multithreading - 如何将父异步与多个子异步链接

Question

async package的文档将withAsync函数描述为:

Spawn an asynchronous action in a separate thread, and pass its Async handle to the supplied function. When the function returns or throws an exception, uninterruptibleCancel is called on the Async. This is a useful variant of async that ensures an Async is never left running unintentionally.

在过去的两个小时中，我一直在盯着它看，并且一直无法弄清楚如何启动一个监视线程，该监视线程产生了多个工作线程，例如:

如果监视器线程死亡，则应终止所有工作线程，

但是，如果任何工作线程死亡，其他工作线程都不应该受到影响。 应该通知监视器，它应该能够重新启动工作线程。

Answer 1

似乎我们需要两个功能:一个启动所有异步任务，另一个监视它们并在它们死时重新启动它们。

第一个可以这样写:

withAsyncMany :: [IO t] -> ([Async t] -> IO b) -> IO b
withAsyncMany []     f = f []
withAsyncMany (t:ts) f = withAsync t $ \a -> withAsyncMany ts (f . (a:))

如果我们使用 managed包，我们也可以这样写:

import Control.Monad.Managed (with,managed)

withAsyncMany' :: [IO t] -> ([Async t] -> IO b) -> IO b
withAsyncMany' = with . traverse (\t -> managed (withAsync t))

重新启动功能将循环异步列表，轮询它们的状态，并在它们失败时进行更新:

{-# language NumDecimals #-}
import Control.Concurrent (threadDelay)

resurrect :: IO t -> [Async t] -> IO ()
resurrect restartAction = go []
    where
    go ts [] = do
        threadDelay 1e6    -- wait a little before the next round of polling
        go [] (reverse ts)
    go past (a:pending) = do
        status <- poll a   -- has the task died, or finished?
        case status of
            Nothing -> go (a:past) pending
            Just _  -> withAsync restartAction $ \a' -> go (a':past) pending

但是，我担心许多嵌套的 withAsyncs导致某种类型的资源泄漏的可能性(因为必须在每种 withAsync上安装某种异常处理程序，以在父线程死亡的情况下通知 child )。

因此，在这种情况下，最好使用简单的 async生成工作程序，将 Async的集合存储到某种可变引用中，并在监视器线程中安装单个异常处理程序，该异常处理程序将遍历终止每个任务的容器。