如何定义一个单子变压器，即两个任意单子变压器的组成？

发布于 2025-01-22 12:00:32 字数 478 浏览 2 评论 0原文

我想写一些类似于以下内容的内容：

newtype FooT c d m a = FooT { unFooT :: (c (d m)) a }

instance (MonadTrans c, MonadTrans d) => MonadTrans (FooT c d) where
  lift = FooT . lift . lift

但是，此片段不会编译：

Could not deduce (Monad (d m)) arising from a use of ‘lift’

我明白为什么这不会编译；我们不知道任意变压器的应用d M本身就是一个单月。但是，我不确定最好的方法。有没有干净的方法来制作这样的工作？据推测，如果我可以沿monad（dm）的行添加一个约束在实例声明的左侧，但是我不知道该怎么做，因为<代码> m 不绑定。

原文

I want to write something similar to the following:

newtype FooT c d m a = FooT { unFooT :: (c (d m)) a }

instance (MonadTrans c, MonadTrans d) => MonadTrans (FooT c d) where
  lift = FooT . lift . lift

However, this snippet will not compile:

Could not deduce (Monad (d m)) arising from a use of ‘lift’

I understand why this won't compile; we don't know that the application of an arbitrary transformer d m is itself a monad. However, I'm not sure of the best way to proceed.
Is there a clean way to make something like this work? Presumably it would go through if I could add a constraint along the lines of Monad (d m) to the left-hand-side of the instance declaration, but I don't know how to do so since m is not bound.

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满身野味 2025-01-29 12:00:32

使用endenifiedConstraints GHC扩展程序，这是

{-# LANGUAGE QuantifiedConstraints #-}

instance (MonadTrans c, MonadTrans d, forall m. Monad m => Monad (d m)) =>
         MonadTrans (FooT c d) where
  lift = FooT . lift . lift

约束中的m与m中的m不像lift中的m。量化的约束仅表示它说的内容（“对于任何m :: type - ＆gt; type，如果monad m requient requiend monad（dm）> ”），在lift中，正在与特定m作为参数传递给lift的特定m实例化。因此，提升's m不会逃脱其范围。

With the QuantifiedConstraints GHC extension, this is

{-# LANGUAGE QuantifiedConstraints #-}

instance (MonadTrans c, MonadTrans d, forall m. Monad m => Monad (d m)) =>
         MonadTrans (FooT c d) where
  lift = FooT . lift . lift

m in the constraint is not the same m as in lift. The quantified constraint simply means what it says ("for any m :: Type -> Type, if Monad m require Monad (d m)"), and in lift that universal statement is being instantiated with the particular m being passed as argument to lift. Thus lift's m does not escape its scope.

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陌生 2025-01-29 12:00:32

因为 promensioners 0.6 monadtrans 类型类有一个要求它可以保留它。

这意味着 monadtrans是：

type Lifting cls trans = forall m. cls m => cls (trans m)

class Lifting Monad trans => MonadTrans trans where
  lift :: Monad m => m ~> trans m

变形金刚的组成（composet），您调用foot不需要指定提升，因此代码代码您的问题中提供的版本应适用于 0.6+。

已经存在于 derving-trans 中

newtype Ok m a = Ok (Int -> Bool -> m a)
  deriving
    ( Functor, Applicative, Alternative, Contravariant
    , Monad, MonadPlus, MonadCont, MonadIO, MonadFix,
    , MonadFail, MonadZip
    )
  via ReaderT Int (ReaderT Bool m)

  deriving MonadTrans
  via ComposeT (ReaderT Int) (ReaderT Bool)

Since transformers 0.6 the MonadTrans type class has had a requirement that it preserves Monad.

This means the definition of MonadTrans is:

type Lifting cls trans = forall m. cls m => cls (trans m)

class Lifting Monad trans => MonadTrans trans where
  lift :: Monad m => m ~> trans m

The composition of transformers (ComposeT), which you call FooT doesn't need to specify the lifting, so the code you supplied in your question should work for versions 0.6+.

ComposeT already exists in deriving-trans

newtype Ok m a = Ok (Int -> Bool -> m a)
  deriving
    ( Functor, Applicative, Alternative, Contravariant
    , Monad, MonadPlus, MonadCont, MonadIO, MonadFix,
    , MonadFail, MonadZip
    )
  via ReaderT Int (ReaderT Bool m)

  deriving MonadTrans
  via ComposeT (ReaderT Int) (ReaderT Bool)

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