Haskell 函数定义中的代码缩进问题

Question

我试图在 Haskell 中编写合并排序代码，但出了点问题。我觉得下面的代码应该是正确的

-- UPDATE : COMPLETE PROGRAMM'S CODE
merge xs [] = xs
merge [] ys = ys
merge (x:xs) (y:ys) 
    | x <= y = x:merge xs (y:ys)
    | otherwise = y:merge (x:xs) ys

mergeSort xs
    | length xs < 2 = xs
    | otherwise  = merge (mergeSort (myleft xs)) (mergeSort (myright xs))
       where myleft xs = take (half xs) xs
             myright xs = drop (half xs) xs
             where half = \xs -> div (length xs) 2

因为第一个 where 被称为“否则”行，所以缩进得比那行更远。第二个 where 是指所有第一个“where 块”，因此它应该嵌套在第一个中。编译时它给我这个错误

 Variable not in scope: mergeSort
   |
26 |  mergeSort xs
   |  ^^^^^^^^^

所以函数的定义有问题。我尝试以多种不同的方式修复布局，例如

 mergeSort xs
    | length xs < 2 = xs
    | otherwise  = merge (mergeSort (myleft xs)) (mergeSort (myright xs))
    where myleft xs = take (half xs) xs
          myright xs = drop (half xs) xs
          where half = \xs -> div (length xs) 2

但其中 none 个有效。

我也在这里寻找解决方案，我发现了一些与我正在做的非常接近的东西 --> Haskell Merge Sort 它还提供了正确的代码

mergeSort merge xs
        | length xs < 2 = xs
        | otherwise = merge (mergeSort merge first) (mergeSort merge second)
        where first = take half xs 
              second = drop half xs 
              half = length xs `div` 2

但这为什么有效？是不是需要一秒钟在哪里指定“一半”？为什么不存在？无法理解为什么我的仍然无法正常工作。

Answer 1

在您的代码中 half 仅在 myright 定义中“可见”。

删除第二个 where 有效：

merge :: [a] -> [a] -> [a]
merge = undefined

mergeSort :: [a] -> [a]
mergeSort xs
  | length xs < 2 = xs
  | otherwise  = merge (mergeSort (myleft xs)) (mergeSort (myright xs))
     where myleft xs = take (half xs) xs
           myright xs = drop (half xs) xs
           half = \xs -> div (length xs) 2

Haskell 函数定义中的代码缩进问题

Haskell's code indentation issue in function's definition

layout

mergesort

haskell

syntax-error