variables - 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出现以下错误：

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

任何人都可以帮忙吗？

Answer 1

这完全符合预期。AnSTRef仅在一次运行中有效runST。并且您尝试将外部STRef放入新的runST. 那是无效的。这将允许纯代码中的任意副作用。

所以，你尝试的东西是不可能实现的。按设计！

Answer 2

您需要留在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typeclass Ord，它Num不提供。）