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
有人可以帮忙吗?
答案 0 :(得分:12)
这完全符合预期。 STRef
仅在runST
的一次运行中有效。并且您尝试将外部STRef
放入runST
的新版本中。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!
答案 1 :(得分:7)
您需要保持在ST
范围内:
(.>=.) :: 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>=
,您需要Ord
类型类Num
未提供的类型类。)