Question

我在翻译do声明时遇到了一些麻烦：

ex10 :: [Int]
ex10 = do
    num <- [1..20]
    guard (even num)
    guard (num `mod` 3 == 0)
    return num

采用某种monad语法，如下所示：

ex10' :: [Int]
ex10' = [1..20] >>= (guard . even) >>= (guard . (==0) . (mod 3)) >>= \r -> return r

这不起作用，我不太清楚为什么。我想我可能误解了guard，但不确定以什么方式。

Answer 1

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MonadPlus m => Bool -> m ()

收到(guard . (==0) . (mod 3))类型的值，这不是您想要的。在()表示法

do

被贬低为类似

的东西

do
  guard (even num)
  guard (num `mod` 3 == 0)

因此您可以将guard (even num) >> guard (num `mod` 3 == 0)实施为

ex10'

Desugaring'do'块

1 个答案: