Question

我在AsyncDemoViewModel中打印出Messages属性字符串列表。

如果我在GetVideosSlowlyAsync（）中使用Thread.Sleep（1000），则输出为

Time spent: 4001 milliseconds 

 - Started on thread 18, finished on thread 18  
 - Started on thread 18, finished on thread 18  
 - Started on thread 18, finished on thread 18 
 - Started on thread 18, finished on thread 18

如果我在GetVideosSlowlyAsync（）中使用等待Task.Delay（1000），则输出为

Time spent: 4053 milliseconds 


 - Started on thread 12, finished on thread 19 
 - Started on thread 19, finished on thread 16 
 - Started on thread 16, finished on thread 19 
 - Started on thread 19, finished on thread 9

为什么await调用没有释放调用线程？我期待await版本在完成时间上快4倍。

控制器代码：

public class AsyncDemoController : Controller
{
    private DepartmentDb _db;

    public AsyncDemoController(DepartmentDb db)
    {
        _db = db; 
    }

    public async Task<ActionResult> Index()
    {

        var sw = Stopwatch.StartNew();
        var v1 = await GetVideosSlowlyAsync();
        var v2 = await GetVideosSlowlyAsync();
        var v3 = await GetVideosSlowlyAsync();
        var v4 = await GetVideosSlowlyAsync();

        var vm = new AsyncDemoViewModel() {Videos = v1.Item2, Messages = new List<string>()};
        sw.Stop();
        vm.Messages.Add(string.Format("Time spent: {0} milliseconds", sw.ElapsedMilliseconds));
        vm.Messages.Add(v1.Item1);
        vm.Messages.Add(v2.Item1);
        vm.Messages.Add(v3.Item1);
        vm.Messages.Add(v4.Item1);

        return View(vm);
    }

    private async Task<Tuple<string, IEnumerable<Video>>> GetVideosSlowlyAsync()
    {
        var t1 = Thread.CurrentThread.ManagedThreadId;
        await Task.Delay(1000); // Thread.Sleep(1000);
        var t2 = Thread.CurrentThread.ManagedThreadId;
        return Tuple.Create(string.Format("Started on thread {0}, finished on thread {1}", t1, t2), _db.Videos.AsEnumerable());
    }

}

Answer 1

await方法等待GetVideosSlowlyAsync方法完成。您需要将await移动到需要操作结果的位置：

public async Task<ActionResult> Index()
{

    var sw = Stopwatch.StartNew();
    var v1 = GetVideosSlowlyAsync();
    var v2 = GetVideosSlowlyAsync();
    var v3 = GetVideosSlowlyAsync();
    var v4 = GetVideosSlowlyAsync();

    var vm = new AsyncDemoViewModel() {Videos = (await v1).Item2, Messages = new List<string>()};
    sw.Stop();
    vm.Messages.Add(string.Format("Time spent: {0} milliseconds", sw.ElapsedMilliseconds));
    vm.Messages.Add((await v1).Item1);
    vm.Messages.Add((await v2).Item1);
    vm.Messages.Add((await v3).Item1);
    vm.Messages.Add((await v4).Item1);

    return View(vm);
}

Answer 2

您正在调用await GetVideosSlowlyAsync()，它将返回调用方法，直到任务完成。如果你想看到你期望的结果，你需要删除它等待。这样，当GetVideosSlowlyAsync()方法在线程池上休眠时，它将一个接一个地启动所有4 Task.Delay()。

尝试这样的事情，

List<Task> taskList = new List<Task>();
taskList.Add(GetVideosSlowlyAsync());
taskList.Add(GetVideosSlowlyAsync());
taskList.Add(GetVideosSlowlyAsync());
taskList.Add(GetVideosSlowlyAsync());

Task.WaitAll(taskList.ToArray())
foreach (Task task in taskList)
    vm.Messages.Add(task.Result.Item1);

等待不在异步方法中释放调用线程

2 个答案: