Question

我有一个小程序，它定义一个帐户，一个提款功能并尝试从中提款。但是，它不能编译到期并引发以下错误：

Couldn't match expected type ‘(STM a0 -> IO a0)
                                    -> STM () -> IO ()’
                  with actual type ‘IO ()’

似乎编译器无法识别从STM到IO的转换。任何指针都很棒。

import System.IO
import Control.Concurrent.STM

type Account = TVar Int

withdraw :: Account -> Int -> STM ()
withdraw acc amount = do
    bal <- readTVar acc    
    writeTVar acc (bal - amount)

good :: Account -> IO ()
good acc = do
    hPutStr stdout "Withdrawing..."
    {-hi-}atomically{-/hi-} (withdraw acc 10)

main = do
    acc <- atomically (newTVar 200)
    good acc 
    hPutStr stdout "\nDone!\n"

Answer 1

{-hi-}和{-/hi-}注释导致automically缩进，因此，您编写了hPutStr stdout "Withdrawing..." atomically (withdraw acc 10)。例如，如果您编写：

good :: Account -> IO ()
good acc = do
        hPutStr stdout "Withdrawing..."
 {-hi-} atomically (withdraw acc 10)

它工作正常，因为“噪声”（{-hi-}注释）不会导致内联atomically函数。

注释的确在语义上没有任何作用，但是您可以考虑将其替换为空格。

Haskell中的软件事务存储器：无法将预期的STM a0类型与实际的IO类型进行匹配（）

1 个答案: