无法使用“Reader”作为字段来“强制”数据类型

She*_*rsh 10 haskell coercion coerce newtype

我有以下 Haskell 代码,可以完美编译:

import Control.Monad.Reader (Reader (..))
import Data.Coerce (Coercible, coerce)

data Flow i o = Flow (i -> o) (o -> i)

coerceFlow
    :: (Coercible i i', Coercible o o')
    => Flow i o
    -> Flow i' o'
coerceFlow = coerce
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但是,如果我将Flow类型的定义更改为以下内容:

data Flow i o = Flow (i -> Reader Int o) (o -> i)
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我开始看到一个奇怪的错误:

Coerce.hs:10:14: error:
    • Couldn't match type ‘o’ with ‘o'’ arising from a use of ‘coerce’
      ‘o’ is a rigid type variable bound by
        the type signature for:
          coerceFlow :: forall i i' o o'.
                        (Coercible i i', Coercible o o') =>
                        Flow i o -> Flow i' o'
        at Coerce.hs:(6,1)-(9,17)
      ‘o'’ is a rigid type variable bound by
        the type signature for:
          coerceFlow :: forall i i' o o'.
                        (Coercible i i', Coercible o o') =>
                        Flow i o -> Flow i' o'
        at Coerce.hs:(6,1)-(9,17)
    • In the expression: coerce
      In an equation for ‘coerceFlow’: coerceFlow = coerce
    • Relevant bindings include
        coerceFlow :: Flow i o -> Flow i' o' (bound at Coerce.hs:10:1)
   |
10 | coerceFlow = coerce
   |              ^^^^^^
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据我了解,我的数据类型不再是Coercible自动的。有没有办法告诉 GHC 我可以Flow自动强制类型值?我可以coerce手动每个字段,但我想coerce一次完成整个数据类型(这是DerivingVia工作所必需的)。

我尝试使用这样的RoleAnnotations扩展:

type role Flow representational representational
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但我看到一个错误:

Coerce.hs:6:1: error:
    • Role mismatch on variable o:
        Annotation says representational but role nominal is required
    • while checking a role annotation for ‘Flow’
  |
6 | type role Flow representational representational
  | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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chi*_*chi 5

我们来调查一下:

> :info Reader
type Reader r = ReaderT r Data.Functor.Identity.Identity :: * -> *
        -- Defined in `Control.Monad.Trans.Reader'
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所以,Reader是根据 来定义的ReaderT

> :info ReaderT
type role ReaderT representational representational nominal
newtype ReaderT r (m :: k -> *) (a :: k)
  = ReaderT {runReaderT :: r -> m a}
        -- Defined in `Control.Monad.Trans.Reader'
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...并且ReaderTnominal在它的第三个参数,导致Readernominal在其第二个参数,使您的威逼失败。您不能使用您的Flow类型的角色注释来颠覆这一点,因为这将处理之前的ReaderT.

现在,您可能想知道为什么ReaderTnominal第三个参数。要理解这一点,请考虑其定义:

newtype ReaderT r m a = ReaderT (r -> m a)
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a上面应该起什么作用?这要看情况。如果m :: * -> *representational在它的论点上,那么ReaderT在 上就是这样a。这同样适用于nominalphantom。在这里表达角色的“最佳”方式是使用角色多态性,如

type role forall r .
     ReaderT representational (representational :: (* with role r) -> *) r
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其中第三个参数的作用取决于第二个更高级的参数。

唉,GHC 不支持像上面那样的角色多态性,所以我们只能使用最严格的角色:nominal.