[Haskell-cafe] Re: Foldr tutorial, Inspired by Getting a Fix from a Fold

[apfelmus at quantentunnel.de]

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