通过量化约束,我可以推导出 Eq (A f)
正好?但是,当我尝试导出 Ord (A f) 时,它失败了。当约束类具有父类(super class)时,我不明白如何使用量化约束。我如何得出 Ord (A f)
和其他具有父类(super class)的类?
> newtype A f = A (f Int)
> deriving instance (forall a. Eq a => Eq (f a)) => Eq (A f)
> deriving instance (forall a. Ord a => Ord (f a)) => Ord (A f)
<interactive>:3:1: error:
• Could not deduce (Ord a)
arising from the superclasses of an instance declaration
from the context: forall a. Ord a => Ord (f a)
bound by the instance declaration at <interactive>:3:1-61
or from: Eq a bound by a quantified context at <interactive>:1:1
Possible fix: add (Ord a) to the context of a quantified context
• In the instance declaration for 'Ord (A f)'
PS。我还检查了ghc proposals 0109-quantified-constraints .使用 ghc 8.6.5
最佳答案
问题是 Eq
是 Ord
的父类(super class), 和约束 (forall a. Ord a => Ord (f a))
不包含父类(super class)约束 Eq (A f)
这是声明 Ord (A f)
所必需的实例。
(forall a. Ord a => Ord (f a))
Eq (A f)
,即 (forall a. Eq a => Eq (f a))
,我们所拥有的并不暗示这一点。 解决方法:添加
(forall a. Eq a => Eq (f a))
到Ord
实例。(我实际上不明白 GHC 给出的错误信息与问题有何关系。)
{-# LANGUAGE QuantifiedConstraints, StandaloneDeriving, UndecidableInstances, FlexibleContexts #-}
newtype A f = A (f Int)
deriving instance (forall a. Eq a => Eq (f a)) => Eq (A f)
deriving instance (forall a. Eq a => Eq (f a), forall a. Ord a => Ord (f a)) => Ord (A f)
或者更整洁一点:
{-# LANGUAGE ConstraintKinds, RankNTypes, KindSignatures, QuantifiedConstraints, StandaloneDeriving, UndecidableInstances, FlexibleContexts #-}
import Data.Kind (Constraint)
type Eq1 f = (forall a. Eq a => Eq (f a) :: Constraint)
type Ord1 f = (forall a. Ord a => Ord (f a) :: Constraint) -- I also wanted to put Eq1 in here but was getting some impredicativity errors...
-----
newtype A f = A (f Int)
deriving instance Eq1 f => Eq (A f)
deriving instance (Eq1 f, Ord1 f) => Ord (A f)
关于haskell - 使用量化约束导出 Ord (forall a. Ord a => Ord (f a)),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59743170/