asum (x:xs) = x <|> asum xs

On Friday, June 10, 2016, Tom Ellis <tom-lists-haskell-cafe-2013@jaguarpaw.co.uk> wrote:

On Fri, Jun 10, 2016 at 05:37:41PM +0300, Юрий Сыровецкий (Yuriy Syrovetskiy) wrote:
> It is so because in base library
>
> asum = foldr (<|>) empty
>
> What if it was defined
>
> asum [] = empty
> asum [x:xs] = x <|> asum xs

Those are the same (assuming you mean (x:xs) instead of [x:xs]).

Tom
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