为什么这个任意单子代码不能输入检查？

Question

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

在 'baby.hs' 这个代码断枝造成可怕的编译错误：

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. 

---

为什么出现这种情况？我怎么能让这个代码编译？ 感谢您的任何帮助〜

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......再更多的问题：

instance Monad (Either a) 

使得“无论是”一个单子，我得到了“回报=右” ...我怎么能得到“回报=左”？ 我已经尝试过这一点，但失败：

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

或： 实例单子（\ X - >中X一）

不编译了！

Answer 1

1. 返回应该有类型forall b. b -> Either a b，但左有类型forall c. a -> Either a c。你可能想在这里。
2. fail应该有类型forall b. String -> Either a b，但是右键有forall b. b -> Either a b，所以如果b=String那使得String -> Either a String哪个不适合。
3. >>=的类型应该是Either a b -> (b -> Either a c) -> Either a c然而Right x >>= _ = Right x总是返回类型Either a b，不Either a c的值。
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。

Answer 2

大部分的困惑都来自左侧和右侧的事实。如果仅考虑回国的类型，从单子类型类的类型如下：

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

你试图定义m = Either a一个实例，所以返回的类型应该是：

return :: b -> Either a b 

你将其定义为左，其类型为：

Left :: a -> Either a b 

注意如何->不同的左手边。