haskell - 用do-notation重写haskell列表理解

Question

我已阅读 Learn you a Haskell ，Haskell 中的列表推导可以重写为 monadic join 或(实际上是相同的)do-notation。

但是，当我尝试重写以下代码时(生成所有可能的列表，其中每个元素都来自给定列表之一):

c :: [[a]] -> [[a]]
c []     = [[]]
c (x:xs) = [a:b | a <- x, b <- c xs]

以这样的方式:

d :: [[a]] -> [[a]]
d []     = [[]]
d (x:xs) = do
              a <- x
              b <- d xs
              return a:b

我收到以下错误:

Couldn't match type `a' with [a]
    `a' is a rigid type variable bound by
        the type signature for d :: [[a]] -> [[a]] 
Expected type: [[a]]
  Actual type: [a] 
In the second argument of `(:)', namely `b' 
In a stmt of a 'do' block: return a : b

如果我更改 do 的最后一行对此: return a:[b] ，我没有得到错误，但结果显然是不正确的:

ghci> c [[1, 2], [3, 4]] 
[[1,3],[1,4],[2,3],[2,4]]

ghci> d [[1, 2], [3, 4]]
[[1],[3],[1],[],[1],[4],[1],[],[2],[3],[2],[],[2],[4],[2],[]]

所以问题是:

我怎样才能重写这个列表理解？

列表理解和 do-notation 是否可以互换，一般如何替换？

Answer 1

仔细查看错误信息:

Couldn't match type `a' with [a]    `a' is a rigid type variable bound by        the type signature for d :: [[a]] -> [[a]] Expected type: [[a]]  Actual type: [a] 
In the second argument of `(:)', namely `b' In a stmt of a 'do' block: return a : b

This means that it was parsed as

(return a) : b

因此 b成为 (:) 的第二个参数那里;但你打算这样做

return (a : b)