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] 847288609443 Prelude> foldl (^) 2 [3,5] 32768

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.

Cheers, John

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners