我可以在 WPF 中使用 Monitor.Enter/Exit (c# lock) 而不必担心重入错误吗？
如果我在 WPF 应用程序中使用 Monitor.Enter/Exit（通过 C# 锁定语法），调度程序会导致重新进入吗？
在下面的示例中，假设当文本框中的文本发生更改时调用 OnTextChanged，是否会错误地调用 _worker.RunWorkerAsync() ？
public class SomeClass
{
private object _locker = new object();
private bool _running = false;
private BackgroundWorker _worker;
public void SomeClass()
{
// initialize worker...
}
void _worker_RunWorkerCompleted(object sender, RunWorkerCompletedEventArgs e)
{
lock (_locker)
_running = false;
}
void _worker_DoWork(object sender, DoWorkEventArgs e)
{
// ... do something time consuming ...
}
private void OnTextChanged()
{
lock(_locker)
{
if (!_running)
{
_worker.RunWorkerAsync();
_running = true;
}
}
}
}
我相信这是可能的，但我无法重现这一点。 WPF 是否以某种方式阻止调度程序在等待监视器时调用等待任务？
If I use Monitor.Enter/Exit (through the c# lock syntax) in a WPF application, can the dispatcher cause re-entrance?
In the sample below, presuming OnTextChanged is called when the text in a textbox changes, could the call to _worker.RunWorkerAsync() be called incorrectly?
public class SomeClass
{
private object _locker = new object();
private bool _running = false;
private BackgroundWorker _worker;
public void SomeClass()
{
// initialize worker...
}
void _worker_RunWorkerCompleted(object sender, RunWorkerCompletedEventArgs e)
{
lock (_locker)
_running = false;
}
void _worker_DoWork(object sender, DoWorkEventArgs e)
{
// ... do something time consuming ...
}
private void OnTextChanged()
{
lock(_locker)
{
if (!_running)
{
_worker.RunWorkerAsync();
_running = true;
}
}
}
}
I believe it's possible, but I've not been able to reproduce this. Does WPF somehow prevent the dispatcher from invoking waiting tasks when waiting on monitor?
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不确定你害怕什么。 OnTextChanged 和 RunWorkerCompleted 都在 UI 线程上运行。它不会是可重入的，你也不需要锁。这两种方法都只能在 UI 线程空闲时开始运行，从而推动消息循环。
Not sure what you fear. Both OnTextChanged and RunWorkerCompleted run on the UI thread. It won't be re-entrant, you don't need the lock either. Either method can only start running when the UI thread is idle, pumping the message loop.
虽然与您的问题没有直接关系，但如果您不将_running标记为volatile.实际上这并不完全正确，因为您没有使用双重检查锁。无论如何，我把与 volatile 相关的信息留在那里，供你参考。
While not directly related to your question, you could run into register caching issues if you don't mark_runningasvolatile.Actually this isn't strictly true, as you are not using a double-checked lock. I've left the information related to volatile there anyway, for your reference.