[Haskell-cafe] ~ operator ?
Mateusz Kowalczyk
fuuzetsu at fuuzetsu.co.uk
Mon Feb 18 20:13:13 CET 2013
On 18/02/13 19:02, briand at aracnet.com wrote:
> Hi all,
>
> I was creating "bigger" uncurries which I am simply extending from an existing uncurry I found some where, e.g.
>
> uncurry4 :: (a -> b -> c -> d -> e) -> ((a, b, c, d) -> e)
> uncurry4 f ~(a,b,c,d) = f a b c d
>
> when I realized, what's the "~" for ?
>
> I've only been able to find a partial explanation that it involves preserving laziness, or something, maybe ?
>
> I was hoping someone could enlighten me.
>
> Thanks
>
> Brian
>
>
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[1] explains it in section 4.4 — ‘Lazy patterns’.
[1] - http://www.haskell.org/tutorial/patterns.html
--
Mateusz K.
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