为什么此蒙加多代码不键入检查？

Question

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

此代码在“ baby.hs”中引起了可怕的汇编错误：

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？我该如何编译该代码？感谢您的帮助〜

我知道了。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

它成功编译！但是...还有更多问题：

instance Monad (Either a)

使“一个'a monad和我得到'return =正确'...我怎么能得到'return = weled'？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：实例monad（ x-> XA）

根本不编译！

Answer 1

1. 返回应该有类型 forall b. b -> Either a b, 但是，左有类型 forall c. a -> Either a c. 。您可能想在这里。
2. fail 应该有类型 forall b. String -> Either a b, ，但是正确的类型 forall b. b -> Either a b, ， 因此，如果 b=String 这使得 String -> Either a String 这不合适。
3. >>= 应该有类型 Either a b -> (b -> Either a c) -> Either a c 然而 Right x >>= _ = Right x 总是返回类型的值 Either a b, ， 不是 Either a c.
4. Left x >>= f = f x 因为x有类型，所以不起作用 a, ， 但 f 有类型 b -> c.

Answer 2

大多数混乱源于左右事实。仅考虑返回的类型，其单本类型的类型如下：

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

您正在尝试定义一个实例 m = Either a, ，因此返回应该具有类型：

return :: b -> Either a b

您将其定义为左侧，其中具有类型：

Left :: a -> Either a b

注意如何 -> 有所不同。