需要 MonadPlus (ST a) 实例
我正在阅读论文 Haskell 中的类型化逻辑变量< /em>,但我无法理解最终实现的细节。特别是第 4 节中介绍的回溯状态转换器。出于某种我不知道的原因,GHC 认为我需要函数 (ST a)
的 MonadPlus
实例。 code>unify,如下:
newtype BackT m a = BT { run :: forall b . (a -> m [b]) -> m [b] }
instance (Monad m) => Monad (BackT m) where
return a = BT (\k -> k a)
BT m >>= f = BT (\k -> m (\a -> run (f a) k))
instance (MonadPlus m) => MonadPlus (BackT m) where
mzero = BT (\s -> mzero)
f `mplus` g = BT (\s -> (run f) s `mplus` (run g) s)
type LP a = BackT (ST a)
type LR = STRef
type Var s a = LR s (Maybe a)
data Atom s = VarA (Var s (Atom s)) | Atom String
class Unify b a | a -> b where
var :: a -> Maybe (Var b a)
unify :: a -> a -> LP s ()
instance Unify s (Atom s) where
var (VarA a) = Just a
var _ = Nothing
unify (Atom a) (Atom b) | a == b = return () -- type checks
unify _ _ = mzero -- requires MonadPlus (ST a) instance
我不确定问题是什么以及如何解决它。到目前为止,我的印象是我理解了前面的讨论和代码,但显然我错了。如果有人可以指出出了什么问题 - 我是否需要 MonadPlus (ST a)
实例? - 这会很有帮助。
[编辑:澄清]我应该指出作者似乎声称mzero
,或者mzero
的一些变体, 是适当的函数。我只是不知道合适的功能是什么。我想知道我是否应该创建一个 MonadPlus (ST a)
实例,或者我没有使用正确的函数,并且误读了一些内容。
I'm reading the paper Typed Logical Variables in Haskell, but I'm failing to understand the details of the ultimate implementation. In particular, the backtracking state transformer introduced in section 4. For some reason, unknown to me, GHC believes I require a MonadPlus
instance for (ST a)
in the function unify
, below:
newtype BackT m a = BT { run :: forall b . (a -> m [b]) -> m [b] }
instance (Monad m) => Monad (BackT m) where
return a = BT (\k -> k a)
BT m >>= f = BT (\k -> m (\a -> run (f a) k))
instance (MonadPlus m) => MonadPlus (BackT m) where
mzero = BT (\s -> mzero)
f `mplus` g = BT (\s -> (run f) s `mplus` (run g) s)
type LP a = BackT (ST a)
type LR = STRef
type Var s a = LR s (Maybe a)
data Atom s = VarA (Var s (Atom s)) | Atom String
class Unify b a | a -> b where
var :: a -> Maybe (Var b a)
unify :: a -> a -> LP s ()
instance Unify s (Atom s) where
var (VarA a) = Just a
var _ = Nothing
unify (Atom a) (Atom b) | a == b = return () -- type checks
unify _ _ = mzero -- requires MonadPlus (ST a) instance
I'm unsure what the problem is, and how to fix it. I was under the impression that I understood the preceding discussion and code until this point, but apparently I was mistaken. If someone could point out what's going awry - do I need a MonadPlus (ST a)
instance or not? - it would be very helpful.
[EDIT: Clarification] I should have pointed out that the authors appear to claim that mzero
, or some variation on mzero
, is the appropriate function. I just don't know what the appropriate function is. What I'm wondering is whether I am supposed to make a MonadPlus (ST a)
instance or I'm not using the correct function, and have misread something.
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mzero
是类型类MonadPlus
的成员。特别是,用于您的函数
unify
的 monad 是LP
,它实际上是用ST
实例化BackT
的。您还可以为BackT
定义一个MonadPlus
实例,它依赖于底层 monad 的此类实例。由于ST
没有这样的实例,GHC 会嘲笑你。这是重要的部分:
用简单的英语来说:这是
BackT m
的MonadPlus
实例,前提是m
也是MonadPlus
的实例。代码>MonadPlus。由于m
是使用ST
实例化的,因此您需要该ST
实例。我想知道如何在没有委托的情况下定义一个合理的MonadPlus
实例。我有一个想法:这个实例基本上连接了两个输出列表。我希望它适合您的需求。
mzero
is a member of the typeclassMonadPlus
. In particularThe monad that is used for your function
unify
isLP
, which is actuallyBackT
instantiated withST
. You furthermore define an instance ofMonadPlus
forBackT
, that depends on such an instance for the underlying monad. SinceST
has no such instance, GHC mocks you.This is the important part:
In plain english: This is an instance of
MonadPlus
forBackT m
, provided thatm
is also an instance ofMonadPlus
. Sincem
is instanciated withST
, you need that instance forST
. I wonder how you could define a sensible instance ofMonadPlus
without delegation. I have an idea:This instance basically concatenates the two output lists. I hope it suits your needs.