推迟任务执行

Question

我正在玩 Tasks，我想推迟执行任务。

我有一个这样的示例方法：

private async Task<bool> DoSomething(string name, int delayInSeconds)
{
    Debug.WriteLine($"Inside task named: {name}");
    await Task.Delay(TimeSpan.FromSeconds(delayInSeconds));
    Debug.WriteLine($"Finishing task named: {name}");
    return true;
}

我想先创建几个任务，然后执行一些工作，然后运行这些任务。由于行 Task<bool> myTask = DoSomething("Name", 4); 立即触发任务，我想到了这样的事情：

string[] taskNames = new string[2];

Task<Task<bool>>[] myTasks = new Task<Task<bool>>[2];
myTasks[0] =  new Task<Task<bool>>(async () => await DoSomething(taskNames[0], taskNames[0].Length));
myTasks[1] =  new Task<Task<bool>>(async () => await DoSomething(taskNames[1], taskNames[1].Length));

// I think I can declare it also like this, but this will create tasks later
//IEnumerable<Task<Task<bool>>> myTasks = taskNames.Select(x => new Task<Task<bool>>(async () => await DoSomething(x, x.Length)));

taskNames[0] = "First";
taskNames[1] = "Second";

Debug.WriteLine($"Tasks created");

var results = await Task.WhenAll(myTasks.Select(x => { x.Start(); return x.Unwrap(); }));
Debug.WriteLine($"Finishing: {results.Select(x => x.ToString()).Aggregate((a,b) => a + "," + b) }");

这可以用不同的方式完成吗，不包装任务？

Answer 1

您可以只使用 Task 生成委托来稍微简化一下：

string[] taskNames = new string[2];

Func<Task<bool>>[] myTasks = new Func<Task<bool>>[2];
myTasks[0] = new Func<Task<bool>>(async () => await DoSomething(taskNames[0], taskNames[0].Length));
myTasks[1] = new Func<Task<bool>>(() => DoSomething(taskNames[1], taskNames[1].Length)); // Shorter version, near-identical functionally.

// I think I can declare it also like this, but this will create tasks later
//IEnumerable<Task<Task<bool>>> myTasks = taskNames.Select(x => new Task<Task<bool>>(async () => await DoSomething(x, x.Length)));

taskNames[0] = "First";
taskNames[1] = "Second";

Debug.WriteLine($"Tasks created");

var results = await Task.WhenAll(myTasks.Select(x => x()));
Debug.WriteLine($"Finishing: {results.Select(x => x.ToString()).Aggregate((a, b) => a + "," + b) }");

警告：当您调用这些委托时，DoSomething 将同步执行到第一个 await，因此行为相似，但完全相同。

或者，您基于 IEnumerable 的解决方案也可以正常工作。只需在启动时编写迭代器方法和 yield return 任务。

就我个人而言，我会这样做：

string[] taskNames = new string[2];

taskNames[0] = "First";
taskNames[1] = "Second";

var results = await Task.WhenAll(taskNames.Select(n => DoSomething(n, n.Length)));
Debug.WriteLine($"Finishing: {results.Select(x => x.ToString()).Aggregate((a, b) => a + "," + b) }");

Answer 2

你的例子中不只是 "creating a Task"。您正在调用一种方法 DoSomething，即 returns 和 Task。假设这些是 async 方法，Task 创建和启动发生在编译器生成代码的幕后。

此问题的解决方案很简单：在您准备好方法之前不要调用该方法运行。想象一下您要求的行为在任何其他情况下会有多混乱。

推迟任务执行

Deferring tasks execution

c#

task-parallel-library

async-await