我如何在Haskell中将monad任意数目加在一起？

Question

Let's say I have

x :: Monad a => [a]
x = [item1, item2, item3, ...]

And I want to do for these:

y f = f <$> item1 <*> item2 <*> item3 <*> item4 <*> ...

Previously I've tried this for doing it with ZipLists:

zipf' x (y:z) = getZipList $ foldl (<*>) (x <$> y) z

But it unfortunately returned this:

<interactive>:94:36: error:
• Occurs check: cannot construct the infinite type: a1 ~ a -> a1
  Expected type: ZipList a1 -> ZipList a -> ZipList a1
    Actual type: ZipList (a -> a1) -> ZipList a -> ZipList a1
• In the first argument of ‘foldl’, namely ‘(<*>)’
  In the second argument of ‘($)’, namely ‘foldl (<*>) (x <$> y) z’
  In the expression: getZipList $ foldl (<*>) (x <$> y) z
• Relevant bindings include
    z :: [ZipList a] (bound at <interactive>:94:12)
    y :: ZipList a (bound at <interactive>:94:10)
    x :: a -> a1 (bound at <interactive>:94:7)
    zipf' :: (a -> a1) -> [ZipList a] -> [a1]
      (bound at <interactive>:94:1)

Answer 1

The type

x :: Monad a => [a]

is impossible: you cannot have such a thing, because Monad needs a type parameter. You probably mean

x :: Monad m => [m a]

After that, the next problem is: what is the type of y ? It takes a function f , and that f somehow accepts any number of arguments? That doesn't sound right, which is why GHC is complaining about the infinite type: f 's type must contain as a subterm the type of f itself.

Perhaps you mean something more like foldM ? Whether or not foldM is what you need after all, your search will be easier once you nail down the correct types of the functions you're after.