Question

我的GADT定义如（缩写），

{-# LANGUAGE StandaloneDeriving #-}
data D t where
    C :: t -> D t
    R :: D b -> D (Either a b)
deriving instance Show t => Show (D t)

编译器正确地抱怨它不能为R派生Show。在这种情况下，我们有（或者a b）是Show类，但是如果f是Show类，我们就不知道这是真的。错误消息是，

Could not deduce (Show b) from the context (t ~ Either a b)
  arising from a use of `showsPrec' at tmp.hs:37:0-37
Possible fix: add (Show b) to the context of the constructor `R'
In the second argument of `(.)', namely `(showsPrec 11 b1)'
In the second argument of `showParen', namely
    `((.) (showString "R ") (showsPrec 11 b1))'
In the expression:
    showParen ((a >= 11)) ((.) (showString "R ") (showsPrec 11 b1))
When typechecking a standalone-derived method for `Show (D t)':
  showsPrec a (C b1)
              = showParen ((a >= 11)) ((.) (showString "C ") (showsPrec 11 b1))
  showsPrec a (R b1)
              = showParen ((a >= 11)) ((.) (showString "R ") (showsPrec 11 b1))

似乎需要能够对存在主义类型进行限定，对存在类型b说“显示b =＆gt;显示（D（任一ab））”，或升级暗示“（显示a，显示b）=＆gt;显示（任一ab）“因此它是双向的。

非常感谢！

（请随时清理标题或说明。）

Answer 1

将Show约束添加到存在主义

data D t where
    C :: t -> D t
    R :: Show b => D b -> D (Either a b)

你正在做生意。

Prelude> :r
[1 of 1] Compiling A                ( A.hs, interpreted )
Ok, modules loaded: A.
*A> R (C 7)
R (C 7)

Haskell - 双向类实例类型含义还是GADT存在类型资格？

1 个答案: