作为一项学习练习，我正在尝试在Haskell中实现一个heapsort。我认为State monad是正确的选择，因为堆积非常依赖于在单个结构内部移动数据（并且do符号会很有用）。此外，我希望能够巩固我对monad的理解。
了解你一个Haskell 中的examples on the State monad（以及number other的tutorials），说State是定义为：
newtype State s a = State { runState :: s -> (a,s) }
我应该将s -> (a,s)类型的函数（在其他参数中可能会或可能不会被传递）传递给State值构造函数。所以我的函数看起来像这样：
pop :: Ord a => State (Heap a) a
pop = State pop'
pop' :: Ord a => Heap a -> (a, Heap a)
-- implementation of pop' goes here
push :: Ord a => a -> State (Heap a) ()
push item = State $ push' item
push' :: Ord a => a -> Heap a -> ((), Heap a)
-- implementation of push' goes here
这不能编译，出现以下错误：
Not in scope: data constructor `State'
Perhaps you meant `StateT' (imported from Control.Monad.State)
通过阅读Control.Monad.State的{{3}}，看起来State值构造函数已从模块中删除，因为编写了这些教程。作为初学者，我发现文档远非不言自明。所以我的问题是：
State值构造函数已经消失了？答案 0 :(得分：15)
答案 1 :(得分：13)
以下是有关State Monad的书中所示示例的正确实现：
MONADIC STACK：
-- MonadicStack.hs (Learn You a Haskell for Great Good!)
import Control.Monad.State
type Stack = [Int]
pop :: State Stack Int
-- The following line was wrong in the book:
-- pop = State $ \(x:xs) -> (x,xs)
pop = do
x:xs <- get
put xs
return x
push :: Int -> State Stack ()
-- The following line was wrong in the book:
-- push a = State $ \xs -> ((),a:xs)
push a = do
xs <- get
put (a:xs)
return ()
pop1 = runState pop [1..5]
push1 = runState (push 1) [2..5]
stackManip :: State Stack Int
stackManip = do
push 3
a <- pop
pop
stackManip1 = runState stackManip [5,8,2,1]
stackManip2 = runState stackManip [1,2,3,4]
stackStuff :: State Stack ()
stackStuff = do
a <- pop
if a == 5
then push 5
else do
push 3
push 8
stackStuff1 = runState stackStuff [9,0,2,1,0]
stackStuff2 = runState stackStuff [5,4,3,2,1]
moreStack :: State Stack ()
moreStack = do
a <- stackManip
if a == 100
then stackStuff
else return ()
moreStack1 = runState moreStack [100,9,0,2,1,0]
moreStack2 = runState moreStack [9,0,2,1,0]
stackyStack :: State Stack ()
stackyStack = do
stackNow <- get
if stackNow == [1,2,3]
then put [8,3,1]
else put [9,2,1]
stackyStack1 = runState stackyStack [1,2,3]
stackyStack2 = runState stackyStack [10,20,30,40]
MONADIC RANDOM GENERATORS：
-- MonadicRandomGenerator.hs (Learn You a Haskell for Great Good!)
import System.Random
import Control.Monad.State
randomSt :: (RandomGen g, Random a) => State g a
-- The following line was wrong in the book:
-- randomSt = State random
randomSt = do
gen <- get
let (value,nextGen) = random gen
put nextGen
return value
randomSt1 = (runState randomSt (mkStdGen 1)) :: (Int,StdGen)
randomSt2 = (runState randomSt (mkStdGen 2)) :: (Float,StdGen)
threeCoins :: State StdGen (Bool,Bool,Bool)
threeCoins = do
a <- randomSt
b <- randomSt
c <- randomSt
return (a,b,c)
threeCoins1 = runState threeCoins (mkStdGen 33)
threeCoins2 = runState threeCoins (mkStdGen 2)
-- rollDie and rollNDice are not explained in the book LYAHFGG.
-- But these functions are interesting and complementary:
rollDie :: State StdGen Int
rollDie = do
generator <- get
let (value, newGenerator) = randomR (1,6) generator
put newGenerator
return value
rollDie1 = runState rollDie (mkStdGen 1)
rollDie2 = runState rollDie (mkStdGen 2)
rollNDice :: Int -> State StdGen [Int]
rollNDice 0 = do
return []
rollNDice n = do
value <- rollDie
list <- rollNDice (n-1)
return (value:list)
rollNDice1 = runState (rollNDice 10) (mkStdGen 1)
rollNDice2 = runState (rollNDice 20) (mkStdGen 2)
答案 2 :(得分：3)
我认为state简化了pop的实施，但modify更适合push，因为它会返回单位：
import Control.Monad.State
type Stack a = [a]
pop :: State (Stack a) a
pop = state $ \(a:as) -> (a, as)
push :: a -> State (Stack a) ()
push a = modify (a:)
答案 3 :(得分：2)
DESUGARED MONADIC STACK：
-- DesugaredMonadicStack.hs (Learn You a Haskell for Great Good!)
import Control.Monad.State
type Stack = [Int]
pop :: State Stack Int
pop =
get >>=
\(x:xs) -> put xs >>
return x
push :: Int -> State Stack ()
push a =
get >>=
\xs -> put (a:xs) >>
return ()
pop1 = runState pop [1..5]
push1 = runState (push 1) [2..5]
stackManip :: State Stack Int
stackManip =
push 3 >>
pop >>=
\a -> pop
stackManip1 = runState stackManip [5,8,2,1]
stackManip2 = runState stackManip [1,2,3,4]
stackStuff :: State Stack ()
stackStuff =
pop >>=
\a ->
if a == 5 then
push 5
else
push 3 >>
push 8
stackStuff1 = runState stackStuff [9,0,2,1,0]
stackStuff2 = runState stackStuff [5,4,3,2,1]
moreStack :: State Stack ()
moreStack =
stackManip >>=
\a ->
if a == 100 then
stackStuff
else
return ()
moreStack1 = runState moreStack [100,9,0,2,1,0]
moreStack2 = runState moreStack [9,0,2,1,0]
moreStack3 = runState moreStack [100,5,4,3,2,1]
stackyStack :: State Stack ()
stackyStack =
get >>=
\stackNow ->
if stackNow == [1,2,3] then
put [8,3,1]
else
put [9,2,1]
stackyStack1 = runState stackyStack [1,2,3]
stackyStack2 = runState stackyStack [10,20,30,40]
DESUGARED MONADIC RANDOM GENERATOR：
-- DesugaredMonadicRandomGenerator.hs (Learn You a Haskell for Great Good!)
import System.Random
import Control.Monad.State
randomSt :: (RandomGen g, Random a) => State g a
randomSt =
get >>=
\gen ->
let (value,nextGen) = random gen
in
put nextGen >>
return value
randomSt1 = (runState randomSt (mkStdGen 1)) :: (Int,StdGen)
randomSt2 = (runState randomSt (mkStdGen 2)) :: (Float,StdGen)
threeCoins :: State StdGen (Bool,Bool,Bool)
threeCoins =
randomSt >>=
\a -> randomSt >>=
\b -> randomSt >>=
\c -> return (a,b,c)
threeCoins1 = runState threeCoins (mkStdGen 33)
threeCoins2 = runState threeCoins (mkStdGen 2)
-- rollDie and rollNDice are not explained in the book LYAHFGG.
-- But these functions are interesting and complementary:
rollDie :: State StdGen Int
rollDie =
get >>=
\generator ->
let (value, newGenerator) = randomR (1,6) generator
in
put newGenerator >>
return value
rollDie1 = runState rollDie (mkStdGen 1)
rollDie2 = runState rollDie (mkStdGen 2)
rollNDice :: Int -> State StdGen [Int]
rollNDice 0 = return []
rollNDice n =
rollDie >>=
\value -> rollNDice (n-1) >>=
\list -> return (value:list)
rollNDice1 = runState (rollNDice 10) (mkStdGen 1)
rollNDice2 = runState (rollNDice 20) (mkStdGen 2)