Re: [Haskell-beginners] foldr point free notation

The better question is why would you want to? If you could even manage, the result would be practically incomprehensible. On Wed, Jun 10, 2020 at 07:49 Alexander Chen <alexander@chenjia.nl> wrote:

hi,

--any function foldr myAny'' :: (a-> Bool) -> [a] -> Bool myAny'' f = foldr (\a b -> f a || b) False

this is the foldr notions. How would i make this point free?

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