使用多个参数启动任务的标准做法是什么？

Question

现在我有

class MyParamClass
{
   all the parameters I need to pass to the task
}

MyParamClass myParamObj = new MyParamClass();
myParamObj.FirstParam = xyz;
myParamObj.SecondParam = abc;
mytask = new Task<bool>(myMethod, myParamObj,_cancelToken);
mytask.Start()

bool myMethod(object passedMyParamObj)
{
   MyParamClass myParamObj = passedMyParamObj as MyParamClass;
  //phew! finally access to passed parameters
}

无论如何,我可以做到这一点,而无需创建类MyParamClass？如何在没有这种麻烦的情况下将多个参数传递给任务？这是标准做法吗？谢谢

Answer 1

您可以使用lambda或内联委托执行此操作:

myTask = new Task<bool>(() => MyMethod(xyz, abc), _cancelToken);

Answer 2

使用包装类来处理是执行此操作的标准方法.你唯一能做的就是用a Tuple来避免写作MyParamClass.

mytask = new Task(myMethod, Tuple.Create(xyz, abc), _cancelToken);
mytask.Start();

bool myMethod(object passedTuple)
{
     var myParamObj = passTuple as Tuple<int, string>;
}