我偶然发现的问题与>>=
应用程序有关的样本类型有关:
data ThreeArgs a = ThreeArgs a a a deriving (Show,Eq)
instance Functor ThreeArgs where
fmap f (ThreeArgs a b c) = ThreeArgs (f a) (f b) (f c)
instance Applicative ThreeArgs where
pure x = ThreeArgs x x x
(ThreeArgs a b c) <*> (ThreeArgs p s q) = ThreeArgs (a p) (b s) (c q)
我将声明Monad实例如下:
instance Monad ThreeArgs where
return x = ThreeArgs x x x
(ThreeArgs a b c) >>= f = f ... -- a code I need to complete
是,看起来f
似乎将应用于所有三个ThreeArgs
构造函数自变量。如果我完成最后一行
(ThreeArgs a b c) >>= f = f a
然后编译器没有任何抱怨,而结果是:
*module1> let x = do { x <- ThreeArgs 1 2 3; y <- ThreeArgs 4 6 7; return $ x + y }
*module1> x
ThreeArgs 5 5 5
这意味着求和结果将进入具有相同参数值的上下文,尽管正确的输出应为ThreeArgs 5 8 10
。一旦我编辑到
(ThreeArgs a b c) >>= f = (f a) (f b) (f c)
编译器警报:
Couldn't match expected type `ThreeArgs b
-> ThreeArgs b -> ThreeArgs b -> ThreeArgs b'
with actual type `ThreeArgs b'
因此,我看到一个严重的错误引导着我的理解,但是我仍然很难理解单子类以及Haskell中的另一类东西。想必我要在这里使用递归还是其他?
ThreeArgs
与((->) Ordering)
同构。证人:
to :: ThreeArgs a -> Ordering -> a
to (ThreeArgs x _ _) LT = x
to (ThreeArgs _ y _) EQ = y
to (ThreeArgs _ _ z) GT = z
from :: (Ordering -> a) -> ThreeArgs a
from f = ThreeArgs (f LT) (f EQ) (f GT)
您的Functor
和Applicative
实例与((->) r)
的实例匹配,因此我们只需使其与its Monad
one的实例也匹配,就可以完成。
Monad
顺便说一下,如果想要查找更多有关此数据结构的通用术语,例如instance Monad ThreeArgs where
ThreeArgs x y z >>= f = ThreeArgs x' y' z' where
ThreeArgs x' _ _ = f x
ThreeArgs _ y' _ = f y
ThreeArgs _ _ z' = f z
,则为“ representable functor”。
这看起来像是ThreeArgs
但不是Applicative
的示例。请注意,它与标准Monad
类型有点类似,只不过它限于固定大小3的列表。如果查看ZipList
,您会发现ZipList
documentation有一个ZipList
实例,但没有ZipList
]实例。
底线:并非所有事物都是单子。这就是我们拥有应用程序的原因之一-有些有用的对象是应用程序,但不是monad。