Haskell - 如何将两个单子(monad) Maybe 函数组合成一个函数

Question

假设我有这两个数据记录 X 和 Y 以及以下两个函数:

f1 :: IO (Maybe [X])
f2 :: X -> IO (Maybe Y)

我需要首先调用f1，然后对于返回列表的每个元素(存储在IO(可能)中)调用f2 >，这会导致类似 IO (Maybe [IO (Maybe Y)]) 的结果。我怎样才能将它们组合起来以获得有意义的东西，例如

result :: Maybe (IO [Y])

或

result :: IO (Maybe [Y])

？

非常感谢您的帮助:-)

Answer 1

本质上它正在使用 fmap sequenceA . sequenceA . fmap f2 。

使用do语法并一次一步地分解它:

result和result'给你相同的结果。

data X = X
data Y = Y

f1 :: IO (Maybe [X])
f1 = undefined

f2 :: X -> IO (Maybe Y)
f2 = undefined

result :: IO (Maybe [Y])
result = do
 f1' <- f1
 case f1' of
   Just f1'' -> do
     let a = fmap f2 f1'' :: [IO (Maybe Y)]
     let b = sequenceA a :: IO [Maybe Y]
     fmap sequenceA b :: IO (Maybe [Y])
   Nothing -> pure Nothing

result' :: IO (Maybe [Y])
result' = do
 f1' <- f1
 case f1' of
   Just f1'' -> do
     fmap sequenceA . sequenceA . fmap f2 $ f1''
   Nothing -> pure Nothing