Haskell语言练习
Monad
inc n = Just (n + 1)
add1 n = [n + 1]
main = do
print $ Nothing >> (Just 0) -- Nothing
print $ (Just 0) >> (Nothing :: Maybe Int) -- Nothing
print $ (Just 0) >> Nothing >> (Just 1) -- Nothing
print $ (Just 0) >> (Just 1) >> (Just 2) -- Just 2
print $ Nothing >>= inc >>= inc >>= inc -- Nothing
print $ (Just 0) >>= inc >>= inc >>= inc -- Just 3
print $ [] >> [1, 2] -- []
print $ [1, 2] >> ([] :: [Int]) -- []
print $ [1] >> [3, 4, 5] -- [3,4,5]
print $ [1, 2] >> [3, 4, 5] -- [3,4,5,3,4,5]
print $ [1, 2, 3] >> [3, 4, 5] -- [3,4,5,3,4,5,3,4,5]
print $ [] >>= add1 >>= add1 >>= add1 -- []
print $ [1, 2, 3] >>= add1 -- [2,3,4]
print $ [1, 2, 3] >>= add1 >>= add1 -- [3,4,5]
print $ [1, 2, 3] >>= add1 >>= add1 >>= add1 -- [4,5,6]
根据 Monad Maybe 的定义、
instance Monad Maybe where
(Just x) >>= k = k x
Nothing >>= _ = Nothing
Just _m1 >> m2 = m2
Nothing >> _m2 = Nothing
可得
Nothing >>= inc = Nothing
(Just 0) >>= inc
= (Just 0) >>= (\n -> Just (n + 1))
= (\n -> Just (n + 1)) 0
= Just (0 + 1) = Just 1
Nothing >> (Just 0) = Nothing
(Just 0) >> (Nothing :: Maybe Int) = Nothing
(Just 1) >> (Just 2) = Just 2
根据 Monad [] 的定义、
instance Monad [] where
xs >>= f = [y | x <- xs, y <- f x]
xs >> ys = [y | _ <- xs, y <- ys]
可得
[1, 2, 3] >>= add1
= [1, 2, 3] >>= (\n -> [n + 1])
= [y | x <- [1, 2, 3], y <- [x + 1]]
= [2,3,4]
[1, 2, 3] >> [3, 4, 5]
= [y | _ <- [1, 2, 3], y <- [3, 4, 5]]
= [3,4,5,3,4,5,3,4,5]
State Monad
import Control.Monad.State
inc :: State Int Int
inc = do
n <- get
put (n + 1)
return n
incBy :: Int -> State Int Int
incBy x = do
n <- get
modify (+x)
return n
main = do
print $ evalState inc 1 -- 1
print $ execState inc 1 -- 2
print $ runState inc 1 -- (1,2)
print $ runState (withState (+3) inc) 1 -- (4,5)
print $ runState (mapState (\(a, s) -> (a + 3, s + 4)) inc) 1 -- (4,6)
print $ runState (incBy 5) 10 -- (10,15)
关于 State Monad
get 将结果值设置为状态值 s,状态值 s 保持不变。
put s 将结果值设为空,将状态值设为 s。
return a 将结果值设为 a,状态值 s 保持不变。
modify f 将结果值设为空,将状态值设为 f s。
gets f 将结果值设为 f s,状态值 s 保持不变。
由此可得
-- 假设初始状态值为 s
inc = do
n <- get -- (s,s)
put (n + 1) -- ((),s + 1)
return n -- (s,s + 1)
-- 假设初始状态值为 s
incBy x = do
n <- get -- (s,s)
modify (+x) -- ((),s + x)
return n -- (s,s + x)
关于 State Monad
evalState s 针对 State Monad 利用初始状态值 s 进行状态计算,然后返回最终结果值 a’。
execState s 针对 State Monad 利用初始状态值 s 进行状态计算,然后返回最终状态值 s’。
runState s 针对 State Monad 利用初始状态值 s 进行状态计算,然后返回最终结果值和最终状态值组成的一对值 (a’, s')。
mapState f 针对 State Monad 进行状态计算之后,对最终结果值和状态值调用函数 f。
withState f 针对 State Monad 进行状态计算之前,对初始状态值调用函数 f。
由此可得
runState inc 1 = (1,1 + 1) = (1,2)
evalState inc 1 = 1
execState inc 1 = 2
runState (incBy 5) 10 = (10,10 + 5) = (10,15)
runState (withState (+3) inc) 1
= runState inc ((+3) 1)
= runState inc 4
= (4,5)
runState (mapState (\(a, s) -> (a + 3, s + 4)) inc) 1
= (\(a, s) -> (a + 3, s + 4)) (runState inc 1)
= (\(a, s) -> (a + 3, s + 4)) (1,2)
= (4,6)
Reader Monad
import Control.Monad.Reader
data Environment = Environment { text1 :: String, text2 :: String }
getText :: Reader Environment String
getText = do
t1 <- asks text1 -- Hello
t2 <- asks text2 -- world!
t3 <- withReader text1 ask -- Hello
t4 <- mapReader text2 ask -- world!
return $ t1 ++ ", " ++ t2 ++ ", " ++ t3 ++ ", " ++ t4
main = print $ runReader getText $ Environment "Hello" "world!"
Writer Monad
import Control.Monad.Writer
write :: Int -> Writer [Int] String
write n = do
tell [1..n]
return "Done"
main = do
print $ runWriter $ write 10 -- ("Done",[1,2,3,4,5,6,7,8,9,10])
print $ execWriter $ write 10 -- [1,2,3,4,5,6,7,8,9,10]