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

[Lennart Augustsson]

I thought solution one was missing the ~ ?

On Feb 12, 2007, at 22:07 , apfelmus at quantentunnel.de 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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