
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,

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
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

Hi Alexander, I found this tool online[1], that converts your function to a point free style. myAny'' f = foldr (\a b -> f a || b) False is transformed to myAny'' = flip foldr False . ((||) .) Again as Bob mentions, the point free style comes at a cost of unreadability and hence unmaintainable. Cheers! Apoorv [1]: http://pointfree.io/ http://pointfree.io/
On Jun 10, 2020, at 10:00, Bob Ippolito
wrote: 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
mailto: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 mailto:Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
participants (3)
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Alexander Chen
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Apoorv Ingle
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Bob Ippolito