Re: [Haskell-cafe] Foldr tutorial, Inspired by Getting a Fix from a Fold

Guess this is a tricky choice for a foldr intro, since it requires a "paramorphism" (see bananas lenses wires etc.) para :: (a -> [a] -> b -> b) -> b -> [a] -> b para f e [] = e para f e (x:xs) = f x xs (para f e xs) -- note that the original tail of the list (i.e. xs and not xs') is used in the else-branch dropWhile' p = para (\x xs xs' -> if p x then xs' else (x:xs)) [] Prelude> dropWhile' (<5) [1,2,3,4,5,6,7,8] [5,6,7,8] Prelude> dropWhile' (<5) [1,2,3,4,5,6,7,1] [5,6,7,1] Prelude> :m + Test.QuickCheck Prelude Test.QuickCheck> :m + Text.Show.Functions Prelude Test.QuickCheck Text.Show.Functions> quickCheck $ \p l -> dropWhile p (l :: [Int]) == dropWhile' p l Loading package QuickCheck-1.0 ... linking ... done. OK, passed 100 tests. On 2/12/07, Donald Bruce Stewart wrote:

pixel:

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Chris Moline writes:

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dropWhile p = foldr (\x l' -> if p x then l' else x:l') []

invalid: dropWhile (< 5) [1, 10, 1] should return [10, 1]

Prelude Test.QuickCheck Text.Show.Functions> quickCheck $ \p xs -> dropWhile p xs == foldr (\x l' -> if p x then l' else x:l') [] (xs :: [Int])

Falsifiable, after 4 tests: <function> [-1,-3,1]

If in doubt, do a quick check!

-- Don _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe