不完整的类型签名

Question

让我们说我们有一个像f这样的函数,返回一个monad.但是,你看Int,假装它是一个非常复杂的类型.

f :: (Monad m) => m Int -- Pretend this isn't Int but something complicated
f = return 42

现在让我们说我们想强迫它进入Maybemonad.我们不需要编写完整类型f来执行此操作,我们可以执行以下操作:

g :: Maybe a -> Maybe a
g = id

main = print $ (g f)

虚拟函数g强制f成为Maybe.

我认为上面的内容相当混乱.我宁愿写的是这样的:

main = print $ (f :: Maybe a)

但它失败并出现以下错误:

Couldn't match expected type `a' against inferred type `Int'
  `a' is a rigid type variable bound by
      the polymorphic type `forall a. Maybe a' at prog.hs:7:16
  Expected type: Maybe a
  Inferred type: Maybe Int
In the second argument of `($)', namely `(f :: Maybe a)'
In the expression: print $ (f :: Maybe a)

有没有办法以g一种不那么混乱的方式做上面的事情,而不涉及创建一个新的功能？我不想写f :: Maybe Int,因为如果返回类型改变它会成为一个维护问题.GHC扩展在答案中是可以的.

Answer 1

使用asTypeOf.它返回第一个参数,同时将其类型与第二个参数统一.它只是一个类型限制版本const,但对于这种情况很有用.

main = print $ f `asTypeOf` (undefined :: Maybe a)

Answer 2

另一种方法是限制以下类型print:

main = (print :: Show a => Maybe a -> IO ()) f