GHC拒绝ST monad代码无法统一类型​​变量？

Question

我写了以下函数:

(.>=.) :: Num a => STRef s a -> a -> Bool
r .>=. x = runST $ do
 v <- readSTRef r
 return $ v >= x

但是当我尝试编译时,我收到以下错误:

Could not deduce (s ~ s1)
from the context (Num a)
  bound by the type signature for
             .>=. :: Num a => STRef s a -> a -> Bool
  at test.hs:(27,1)-(29,16)
  `s' is a rigid type variable bound by
      the type signature for .>=. :: Num a => STRef s a -> a -> Bool
      at test.hs:27:1
  `s1' is a rigid type variable bound by
       a type expected by the context: ST s1 Bool at test.hs:27:12
Expected type: STRef s1 a
  Actual type: STRef s a
In the first argument of `readSTRef', namely `r'
In a stmt of a 'do' expression: v <- readSTRef r

有人可以帮忙吗？

Answer 1

这完全符合预期.一个STRef只在一次运行中有效runST.而你试图把一个外部STRef投入到新的运行中runST.那是无效的.这将允许纯代码中的任意副作用.

所以,你尝试的是不可能实现的.按设计!

Answer 2

您需要保持在ST上下文中:

(.>=.) :: Ord a => STRef s a -> a -> ST s Bool
r .>=. x = do
 v <- readSTRef r
 return $ v >= x

(正如hammar指出的那样,要使用>=你需要Ord类型规范,它Num不提供.)