如何获得传递到异步方法的参数回调（不拉姆达）

Question

可能重复：
How to get the parameters passed to the asynchronous method in the callback

我需要这个拉姆达转换方法回调

var sendRegistrationDelegate = 
    new AsyncSendRegistrationDelegate(AsyncSendRegistrationMethod); 

sendRegistrationDelegate.BeginInvoke(registrationToUser, label, ar => 
{ 
    var responceFromServer = sendRegistrationDelegate.EndInvoke(ar); 

    if (responceFromServer.IsError) 
    { 
     SetText(label, registrationToUser.Name + @" " + 
      responceFromServer.ErrorMessage); 
    } 
    else 
    { 
     SetText(label, registrationToUser.Name + @" " + 
      responceFromServer.Data); 
    } 
}, null); 

Answer 1

首先，您是否掌握了lambda和匿名代表？

在这个片断：

sendRegistrationDelegate.BeginInvoke(registrationToUser, label, ar => 
    // start of method 
    { 
     var responceFromServer = sendRegistrationDelegate.EndInvoke(ar); 

     if (responceFromServer.IsError) 
     { 
      SetText(label, registrationToUser.Name + @" " + 
       responceFromServer.ErrorMessage); 
     } 
     else 
     { 
      SetText(label, registrationToUser.Name + @" " + 
       responceFromServer.Data); 
     } 
    } 
    // end of method 
    , null); 

...开闭{ }标记的开头和方法的结束，像这样：

void AsyncCallbackMethod(IAsyncResult ar) 
{ 
    // method body 
} 

你BeginInvoke方法会是什么样子：

sendRegistrationDelegate.BeginInvoke(registrationToUser, label, new AsyncCallback(AsyncCallbackMethod), null);