这是一个坏主意。 这可能会启动太multithreading。

更好的方法是简单地使用Parallel.ForEach()如下所示：

 using System; using System.IO; using System.Threading.Tasks; namespace Demo { static class Program { static void Main() { string filename = @"Your test filename goes here"; Parallel.ForEach(File.ReadLines(filename), process); } private static void process(string line) { Console.WriteLine(line); } } } 

但是，这不使用async / await。 但是如果你愿意的话，你可以在一个任务中将整个调用包装到Parallel.ForEach()中。

或者，如果要使用任务并行库 （Microsoft NuGet包），您可以执行以下操作：

 using System; using System.IO; using System.Threading; using System.Threading.Tasks; using System.Threading.Tasks.Dataflow; namespace Demo { static class Program { static void Main() { Task.Run(test).Wait(); } static async Task test() { string filename = @"Your filename goes here"; await processFile(filename); } static async Task processFile(string filename) { var options = new ExecutionDataflowBlockOptions { MaxDegreeOfParallelism = 8, BoundedCapacity = 100 }; var action = new ActionBlock(s => process(s), options); foreach (var line in File.ReadLines(filename)) await action.SendAsync(line); action.Complete(); await action.Completion; } static void process(string line) { Thread.Sleep(100); // Simulate work. Console.WriteLine(Thread.CurrentThread.ManagedThreadId + " " + line); } } } 

这为您提供了async支持。

附录 ：线程池限制的演示。

（这是对shay __的评论的回应。）

如果您启动了许多长时间运行的任务，其中任务运行的时间比一秒钟大，那么您可能会看到线程池限制。

如果当前进程的ThreadPool.GetMinThreads(out workers, out ports);数等于或超过调用ThreadPool.GetMinThreads(out workers, out ports);返回的worker数，则会发生这种情况ThreadPool.GetMinThreads(out workers, out ports); 。

如果发生这种情况，在创建新的线程池线程之前，新的线程池线程的启动将延迟一小段时间（在我的系统上一秒钟）。 通常，这将允许另一个线程池线程变为可用，并且将使用它（当然这是限制的主要原因）。

以下代码演示了此问题：

 int workers, ports; ThreadPool.GetMinThreads(out workers, out ports); Console.WriteLine("Min workers = " + workers); // Prints 8 on my system. var sw = Stopwatch.StartNew(); for (int i = 0; i < 100; ++i) { Task.Run(() => { Console.WriteLine($"Thread {Thread.CurrentThread.ManagedThreadId} started at time {sw.Elapsed}"); Thread.Sleep(10000); }); } Console.ReadLine(); 

在我的系统上，这将打印以下内容：

 Min workers = 8 Thread 3 started at time 00:00:00.0098651 Thread 6 started at time 00:00:00.0098651 Thread 8 started at time 00:00:00.0099841 Thread 5 started at time 00:00:00.0099680 Thread 7 started at time 00:00:00.0099918 Thread 4 started at time 00:00:00.0098739 Thread 10 started at time 00:00:00.0100828 Thread 9 started at time 00:00:00.0101833 Thread 11 started at time 00:00:01.0096247 Thread 12 started at time 00:00:02.0098105 Thread 13 started at time 00:00:03.0099824 Thread 14 started at time 00:00:04.0100671 Thread 15 started at time 00:00:05.0098035 Thread 16 started at time 00:00:06.0099449 Thread 17 started at time 00:00:07.0096293 Thread 18 started at time 00:00:08.0106774 Thread 19 started at time 00:00:09.0098193 Thread 20 started at time 00:00:10.0104156 Thread 3 started at time 00:00:10.0109315 Thread 8 started at time 00:00:10.0112171 Thread 7 started at time 00:00:10.0112531 Thread 9 started at time 00:00:10.0117256 Thread 4 started at time 00:00:10.0117920 Thread 10 started at time 00:00:10.0117298 Thread 6 started at time 00:00:10.0109381 Thread 5 started at time 00:00:10.0112276 Thread 21 started at time 00:00:11.0095859 Thread 11 started at time 00:00:11.0101189 Thread 22 started at time 00:00:12.0095421 Thread 12 started at time 00:00:12.0111173 Thread 23 started at time 00:00:13.0095932 ... 

注意前8个线程如何快速启动，但随后新线程被限制为每秒一个，直到第一批线程终止然后可以重用。

另请注意，仅当线程需要相对较长的时间才能终止时，才会出现此效果。

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