Question

newtype MyNewtype1 f v = MyNewtype1 { getVal1 :: f v } deriving Eq -- OK
newtype MyNewtype2 f maybe v = MyNewtype2 { getVal2 :: f (maybe v) } deriving Eq --OK
newtype MyNewtype3 f v = MyNewtype3 { getVal3 :: f (Maybe v) } --  OK
newtype MyNewtype4 f v = MyNewtype4 { getVal4 :: f (Maybe v) } deriving Eq --NOT OK

我有这些新类型。前三个按预期工作，但第四个给出：

    • No instance for (Eq (f (Maybe v)))
        arising from the 'deriving' clause of a data type declaration
      Possible fix:
        use a standalone 'deriving instance' declaration,
          so you can specify the instance context yourself
    • When deriving the instance for (Eq (MyNewtype4 f v))

我无法理解问题出在哪里。在我看来，第二个新类型严格更普遍，因此必须遇到同样的问题。因此，如果MyNewtype2可以导出Eq，为什么MyNewtype2不能？这可能让我最困惑。有人可以向我解释一下吗？另外，如果我想要这样的新类型，首选的解决方案是什么？

Answer 1

我不知道为什么2个有效但4个没有。可能与灵活的上下文有关（我不太清楚，不能解释）。

要回答第二个问题......首选的解决方案，如果你想要一个类似的新类型，就像GHC建议的那样：使用一个独立的派生实例。

{-# LANGUAGE StandaloneDeriving, FlexibleContexts, UndecidableInstances #-}

newtype MyNewtype4 f v = MyNewtype4 { getVal4 :: f (Maybe v) }
deriving instance (Eq (f (Maybe v))) => Eq (MyNewtype4 f v)

Answer 2

The third [generated instance] is not Haskell 98, and risks losing termination of instances. GHC takes a conservative position: it accepts the first two, but not the third. The rule is this: each constraint in the inferred instance context must consist only of type variables, with no repetitions. This rule is applied regardless of flags. If you want a more exotic context, you can write it yourself, using the standalone deriving mechanism.

我认为MyNewType4被认为是不安全的，所以不允许派生，因为它违反了第二个帕特森条件：

2.The constraint has fewer constructors and variables (taken together and counting repetitions) than the head

MyNewtype4 f v有3个，f (Maybe v)也有。 MyNewtype2 f maybe v有4个，所以instance (Eq (f (maybe v))) => MyNewtype2 f maybe v遵守条件（根据开头引用的规则也是如此）。

使用更多类型变量导出newtype

2 个答案: