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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