They're two different 'env's, which can be observed by desugaring the do-notation: do env <- liftIO (readIORef envRef) env <- return (filter (\(_id, _) -> _id /= id) env) ... Desugaring do-notation gets us: liftIO (readIORef envRev) >>= \ env -> return (filter (\(_id, _) -> _id /= id) env) >>= \ env -> ... Sometimes people use different names to make this obvious, e.g. do env <- liftIO $ readIORef envRef env' <- return (filter ... env) Also note that you're doing a pure operation here, so you don't need two bindings. You could instead do: do env <- filter (\(_id, _) -> _id /= id) <$> readIORev envRef ... (<$> is from the supremely useful Control.Applicative, and is equivalent to fmap from Functor, or liftM from Monad) or: do env <- liftIO $ readIORef envRef let env' = filter ... env Using let notation here makes it somewhat more obvious that that line doesn't have any side effects. -Ross On Nov 15, 2009, at 2:05 AM, zaxis wrote:
defineVar :: Env -> (Id, Val) -> IOThrowsError Val defineVar envRef (id, val) = do { env <- liftIO $ readIORef envRef; env <- return $ filter (\(_id, _) -> _id/=id) env; -- clear the current scope valRef <- liftIO $ newIORef val; liftIO $ writeIORef envRef $ ((id, valRef):env); return val; }
In haskell, the variable canot change its value , right? If so, why can the `env` be assigned value twice?
Sincerely!
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