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我正在从一个函数开始一个新任务,但我不希望它在同一个线程上运行。我不关心它在哪个线程上运行,只要它是不同的线程即可(因此 this question 中提供的信息没有帮助)。
我能保证下面的代码在允许 Task t
再次进入之前总是退出 TestLock
吗?如果不是,推荐的防止重入的设计模式是什么?
object TestLock = new object();
public void Test(bool stop = false) {
Task t;
lock (this.TestLock) {
if (stop) return;
t = Task.Factory.StartNew(() => { this.Test(stop: true); });
}
t.Wait();
}
编辑:根据 Jon Skeet 和 Stephen Toub 的以下回答,一种确定性地防止重入的简单方法是传递 CancellationToken,如本扩展方法所示:
public static Task StartNewOnDifferentThread(this TaskFactory taskFactory, Action action)
{
return taskFactory.StartNew(action: action, cancellationToken: new CancellationToken());
}
最佳答案
我给 Stephen Toub 发了邮件 - PFX Team 的成员- 关于这个问题。他很快就回复了我,提供了很多细节——所以我只是复制并粘贴他的文字在这里。我没有全部引用,因为阅读大量引用的文字最终会比普通的黑白相间更不舒服,但实际上,这是斯蒂芬 - 我不知道这么多东西 :) 我做了这个答案社区 wiki 反射(reflect)了下面所有的优点并不是我真正的内容:
If you call
Wait()
on a Task that's completed, there won't be any blocking (it'll just throw an exception if the task completed with a TaskStatus other thanRanToCompletion
, or otherwise return as a nop). If you callWait()
on a Task that's already executing, it must block as there’s nothing else it can reasonably do (when I say block, I'm including both true kernel-based waiting and spinning, as it'll typically do a mixture of both). Similarly, if you callWait()
on a Task that has theCreated
orWaitingForActivation
status, it’ll block until the task has completed. None of those is the interesting case being discussed.The interesting case is when you call
Wait()
on a Task in theWaitingToRun
state, meaning that it’s previously been queued to a TaskScheduler but that TaskScheduler hasn't yet gotten around to actually running the Task's delegate yet. In that case, the call toWait
will ask the scheduler whether it's ok to run the Task then-and-there on the current thread, via a call to the scheduler'sTryExecuteTaskInline
method. This is called inlining. The scheduler can choose to either inline the task via a call tobase.TryExecuteTask
, or it can return 'false' to indicate that it is not executing the task (often this is done with logic like...return SomeSchedulerSpecificCondition() ? false : TryExecuteTask(task);
The reason
TryExecuteTask
returns a Boolean is that it handles the synchronization to ensure a given Task is only ever executed once). So, if a scheduler wants to completely prohibit inlining of the Task duringWait
, it can just be implemented asreturn false;
If a scheduler wants to always allow inlining whenever possible, it can just be implemented as:return TryExecuteTask(task);
In the current implementation (both .NET 4 and .NET 4.5, and I don’t personally expect this to change), the default scheduler that targets the ThreadPool allows for inlining if the current thread is a ThreadPool thread and if that thread was the one to have previously queued the task.
Note that there isn't arbitrary reentrancy here, in that the default scheduler won’t pump arbitrary threads when waiting for a task... it'll only allow that task to be inlined, and of course any inlining that task in turn decides to do. Also note that
Wait
won’t even ask the scheduler in certain conditions, instead preferring to block. For example, if you pass in a cancelable CancellationToken, or if you pass in a non-infinite timeout, it won’t try to inline because it could take an arbitrarily long amount of time to inline the task's execution, which is all or nothing, and that could end up significantly delaying the cancellation request or timeout. Overall, TPL tries to strike a decent balance here between wasting the thread that’s doing theWait
'ing and reusing that thread for too much. This kind of inlining is really important for recursive divide-and-conquer problems (e.g. QuickSort) where you spawn multiple tasks and then wait for them all to complete. If such were done without inlining, you’d very quickly deadlock as you exhaust all threads in the pool and any future ones it wanted to give to you.Separate from
Wait
, it’s also (remotely) possible that the Task.Factory.StartNew call could end up executing the task then and there, iff the scheduler being used chose to run the task synchronously as part of the QueueTask call. None of the schedulers built into .NET will ever do this, and I personally think it would be a bad design for scheduler, but it’s theoretically possible, e.g.:protected override void QueueTask(Task task, bool wasPreviouslyQueued)
{
return TryExecuteTask(task);
}The overload of
Task.Factory.StartNew
that doesn’t accept aTaskScheduler
uses the scheduler from theTaskFactory
, which in the case ofTask.Factory
targetsTaskScheduler.Current
. This means if you callTask.Factory.StartNew
from within a Task queued to this mythicalRunSynchronouslyTaskScheduler
, it would also queue toRunSynchronouslyTaskScheduler
, resulting in theStartNew
call executing the Task synchronously. If you’re at all concerned about this (e.g. you’re implementing a library and you don’t know where you’re going to be called from), you can explicitly passTaskScheduler.Default
to theStartNew
call, useTask.Run
(which always goes toTaskScheduler.Default
), or use aTaskFactory
created to targetTaskScheduler.Default
.
编辑:好的,看起来我完全错了,当前正在等待任务的线程可以被劫持。这是发生这种情况的一个更简单的示例:
using System;
using System.Threading;
using System.Threading.Tasks;
namespace ConsoleApplication1 {
class Program {
static void Main() {
for (int i = 0; i < 10; i++)
{
Task.Factory.StartNew(Launch).Wait();
}
}
static void Launch()
{
Console.WriteLine("Launch thread: {0}",
Thread.CurrentThread.ManagedThreadId);
Task.Factory.StartNew(Nested).Wait();
}
static void Nested()
{
Console.WriteLine("Nested thread: {0}",
Thread.CurrentThread.ManagedThreadId);
}
}
}
示例输出:
Launch thread: 3
Nested thread: 3
Launch thread: 3
Nested thread: 3
Launch thread: 3
Nested thread: 3
Launch thread: 3
Nested thread: 3
Launch thread: 4
Nested thread: 4
Launch thread: 4
Nested thread: 4
Launch thread: 4
Nested thread: 4
Launch thread: 4
Nested thread: 4
Launch thread: 4
Nested thread: 4
Launch thread: 4
Nested thread: 4
如您所见,很多时候等待线程被重新用于执行新任务。即使线程已获得锁,也会发生这种情况。讨厌的重入。我感到震惊和担心:(
关于c# - Task.Factory.StartNew() 是否保证使用调用线程之外的另一个线程?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32629268/
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