英文:
How to map a function over multiple constructor arguments working with Monad class in Haskell?
问题
以下是您要翻译的内容：
我卡住的问题涉及到像这样对`>>=`应用到一个示例类型的情况：
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)
我会像下面这样声明一个Monad实例：
instance Monad ThreeArgs where
return x = ThreeArgs x x x
(ThreeArgs a b c) >>= f = f ... -- 我需要完成的代码
是的，看起来`f`要应用到所有三个`ThreeArgs`构造函数参数上。如果我完成最后一行
(ThreeArgs a b c) >>= f = f a
那么编译器不会有任何投诉，但结果是：
*module1> let x = do { x <- ThreeArgs 1 2 3; y <- ThreeArgs 4 6 7; return $ x + y }
*module1> x
ThreeArgs 5 5 5
这意味着求和结果是一个具有相同参数值的上下文，尽管正确的输出应该是`ThreeArgs 5 8 10`。一旦我编辑成
(ThreeArgs a b c) >>= f = (f a) (f b) (f c)
编译器会警告：
无法匹配预期类型`ThreeArgs b
-> ThreeArgs b -> ThreeArgs b -> ThreeArgs b'
实际类型是`ThreeArgs b'
所以，我看到一个严重的错误指导了我的理解，但对于Haskell中的monadic类和其他类似的东西来说，对我来说仍然相当困难。我可能想在这里使用递归还是其他什么？
英文:
The problem I'm stumbled at has to do with >>= application to a sample type like that:
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)
I'd declare a Monad instance as follows:
instance Monad ThreeArgs where
return x = ThreeArgs x x x
(ThreeArgs a b c) >>= f = f ... -- a code I need to complete
Yes, it looks as if f to be applied to all three ThreeArgs contructor arguments. If I complete last line
(ThreeArgs a b c) >>= f = f a
then compiler doesn't have any complaints, whereas result is:
*module1> let x = do { x <- ThreeArgs 1 2 3; y <- ThreeArgs 4 6 7; return $ x + y }
*module1> x
ThreeArgs 5 5 5
it means that summation results into a context with same argument values, although correct output should be ThreeArgs 5 8 10. Once I edit to
(ThreeArgs a b c) >>= f = (f a) (f b) (f c)
compiler alerts:
Couldn't match expected type `ThreeArgs b
-> ThreeArgs b -> ThreeArgs b -> ThreeArgs b'
with actual type `ThreeArgs b'
So, I see a serious mistake guides my comprehension, but it's still rather hard to me to understand monadic class and another such things in Haskell. Presumably, do I want to use recursion here or what else?
答案1
得分: 7
ThreeArgs与((->) Ordering)同构。证明如下：
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)
你的Functor和Applicative实例与((->) r)的工作方式相匹配，所以我们只需要使其与其Monad实例的工作方式相匹配，就完成了。
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
顺便说一下，类似于ThreeArgs这样的数据结构的通用术语是"可表示函子"，如果你想查找更多相关信息，可以搜索这个术语。
英文:
ThreeArgs is isomorphic to ((->) Ordering). Witness:
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)
Your Functor and Applicative instances match how the ones for ((->) r) work, so we can just make it match how its Monad one works too and we're done.
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
By the way, the general term for data structures like ThreeArgs is "representable functor", if you want to look up more about this.
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