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Today's Topics:
1. splitAt implementation (using foldr) and infinite lists
(Dmitriy Matrosov)
2. Re: Tying the knot with binary trees (Michael Schober)
----------------------------------------------------------------------
Message: 1
Date: Mon, 16 Apr 2012 12:52:02 +0400
From: Dmitriy Matrosov <[email protected]>
Subject: [Haskell-beginners] splitAt implementation (using foldr) and
infinite lists
To: [email protected]
Message-ID: <[email protected]>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Hi all.
If i implement take using foldr
take' :: Int -> [a] -> [a]
take' n = foldr (\x z -> if fst x <= n then snd x : z else []) []
. zip [1..]
it'll work fine with infinite lists. But if i implement splitAt similarly
splitAt' :: Int -> [a] -> ([a], [a])
splitAt' n = foldr (\x (z1, z2) -> if fst x <= n then (snd x : z1, z2)
else ([], snd x : z2))
([], [])
. zip [1..]
and call it like this
*Main> fst $ splitAt' 4 [1..]
^CInterrupted.
it will hang. I don't understand, why it tries to compute second list
(z2), if it only needs first?
------------------------------
Message: 2
Date: Mon, 16 Apr 2012 11:17:19 +0200
From: Michael Schober <[email protected]>
Subject: Re: [Haskell-beginners] Tying the knot with binary trees
To: Markus L?ll <[email protected]>
Cc: [email protected]
Message-ID: <[email protected]>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Hi Markus,
> They say that for this or similar reasons you have to use a zipper,
> but I know nothing more about them than this :-)
Thanks for the hint. I hadn't heard of zippers before and after some
reading, they indeed seem to do the trick, although I have yet to
implement it in my solution.
Anyways, here are the resources which worked best for my understanding:
[1] http://en.wikibooks.org/wiki/Haskell/Zippers
[2] http://learnyouahaskell.com/zippers
Best,
Michael
------------------------------
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End of Beginners Digest, Vol 46, Issue 22
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