在Haskell中将函数从LET迁移到WHERE

为了提高我的Haskell技能，我决定阅读一些示例代码，并尝试以不同的方式重写

以下是初始函数：

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  

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

---

非常感谢

您需要匹配

where

子句中表达式的缩进，例如

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

快速排序：：（Ord a）=>[a]->[a]
快速排序[]=[]
快速排序（x:xs）=小排序+++[x]++大排序
其中smallerSorted=快速排序[a | a[a]
快速排序[]=[]
快速排序（x:xs）=小排序+++[x]++大排序
哪里
smallerSorted=quicksort[a | a通常在寻求帮助时，你应该说明你认为它不起作用的原因。如果代码没有编译，你应该粘贴错误消息。如果代码生成错误的输出，你应该给我们输入和输出示例，并告诉我们你认为输出应该是什么。
quicksort :: (Ord a) => [a] -> [a]  
quicksort [] = []  
quicksort (x:xs) = smallerSorted ++ [x] ++ biggerSorted 
    where smallerSorted = quicksort [a | a <- xs, a <= x]  
          biggerSorted = quicksort [a | a <- xs, a > x]  

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