# 99 questions/Solutions/4

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< 99 questions | Solutions(Difference between revisions)

m (NTUA solution ;-)) |
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</haskell> | </haskell> | ||

+ | This is <hask>length</hask> in <hask>Prelude</hask>. | ||

+ | |||

+ | A fancier one! :-) | ||

<haskell> | <haskell> | ||

myLength = foldl (const . (+1)) 0 | myLength = foldl (const . (+1)) 0 | ||

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

− | |||

− |

## Revision as of 13:29, 18 July 2012

(*) Find the number of elements of a list.

myLength :: [a] -> Int myLength [] = 0 myLength (_:xs) = 1 + myLength xs myLength' :: [a] -> Int myLength' list = myLength_acc list 0 -- same, with accumulator where myLength_acc [] n = n myLength_acc (_:xs) n = myLength_acc xs (n + 1)

myLength' = foldl (\n _ -> n + 1) 0 myLength'' = foldr (\_ n -> n + 1) 0 myLength''' = foldr (\_ -> (+1)) 0 myLength'''' = foldr ((+) . (const 1)) 0 myLength''''' = foldr (const (+1)) 0

myLength' xs = snd $ last $ zip xs [1..] -- Just for fun myLength'' = snd . last . (flip zip [1..]) -- Because point-free is also fun myLength''' = fst . last . zip [1..] -- same, but easier

myLength = sum . map (\x -> 1)

length

Prelude

A fancier one! :-)

myLength = foldl (const . (+1)) 0