# 99 questions/Solutions/4

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< 99 questions | Solutions(Difference between revisions)

(Their was a section for the length function in the prelude but it was missing so I took the function form the 2010 report and link to it.) |
(Grouped all fold solutions together) |
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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'''' | + | myLength'''' = foldr ((+) . (const 1)) 0 |

− | myLength''''' = foldr (const (+1)) 0 | + | myLength''''' = foldr (const (+1)) 0 |

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

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

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The prelude for haskell 2010 can be found [http://www.haskell.org/onlinereport/haskell2010/haskellch9.html#x16-1710009 here.] | The prelude for haskell 2010 can be found [http://www.haskell.org/onlinereport/haskell2010/haskellch9.html#x16-1710009 here.] | ||

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## Revision as of 19:21, 30 November 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'''''' = foldl (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 (\_->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.