Question

动机。我正在尝试创建一个monad变换器，其中包含一条特殊指令f <||> g，意思是“重复包含f <||> g的整个块，一次使用f，下次使用g ”。这可以用于DSL转换，但您可以想象其他应用程序。

示例用法。 computation monad表达了不同的可能选择（在这种情况下，是要打印的东西）。 printme函数说明了如何处理每个不同的结果。在这种情况下，我们在运行之前打印“开始计算”，在之后打印“---”。

computation = do
    lift (print "start -- always")
    (lift (print "first choice") <||> lift (print "second choice"))
    lift (print "intermediate -- always")
    (lift (print "third choice") <||> lift (print "fourth choice"))
    lift (print "end -- always")

printme x = do
    putStrLn "=== start computation"
    xv <- x
    putStrLn "---\n"
    return xv

test = runIndep printme computation

输出如下，

=== start computation
"start -- always"
"first choice"
"intermediate -- always"
"third choice"
"end -- always"
---

=== start computation
"start -- always"
"first choice"
"intermediate -- always"
"fourth choice"
"end -- always"
---

=== start computation
"start -- always"
"second choice"
"intermediate -- always"
"third choice"
"end -- always"
---

=== start computation
"start -- always"
"second choice"
"intermediate -- always"
"fourth choice"
"end -- always"
---

问题。使用某种延续传递方式monad变换器有没有一种干净的方法来实现上述行为？我看过Oleg等人的“Backtracking，Interleaving，Terminating Monad Transformers”论文，但似乎无法完全掌握它们的实现（一旦它们继续实现msplit实现）。 / p>

当前实施。我目前的实现是传递一个分支决策列表。 monad将返回它实际选择的分支列表，然后下次我们将切换最后一个可能的分支。代码如下（应该在7.0.3中运行），

import Control.Monad.Trans.Class

data IndepModelT  α = IndepModelT {
    unIndepModelT :: [Bool] ->  (α, [Bool]) }

instance Monad  => Monad (IndepModelT ) where
    return x = IndepModelT $ \choices -> return (x, [])
    (IndepModelT x) >>= f = IndepModelT $ \choices -> do
        (xv, branches) <- x choices
        let choices' = drop (length branches) choices
        (fxv, branches') <- unIndepModelT (f xv) choices'
        return (fxv, branches ++ branches')

instance MonadTrans IndepModelT where
    lift x = IndepModelT $ \c -> liftWithChoice [] x
liftWithChoice cs mx = mx >>= \xv -> return (xv, cs)

(<||>)
  :: Monad  => IndepModelT  α -> IndepModelT  α -> IndepModelT  α
(IndepModelT f) <||> (IndepModelT g) = IndepModelT go where
    go (False:cs) = do
        (fv, branches) <- f cs
        return (fv, False : branches)
    go (True:cs) = do
        (fv, branches) <- g cs
        return (fv, True : branches)

run_inner next_choices k comp@(IndepModelT comp_inner) = do
    (xv, branches) <- k $ comp_inner next_choices
    case (get_next_choices branches) of
        Nothing -> return ()
        Just choices -> run_inner (choices ++ repeat False) k comp
    where
        get_next_choices [] = Nothing
        get_next_choices [True] = Nothing
        get_next_choices [False] = Just [True]
        get_next_choices (c:cs)
            | Just cs' <- get_next_choices cs = Just $ c:cs'
            | c Prelude.== False = Just [True]
            | otherwise = Nothing

runIndep :: Monad  =>
    ( (α, [Bool]) ->  (β, [Bool]))
    -> IndepModelT  α
    ->  ()
runIndep = run_inner (repeat False)

runIndepFirst (IndepModelT comp) = comp (repeat False)

Answer 1

问题在于：这不是一个单子！这种行为甚至没有明确定义。 F.E.在这种情况下应该怎么做：

do
  b <- ...randomly True or False...
  if b then ...some choices... else ...some other choices...

但是，它是Applicative。我们需要的类型是[IO a]，它是2个applicative functor的组合，因此我们可以使用变换器包中的Data.Functor.Compose。这样也可以免费提供Alternative <|>个实例。我们将使用Rebindable Syntax为Applicatives使用do-notation：

{-# LANGUAGE RebindableSyntax #-}
import Prelude hiding ((>>), (>>=))
import Control.Applicative
import Data.Functor.Compose

lift :: Applicative f => g a -> Compose f g a
lift = Compose . pure

(>>) :: Applicative f => f a -> f b -> f b
(>>) = (*>)

computation :: Alternative f => Compose f IO ()
computation = do
    lift (print "start -- always")
    lift (print "first choice") <|> lift (print "second choice")
    lift (print "intermediate -- always")
    lift (print "third choice") <|> lift (print "fourth choice")
    lift (print "end -- always")

printme x = do
    putStrLn "=== start computation"
    x
    putStrLn "---\n"

test = mapM printme $ getCompose computation

Answer 2

到目前为止你的建议不起作用。这是怎么回事：

f <||> g = ContT $ \k -> do
  xs <- runContT f k
  ys <- runContT g k
  return $ xs ++ ys

test = runContT computation (return . (:[]))

但是这不会重新启动每个选择的整个计算，结果如下：

"start -- always"
"first choice"
"intermediate -- always"
"third choice"
"end -- always"
"fourth choice"
"end -- always"
"second choice"
"intermediate -- always"
"third choice"
"end -- always"
"fourth choice"
"end -- always"

我还没有找到一个好的解决方案。

Answer 3

如果您正在寻找专门针对基于延续的方法，那么您将不会比the LogicT paper中的SFKT成功/失败延续实施简单得多。

如果msplit太多（并且它是一个非常微妙的野兽），你可以忽略它为这个应用程序。它的目的是允许公平的连接和分离，如果那些样本输出线要按顺序打印，这不是您的规范的一部分。只需关注第5.1节中的Monad和MonadPlus实现，您就可以全部设置。

更新：正如Sjoerd Visscher所指出的那样，这是不正确的，因为重新启动只发生在mplus而不是整个计算。这比第一次阅读时出现的问题要复杂得多。

如何实现这个monad变换器的延续？

3 个答案: