Haskell：STM writeTVar的嵌套monad错误

2sbarzqh 于 2023-10-19 发布在其他

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我有以下代码

module MyLib (write) where

import Control.Monad.STM
import Control.Concurrent.STM.TVar

write :: STM (TVar Störm) -> Störm -> IO ()
write state' new = atomically $ write' state'
      where
        write' :: STM (TVar Störm) -> STM ()
        write' state' = write'' <$> state'
        write'' :: TVar Störm -> STM ()
        write'' state' = writeTVar state' new

使用依赖关系STM
我收到以下错误

src/MyLib.hs:39:25: error:
    • Couldn't match type ‘STM ()’ with ‘()’
      Expected: TVar Störm -> ()
        Actual: TVar Störm -> STM ()
    • In the first argument of ‘(<$>)’, namely ‘write''’
      In the expression: write'' <$> state'
      In an equation for ‘write'’: write' state' = write'' <$> state'
   |
39 |         write' state' = write'' <$> state'

在我的测试中，它将Monad嵌套到STM (STM ())。却不知道如何解决

haskell

来源：https://stackoverflow.com/questions/76874567/haskell-nested-monad-error-with-stm-writetvar

1条答案

按热度按时间

oxosxuxt1#

问题出在write'' <$> state'项上。在这种情况下，(<$>)具有类型：

(<$>) :: (TVar Störm -> STM ()) -> STM (TVar Störm) -> STM (STM ())

和

write'' <$> state' :: STM (STM ())

但预期类型为STM ()。
要解决这个问题，您可以执行以下操作：

join $ write'' <$> state'

但更简单的是：

write'' =<< state'

或：

state' >>= write''

但是，我会这样写：

write :: STM (TVar Störm) -> Störm -> IO ()
write action new = atomically $ do
    var <- action
    writeTVar var new

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Haskell：STM writeTVar的嵌套monad错误

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