Question

这个编译

foo ma = case ma of
  [Just a] -> newRVar 0 >>= a

view                :: forall a eff. 
                       M.Map String (Linker Number a eff) -> 
                       String -> 
                       Eff (reactive :: Reactive | eff) Unit
view         m yaml = case parseYAML yaml           of
  Right (View xs)  -> case (flip M.lookup) m <$> xs of
    (as)       -> foo as

这不是

foo ma = case ma of
  Just a -> newRVar 0 >>= a

view                :: forall a eff. 
                       M.Map String (Linker Number a eff) -> 
                       String -> 
                       Eff (reactive :: Reactive | eff) Unit
view         m yaml = case parseYAML yaml           of
  Right (View xs)  -> case (flip M.lookup) m <$> xs of
    (as)       -> foo <$> as

出现以下错误：

 Cannot unify Control.Monad.Eff.Eff (reactive :: Control.Reactive.Reactive | u8437) u8438 with Prelude.Unit.

为什么？

Answer 1

错误消息告诉您，您要Control.Monad.Eff.Eff (reactive :: Control.Reactive.Reactive | u8437) u8438放置一个Unit。

foo是否有类型声明？无论你在view做什么，它都可能无法工作，因为目前缺乏类型类约束推理。

在第二种情况下，类型是：

foo :: forall a eff. Maybe a -> Eff (reactive :: Control.Reactive.Reactive | eff) a

(<$>)的类型是

(<$>) :: forall f a b. (Functor f) => (a -> b) -> f a -> f b

as的类型将是Maybe (Linker Number a eff)，所以如果你开始替换类型，你可以很快看到它出错的地方：

a ~ Maybe (Linker Number a1 eff)
b ~ Eff (reactive :: Control.Reactive.Reactive | eff) (Linker Number a1 eff)
f ~ Maybe

f b结果类型foo <$> as是：

Maybe (Eff (reactive :: Control.Reactive.Reactive | eff) (Linker Number a1 eff))

PureScript Type Error很难理解

1 个答案: