Question

你能告诉我我哪里错了吗？我从单子开始，我真的不明白错误。它似乎正在运作

myforM :: (Monad m) => [a] -> (a -> m b) -> m [b]
myforM [] _ = return []
myforM (x:xs) f = (f x) >>= (\r -> myforM xs >>= (\rs -> return (r:rs)))

第一个错误：

 m.hs:110:36:
        Couldn't match type ‘m’ with ‘(->) (a -> [b0])’
          ‘m’ is a rigid type variable bound by
              the type signature for
                myforM :: Monad m => [a] -> (a -> m b) -> m [b]
              at monady1.hs:108:11
        Expected type: m [b]
          Actual type: (a -> [b0]) -> [[b0]]
        Relevant bindings include
          f :: a -> m b (bound at monady1.hs:110:15)
          xs :: [a] (bound at monady1.hs:110:11)
          x :: a (bound at monady1.hs:110:9)
          myforM :: [a] -> (a -> m b) -> m [b] (bound at monady1.hs:109:1)
        Possible cause: ‘myforM’ is applied to too many arguments
        In the first argument of ‘(>>=)’, namely ‘myforM xs’
        In the expression: myforM xs >>= (\ rs -> return (r : rs))

第二个错误：

   m.hs:110:36:
            Couldn't match type ‘b’ with ‘[b0]’
              ‘b’ is a rigid type variable bound by
                  the type signature for
                    myforM :: Monad m => [a] -> (a -> m b) -> m [b]
                  at monady1.hs:108:11
            Expected type: m [b]
              Actual type: (a -> [b0]) -> [[b0]]
            Relevant bindings include
              r :: b (bound at monady1.hs:110:31)
              f :: a -> m b (bound at monady1.hs:110:15)
              myforM :: [a] -> (a -> m b) -> m [b] (bound at monady1.hs:109:1)
            Possible cause: ‘myforM’ is applied to too many arguments
            In the first argument of ‘(>>=)’, namely ‘myforM xs’
            In the expression: myforM xs >>= (\ rs -> return (r : rs))
        Failed, modules loaded: none.

Answer 1

Dagger2

类型myforM xs >>= ...。你在这里遗漏了(a -> m b) -> m [b]。该错误有点误导：它包含问题，但不是正确的原因：

自己执行forM，我找不到错误

1 个答案: