我想将我的Thread配置为后台线程,为什么我的线程中缺少此属性?
ThreadStart starter = delegate { openAdapterForStatistics(_device); };
new Thread(starter).Start();
public void openAdapterForStatistics(PacketDevice selectedOutputDevice)
{
using (PacketCommunicator statCommunicator = selectedOutputDevice.Open(100, PacketDeviceOpenAttributes.Promiscuous, 1000)) //open the output adapter
{
statCommunicator.Mode = PacketCommunicatorMode.Statistics; //put the interface in statstics mode
statCommunicator.ReceiveStatistics(0, statisticsHandler);
}
}
我试过了:
Thread thread = new Thread(openAdapterForStatistics(_device));
但我有2个编译错误:
- 'System.Threading.Thread.Thread(System.Threading.ThreadStart)'的最佳重载方法匹配有一些无效的参数
- 参数1:无法从'void'转换为'System.Threading.ThreadStart'
醇>
我不知道为什么
答案 0 :(得分:1)
关于后台的事情,我没有看到你期望如何设置它,因为你没有保持对线程的引用。应该是这样的:
ThreadStart starter = delegate { openAdapterForStatistics(_device); };
Thread t = new Thread(starter);
t.IsBackground = true;
t.Start();
此
Thread thread = new Thread(openAdapterForStatistics(_device));
将无效,因为您应该传递一个以object
作为参数的方法,而实际上是在传递方法调用的结果。所以你可以这样做:
public void openAdapterForStatistics(object param)
{
PacketDevice selectedOutputDevice = (PacketDevice)param;
using (PacketCommunicator statCommunicator = selectedOutputDevice.Open(100, PacketDeviceOpenAttributes.Promiscuous, 1000)) //open the output adapter
{
statCommunicator.Mode = PacketCommunicatorMode.Statistics; //put the interface in statstics mode
statCommunicator.ReceiveStatistics(0, statisticsHandler);
}
}
和
Thread t = new Thread(openAdapterForStatistics);
t.IsBackground = true;
t.Start(_device);
答案 1 :(得分:0)
您应该使用BackgroundWorker
类,它专门用于您的情况。您希望在后台完成的任务。
答案 2 :(得分:0)
PacketDevice selectedOutputDeviceValue = [some value here];
Thread wt = new Thread(new ParameterizedThreadStart(this.openAdapterForStatistics));
wt.Start(selectedOutputDeviceValue);