我在haskell中编写了一个小程序,使用State Monad和Vector来计算Tree中Int值的所有出现次数:
import Data.Vector
import Control.Monad.State
import Control.Monad.Identity
data Tree a = Null | Node (Tree a) a (Tree a) deriving Show
main :: IO ()
main = do
print $ runTraverse (Node Null 5 Null)
type MyMon a = StateT (Vector Int) Identity a
runTraverse :: Tree Int -> ((),Vector Int)
runTraverse t = runIdentity (runStateT (traverse t) (Data.Vector.replicate 7 0))
traverse :: Tree Int -> MyMon ()
traverse Null = return ()
traverse (Node l v r) = do
s <- get
put (s // [(v, (s ! v) + 1)]) -- s[v] := s[v] + 1
traverse l
traverse r
return ()
但是不可变向量的“更新”是在O(n)复杂度下完成的。我正在寻找O(1)中的更新和O(1)中的访问。据我了解,可变向量可以满足我的需求。要使用它们,我需要使用ST或IO。因为我想执行一些UnitTest,所以我更喜欢ST monad,但是我不想在函数调用中传递该向量。我需要继续使用Monad变形金刚,因为我将添加诸如ErrorT和WriterT之类的变形金刚。
问题:如何使用Monad变压器将可变向量置于状态Monad中?
我想出了以下无法编译的代码:
import Data.Vector
import Control.Monad.State
import Control.Monad.Identity
import qualified Data.Vector.Mutable as VM
import Control.Monad.ST
import Control.Monad.ST.Trans
type MyMon2 s a = StateT (VM.MVector s Int) (STT s Identity) a
data Tree a = Null | Node (Tree a) a (Tree a) deriving Show
main :: IO ()
main = do
print $ runTraverse (Node Null 5 Null)
runTraverse :: Tree Int -> ((),Vector Int)
runTraverse t = runIdentity (Control.Monad.ST.Trans.runST $ do
emp <- VM.replicate 7 0
(_,x) <- (runStateT (traverse t) emp)
v <- Data.Vector.freeze x
return ((), v)
)
traverse :: Tree Int -> MyMon2 s ()
traverse Null = return ()
traverse (Node l v r) = do
d <- get
a <- (VM.read d v)
VM.write d v (a + 1)
put d
return ()
编译错误是:
TranformersExample: line 16, column 16:
Couldn't match type `s'
with `primitive-0.5.2.1:Control.Monad.Primitive.PrimState
(STT s Identity)'
`s' is a rigid type variable bound by
a type expected by the context: STT s Identity ((), Vector Int)
at test/ExecutingTest.hs:15:30
Expected type: STT s Identity (MVector s Int)
Actual type: STT
s
Identity
(MVector
(primitive-0.5.2.1:Control.Monad.Primitive.PrimState
(STT s Identity))
Int)
In the return type of a call of `VM.new'
In a stmt of a 'do' block: emp <- VM.new 7
In the second argument of `($)', namely
`do { emp <- VM.new 7;
(_, x) <- (runStateT (traverse t) emp);
v <- freeze x;
return ((), v) }'
TranformersExample: line 26, column 14:
Couldn't match type `s'
with `primitive-0.5.2.1:Control.Monad.Primitive.PrimState
(StateT (MVector s Int) (STT s Identity))'
`s' is a rigid type variable bound by
the type signature for traverse :: Tree Int -> MyMon2 s ()
at test/ExecutingTest.hs:21:13
Expected type: MVector
(primitive-0.5.2.1:Control.Monad.Primitive.PrimState
(StateT (MVector s Int) (STT s Identity)))
Int
Actual type: MVector s Int
In the first argument of `VM.write', namely `d'
In a stmt of a 'do' block: VM.write d v (a + 1)
In the expression:
do { d <- get;
a <- (VM.read d v);
VM.write d v (a + 1);
put d;
.... }
注意:我知道不检查边界。
使用ST
状态时,您永远不会显式传递矢量(始终隐藏在s
参数中),而只是对其的引用。该引用是不可变的,不会被复制,因此您仅需要State
读者隐式地传递它即可。
import Data.Vector
import Control.Monad.Reader
import qualified Data.Vector.Mutable as VM
import Control.Monad.ST
type MyMon3 s = ReaderT (VM.MVector s Int) (ST s)
data Tree a = Null | Node (Tree a) a (Tree a) deriving Show
main :: IO ()
main = do
print $ runTraverse (Node Null 5 Null)
runTraverse :: Tree Int -> Vector Int
runTraverse t = runST $ do
emp <- VM.replicate 7 0
runReaderT (traverse t) emp
Data.Vector.freeze emp
traverse :: Tree Int -> MyMon3 s ()
traverse Null = return ()
traverse (Node l v r) = do
d <- ask
a <- lift $ VM.read d v
lift $ VM.write d v (a + 1)
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