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”中的这段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.

---

为什么会这样？我怎样才能使这段代码编译？ 感谢您的帮助~

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

编译成功！ 但是......还有更多问题：

instance Monad (Either a)

使 'Either a' 成为一个 monad，我得到了 'return = Right'...我怎么能得到 'return = Left'？ 我试过这个但失败了：

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

或： 实例 Monad (\x -> 任意一个 x a)

根本不编译！

Answer 1

大部分混淆源于 Left 和 Right 倒置使用的事实。仅考虑返回的类型，其来自 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

注意-> 左侧的不同之处。

Answer 2

1. return 的类型应为 forall b. b -> Either a b，但 Left 的类型为 forall c. a -> Either a c。您可能想要就在这里。
2. fail 应该有 forall b. String -> Either a b 类型，但是 Right 有 forall b. b -> Either a b 类型，所以如果 b=String 使得 String -> Either a String 不适合。
3. >>= 应该具有 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。
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。