"