Haskell quicksort:用'where'解析错误

Question

我刚刚开始教自己Haskell出自"为了好的方面了解你一个好的东西"一书,我在第5章中使用where以下内容重写了快速排序:

quicksort :: (Ord a) => [a] -> [a]
quicksort [] = []
quicksort (x:xs) = smaller ++ [x] ++ bigger
    where smaller = quicksort [a | a <- xs, a <= x]
      bigger = quicksort [a |a <- xs, a > x]

但是当我将它加载到GHCi 7.0.3中时,我收到以下错误:

parse error on input '='

书上的原始代码:

quicksort :: (Ord a) => [a] -> [a]  
quicksort [] = []  
quicksort (x:xs) =   
    let smallerSorted = quicksort [a | a <- xs, a <= x]  
        biggerSorted = quicksort [a | a <- xs, a > x]  
    in  smallerSorted ++ [x] ++ biggerSorted

你能帮我找到它为什么不能用where吗？

Answer 1

那是空白规则.您的定义where必须在同一个空格缩进处.这将编译:

quicksort :: (Ord a) => [a] -> [a]
quicksort [] = []
quicksort (x:xs) = smaller ++ [x] ++ bigger
    where smaller = quicksort [a | a <- xs, a <= x]
          bigger = quicksort [a |a <- xs, a > x]

你可能想读这个.