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?

best,
_______________________________________________
Beginners mailing list
Beginners@haskell.org
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners