Haskell：变量不在范围内答案

【问题标题】：Haskell: Variable not in scopeHaskell：变量不在范围内
【发布时间】：2016-09-09 04:17:58
【问题描述】：

我有一个代码：

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

但我收到错误：

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])

不知道为什么？？

有什么想法吗？？

谢谢。

【问题讨论】：

标签： haskell

【解决方案1】：

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

格式在 Haskell 中真的很重要...

也许这样看起来更好：

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;
              }

【讨论】：