99 questions/Solutions/4
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(Difference between revisions)
(Added another two solutions) |
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<haskell> | <haskell> | ||
| - | myLength' | + | myLength' = foldl (\n _ -> n + 1) 0 |
| - | myLength'' | + | myLength'' = foldr (\_ n -> n + 1) 0 |
| - | myLength''' | + | myLength''' = foldr (\_ -> (+1)) 0 |
| - | myLength'''' = foldr ((+) . (const 1)) 0 | + | myLength'''' = foldr ((+) . (const 1)) 0 |
| + | myLength''''' = foldr (const (+1)) 0 | ||
</haskell> | </haskell> | ||
<haskell> | <haskell> | ||
myLength' xs = snd $ last $ zip xs [1..] -- Just for fun | myLength' xs = snd $ last $ zip xs [1..] -- Just for fun | ||
| + | myLength'' = snd . last . (flip zip [1..]) -- Because point-free is also fun | ||
</haskell> | </haskell> | ||
Revision as of 20:38, 21 November 2011
(*) Find the number of elements of a list.
myLength :: [a] -> Int myLength [] = 0 myLength (_:xs) = 1 + myLength xs
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 = sum . map (\x -> 1)
length
Prelude
