Question

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定义为

return  = MaybeT . return . Just

但我读到了"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？

Answer 1

考虑MaybeT的定义：

newtype MaybeT m a = MaybeT { runMaybeT :: m (Maybe a) }

第一个return定义（从下到上读取）：

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 a

Maybe也是一个单子。它的return是Just。这就是为什么第二个MaybeT return定义完全相同的功能的原因：

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 a

Answer 2

在Haskell中，像return这样的函数是多态的。如果你检查return的类型，你会发现它是return :: Monad m => a -> m a，表明它适用于任何monad。事实证明，Maybe是一个monad，所以必须（标准库中的某个地方）instance声明

instance Monad Maybe where
  return = ...
  ...

并且，事实证明return Maybe的实例定义是

instance Monad Maybe where
  return = Just

这表明我们允许用return替换Just的原因，但它并没有解释为什么Haskell实际上会这样做。

事实证明，Haskell使用类型推断来决定多态函数的“实际”类型。所以，要清楚，你的例子中发生的事情是Haskell能够识别最右边的return必须返回Maybe中包含的值，因此知道return专门化return :: a -> Maybe a 1}}然后它使用instance Monad Maybe的{{1}}定义并将其转换为return。

在monad变换器中定义功能“返回”MaybeT

2 个答案: