Question

instance Monad (Either a) where
     return = Left
     fail = Right
     Left x >>= f = f x
     Right x >>= _ = Right x

'baby.hs'中的这段代码片段造成了可怕的编译错误：

Prelude> :l baby
[1 of 1] Compiling Main             ( baby.hs, interpreted )

baby.hs:2:18:
Couldn't match expected type `a1' against inferred type `a'
  `a1' is a rigid type variable bound by
       the type signature for `return' at <no location info>
  `a' is a rigid type variable bound by
      the instance declaration at baby.hs:1:23
In the expression: Left
In the definition of `return': return = Left
In the instance declaration for `Monad (Either a)'

baby.hs:3:16:
Couldn't match expected type `[Char]' against inferred type `a1'
  `a1' is a rigid type variable bound by
       the type signature for `fail' at <no location info>
  Expected type: String
  Inferred type: a1
In the expression: Right
In the definition of `fail': fail = Right

baby.hs:4:26:
Couldn't match expected type `a1' against inferred type `a'
  `a1' is a rigid type variable bound by
       the type signature for `>>=' at <no location info>
  `a' is a rigid type variable bound by
      the instance declaration at baby.hs:1:23
In the first argument of `f', namely `x'
In the expression: f x
In the definition of `>>=': Left x >>= f = f x

baby.hs:5:31:
Couldn't match expected type `b' against inferred type `a'
  `b' is a rigid type variable bound by
      the type signature for `>>=' at <no location info>
  `a' is a rigid type variable bound by
      the instance declaration at baby.hs:1:23
In the first argument of `Right', namely `x'
In the expression: Right x
In the definition of `>>=': Right x >>= _ = Right x
Failed, modules loaded: none.

为什么会这样？我怎么能编译这个代码？谢谢你的帮助〜

我明白了。我调整了代码，看它编译：

instance Monad (Either a) where
     return = Right
     Left a >>= f = Left a
     Right x >>= f = f x

它编译成功！但是......还有一个问题：

instance Monad (Either a)

使'要么'一个单子而且我得到'返回=正确'......我怎么能得到'return = Left'？我试过这个却失败了：

instance Monad (`Either` a) where
     return = Left
     Right a >>= f = Right a
     Left x >>= f = f x

或：实例Monad（\ x - ＆gt;或者x a）

根本不编译！

Answer 1

大多数混淆源于左和右向后使用的事实。仅考虑返回的类型，Monad类型类中的类型如下：

return :: (Monad m) => b -> m b

您正在尝试为m = Either a定义一个实例，因此返回应该有类型：

return :: b -> Either a b

您将其定义为Left，其类型为：

Left :: a -> Either a b

请注意->的左侧如何不同。

Answer 2

return应该有forall b. b -> Either a b类型，但是Left的类型为forall c. a -> Either a c。你可能想要在这里。
fail应该有forall b. String -> Either a b类型，但是右键有forall b. b -> Either a b类型，所以如果b=String使String -> Either a String不合适。
>>=应该有Either a b -> (b -> Either a c) -> Either a c类型，但Right x >>= _ = Right x始终返回Either a b类型的值，而不是Either a c。
Left x >>= f = f x不起作用，因为x的类型为a，但f的类型为b -> c。

为什么这个Either-monad代码没有打字？

2 个答案: