[Haskell-cafe] Re: [Haskell-beginners] folds -- help!
aneumann at inf.fu-berlin.de
Sat Mar 14 02:55:35 EDT 2009
That's why I said "for appropriate g and f". But I see that my
wording was misleading.
However you can say:
> foldl' oplus alpha bs =(foldr f id bs) alpha where
> f a g = \alpha -> g (alpha `oplus` a)
> foldr' oplus alpha bs = (foldl f id bs) alpha where
> f g a = \alpha -> g (a `oplus` alpha)
And it works as long as oplus is strict in both arguments.
Am 10.03.2009 um 21:54 schrieb John Dorsey:
> Adrian Neumann wrote:
>> Notice that there is no difference between
>> foldr g a
>> foldl f a
>> (for appropriate g and f) if g and f are strict in both arguments.
> Be careful... as apfelmus noted elsewhere in this thread, that's
> not (in
> general) true.
> Prelude> foldr (^) 2 [3,5]
> Prelude> foldl (^) 2 [3,5]
> The reason? Integer exponentiation (^) isn't associative and
> commutative. So the first is (3 ^ (5^2)) = 3^25, while the second is
> ((2 ^ 3) ^ 5) = 2^15.
> Beginners mailing list
> Beginners at haskell.org
-------------- next part --------------
A non-text attachment was scrubbed...
Size: 194 bytes
Desc: Signierter Teil der Nachricht
Url : http://www.haskell.org/pipermail/haskell-cafe/attachments/20090314/8bae8007/PGP.bin
More information about the Haskell-Cafe