Haskell: Variable not in scope

Question

I have a code as:

main = interact $ show . maxsubseq . map read . words

maxsubseq :: (Ord a,Num a) => [a] -> (a,[a])
maxsubseq = snd . foldl f ((0,[]),(0,[])) where 
f ((h1,h2),sofar) x = (a,b) where
a = max (0,[]) (h1 + x ,h2 ++ [x]) 
b = max sofar a

But I am getting error:

maxSub.hs:6:17: error: Variable not in scope: h1

maxSub.hs:6:22: error: Variable not in scope: x

maxSub.hs:6:25: error: Variable not in scope: h2 :: [t1]

maxSub.hs:6:32: error: Variable not in scope: x

maxSub.hs:7:9: error: Variable not in scope: sofar :: (t, [t1])

Not able to figure out why??

Any ideas??

Thanks.

Frederick Zhang · Accepted Answer · 2016-09-09 04:26:26Z

16

main = interact $ show . maxsubseq . map read . words

maxsubseq :: (Ord a,Num a) => [a] -> (a,[a])
maxsubseq = snd . foldl f ((0,[]),(0,[])) where
f ((h1,h2),sofar) x = (a,b) where
    a = max (0,[]) (h1 + x ,h2 ++ [x])
    b = max sofar a

Formats really matter in Haskell...

Perhaps this looks better:

main = interact $ show . maxsubseq . map read . words

maxsubseq :: (Ord a,Num a) => [a] -> (a,[a])
maxsubseq = snd . foldl f ((0,[]),(0,[])) where
    f ((h1,h2),sofar) x = (a,b)
        where {
        a = max (0,[]) (h1 + x ,h2 ++ [x]);
        b = max sofar a;
              }

answered Sep 9, 2016 at 4:26

Frederick Zhang

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