99 questions/Solutions/4
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(Grouped all fold solutions together) 

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Revision as of 19:47, 18 January 2014
(*) 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'''''' = foldl (const . (+1)) 0
myLength' xs = snd $ last $ zip xs [1..]  Just for fun myLength'' = snd . last . (flip zip [1..])  Because pointfree is also fun myLength''' = fst . last . zip [1..]  same, but easier
myLength = sum . map (\_>1)
length
Prelude
 length returns the length of a finite list as an Int. length :: [a] > Int length [] = 0 length (_:l) = 1 + length l
The prelude for haskell 2010 can be found here.