j
k
j a
j l
Bernie Pope wrote:
Lennart Augustsson wrote: ...Sure, but we also have para f e xs = snd $ foldr (\ x ~(xs, y) -> (x:xs, f x xs y)) ([], e) xs Nice one.
Lennart Augustsson wrote:
Sure, but we also have para f e xs = snd $ foldr (\ x ~(xs, y) -> (x:xs, f x xs y)) ([], e) xs Nice one.
Sure, but we also have
para f e xs = snd $ foldr (\ x ~(xs, y) -> (x:xs, f x xs y)) ([], e) xs Nice one.
"Nice one" is an euphemism, it's exactly solution one :) Regards, apfelmus
Back to the thread
Back to the list
