12 Feb
2007
12 Feb
'07
5:39 p.m.
I thought solution one was missing the ~ ? On Feb 12, 2007, at 22:07 , apfelmus@quantentunnel.de wrote:
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.
"Nice one" is an euphemism, it's exactly solution one :)
Regards, apfelmus
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