Chris Moline [EMAIL PROTECTED] writes:
dropWhile p = foldr (\x l' - if p x then l' else x:l') []
invalid: dropWhile ( 5) [1, 10, 1] should return [10, 1]
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pixel:
Chris Moline [EMAIL PROTECTED] writes:
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')
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
Nicolas Frisby wrote:
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
Oops; I totally forgot the context of this whole discussion!
I enjoyed your article.
On 2/12/07, Bernie Pope [EMAIL PROTECTED] wrote:
Nicolas Frisby wrote:
Guess this is a tricky choice for a foldr intro, since it requires a
paramorphism (see bananas lenses wires etc.)
para :: (a - [a] - b
Sure, but we also have
para f e xs = snd $ foldr (\ x ~(xs, y) - (x:xs, f x xs y)) ([], e) xs
So I think using para is fine.
-- Lennart
On Feb 12, 2007, at 18:40 , Bernie Pope wrote:
Nicolas Frisby wrote:
Guess this is a tricky choice for a foldr intro, since it requires a
Lennart Augustsson wrote:
Sure, but we also have
para f e xs = snd $ foldr (\ x ~(xs, y) - (x:xs, f x xs y)) ([], e) xs
Nice one.
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I hope the following doesn't come across as
condescending...
The recent issue of The Monad Reader has generated
some excitement, mostly to do with the time travel
article. I, however, would like to discuss a simpler
solution to implementing dropWhile with foldr, which
is discussed in the first
