单变压器MaybeT中函数“返回”的两个定义
MaybeT被定义为
newtype MaybeT m a = MaybeT { runMaybeT :: m (Maybe a) }而且，MaybeT m也是Monad类的一个实例，函数return定义如下
return = MaybeT . return . Just但是我读到了"It would also have been possible (though arguably less readable) to write return = MaybeT . return . return"，这让我很困惑。
return = MaybeT . return . return如何等于return = MaybeT . return . Just？
回答 2
Stack Overflow用户
发布于 2014-11-08 22:54:13
在Haskell中，像return这样的函数是多态的。如果检查return的类型，就会发现它是return :: Monad m => a -> m a，这表明它适用于任何monad。事实证明，Maybe是一个monad，因此必须(在标准库中的某个地方)有一个instance声明，如
instance Monad Maybe where
return = ...
...而且，事实证明，return用于Maybe的实例定义是
instance Monad Maybe where
return = Just这说明了为什么允许用return替换Just，但这并不能解释为什么Haskell会这样做。
结果是，Haskell使用类型推断来决定多态函数的“实际”类型是什么。因此，要明确的是，在您的示例中发生的是，Haskell能够识别出最右边的return必须返回一个以Maybe包装的值，从而知道如何将return专门化为return :: a -> Maybe a，然后使用return的instance Monad Maybe定义并将其转化为Just。
Stack Overflow用户
发布于 2014-11-08 18:57:39
考虑一下MaybeT的定义：
newtype MaybeT m a = MaybeT { runMaybeT :: m (Maybe a) }第一个return定义(自下而上阅读)：
return =
MaybeT . -- put `m (Maybe a)` into `MaybeT`: MaybeT (m (Maybe a))
return . -- put `Maybe a` into external monad: m (Maybe a)
Just -- put value into `Maybe`: Maybe aMaybe也是一个单曲。是return是Just。这就是为什么第二MaybeT的return定义了完全相同的函数：
return =
MaybeT . -- put `m (Maybe a)` into `MaybeT`: MaybeT (m (Maybe a))
return . -- put `Maybe a` into external monad: m (Maybe a)
return -- it's the same as `Just` for `Maybe`: Maybe ahttps://stackoverflow.com/questions/26820186
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