为什么这个 Either-monad 代码不进行类型检查？

Question

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

“baby.hs”中的这个代码片段导致了可怕的编译错误：

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.

---

为什么会发生这种情况？我怎样才能让这段代码编译？ 谢谢你的帮助~

我明白了。我调整了代码以查看它的编译：

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

它编译成功！ 但是......对于进一步的问题：

instance Monad (Either a)

使'Either a'成为单子并且我得到'return = Right'...我怎样才能得到'return = Left'？ 我已经尝试过但失败了：

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

或者： instance Monad (\x -> Either xa)

根本无法编译！

Answer 1

大多数混乱源于“左”和“右”被反向使用的事实。仅考虑 return 的类型，它来自 Monad 类型类的类型如下：

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

您正在尝试为 m = Either a 定义一个实例，因此 return 应该具有类型：

return :: b -> Either a b

您将其定义为 Left，其类型为：

Left :: a -> Either a b

请注意 -> 左侧的不同之处。

Answer 2

1. 对于all b，返回值应具有类型。 b->要么是b，但Left 的类型是所有c。一个->要么是 c。您可能想要就在这里。
2. fail 的类型应为 forall b。字符串 ->要么是b，但是Right 的类型是所有b。 b->要么是 b，所以如果 b=String 使得 String ->;要么是一个不适合的字符串。
3. >>= 的类型应为 ab ->; (b -> 任一 ac) ->要么是 c，要么是 Right x >>>= _ = Right x 始终返回 Either a b 类型的值，而不是Either a c >。
4. 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.