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 …