I thought solution one was missing the ~ ?

On Feb 12, 2007, at 22:07 , [EMAIL PROTECTED] wrote:

Bernie Pope wrote:
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.

"Nice one" is an euphemism, it's exactly solution one :)

Regards,
apfelmus

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