我定义了以下非常普通的树类型：
data YTree f g a = YNode (f a (g (YTree f g a)))
我已经定义了函子实例，如下所示（以证明可以使用）：
instance (Bifunctor f, Functor g) => Functor (YTree f g) where
fmap f (YNode m) = YNode $ bimap f (fmap (fmap f)) m
然后，我成功地导出了一些实例（使用StandaloneDeriving，ExplicitForAll和QuantifiedConstraints）：
deriving instance (Show a, forall x y. (Show x, Show y) => Show (f x y), forall z. Show z => Show (g z)) => Show (YTree f g a)
deriving instance (Read a, forall x y. (Read x, Read y) => Read (f x y), forall z. Read z => Read (g z)) => Read (YTree f g a)
deriving instance (Eq a, forall x y. (Eq x, Eq y) => Eq (f x y), forall z. Eq z => Eq (g z)) => Eq (YTree f g a)
但是，以下deriving instance子句在包含时会引发错误：
deriving instance (Ord a, forall x y. (Ord x, Ord y) => Ord (f x y), forall z. Ord z => Ord (g z)) => Ord (YTree f g a)
即
Data\Tree\Generalized.hs:32:1: error:
* Could not deduce (Ord z)
arising from the superclasses of an instance declaration
from the context: (Ord a,
forall x y. (Ord x, Ord y) => Ord (f x y),
forall z. Ord z => Ord (g z))
bound by the instance declaration
at Data\Tree\Generalized.hs:32:1-119
or from: Eq z
bound by a quantified context at Data\Tree\Generalized.hs:1:1
Possible fix: add (Ord z) to the context of a quantified context
* In the instance declaration for `Ord (YTree f g a)'
|
32 | deriving instance (Ord a, forall x y. (Ord x, Ord y) => Ord (f x y), forall z. Ord z => Ord (g z)) => Ord (YTree f g a)
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
以及另一个错误。
据我所见，实现方式很简单：
instance (Ord a, forall x y. (Ord x, Ord y) => Ord (f x y), forall z. Ord z => Ord (g z)) => Ord (YTree f g a) where
compare (YNode x) (YNode y) = compare x y
这会产生相同的错误。
有可能得出这个吗？为什么现在不起作用？甚至有可能实现吗？我错了吗？
答案 0 :(得分：3)
deriving instance
( Ord a, forall x y. (Ord x, Ord y) => Ord (f x y), forall z. Ord z => Ord (g z)
, Eq a, forall x y. (Eq x, Eq y) => Eq (f x y), forall z. Eq z => Eq (g z)
) => Ord (YTree f g a)
会编译，但我不知道为什么我们需要Eq约束。