我会尽力解释为runST
的类型推理:
runST :: (forall s. ST s a) -> a
,为什么它是不是这样简单的一个:
alternativeRunST :: ST s a -> a
注意,这alternativeRunST
将工作过为你的程序。
alternativeRunST
也已经让我们泄露变量出ST
单子:
leakyVar :: STRef s Int
leakyVar = alternativeRunST (newSTRef 0)
evilFunction :: Int -> Int
evilFunction x =
alternativeRunST $ do
val <- readSTRef leakyVar
writeSTRef leakyVar (val+1)
return (val + x)
然后,你可以去ghci中做:
>>> map evilFunction [7,7,7]
[7,8,9]
evilFunction
不是引用透明!
顺便说一下,尝试一下自己这里来运行上面的代码所需要的“坏ST”的框架:
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
import Control.Monad
import Data.IORef
import System.IO.Unsafe
newtype ST s a = ST { unST :: IO a } deriving Monad
newtype STRef s a = STRef { unSTRef :: IORef a }
alternativeRunST :: ST s a -> a
alternativeRunST = unsafePerformIO . unST
newSTRef :: a -> ST s (STRef s a)
newSTRef = ST . liftM STRef . newIORef
readSTRef :: STRef s a -> ST s a
readSTRef = ST . readIORef . unSTRef
writeSTRef :: STRef s a -> a -> ST s()
writeSTRef ref = ST . writeIORef (unSTRef ref)
真正runST
不允许我们构建这样的“恶”的职能。它是如何做到的?这是有点棘手,见下图:
试图运行:
>>> runST (newSTRef "Hi")
error:
Couldn't match type `a' with `STRef s [Char]'
...
>>> :t runST
runST :: (forall s. ST s a) -> a
>>> :t newSTRef "Hi"
newSTRef "Hi" :: ST s (STRef s [Char])
newSTRef "Hi"
不适合(forall s. ST s a)
。由于可以使用更简单的例子来也看到了,在那里GHC为我们提供了一个相当不错的错误:
dontEvenRunST :: (forall s. ST s a) -> Int
dontEvenRunST = const 0
>>> dontEvenRunST (newSTRef "Hi")
<interactive>:14:1:
Couldn't match type `a0' with `STRef s [Char]'
because type variable `s' would escape its scope
请注意,我们也可以写
dontEvenRunST :: forall a. (forall s. ST s a) -> Int
它相当于省略forall a.
因为我们之前做过。
请注意,a
的范围大于s
的范围,但在newSTRef "Hi"
的情况下,其值应取决于s
。类型系统不允许这样做。
哇,真是太棒了!感谢您的详细解释。 – 2012-07-26 13:57:05