c# MRDS:为什么处理程序是虚拟的?

Question

熟悉microsoft robotics studio的人能否解释一下为什么处理程序操作是虚拟的，而有些设置为非虚拟的?

[ServiceHandler(ServiceHandlerBehavior.Concurrent)]
public virtual IEnumerator<ITask> GetHandler(Get get)
{
    get.ResponsePort.Post(_state);
    yield break;
}

和

[ServiceHandler(ServiceHandlerBehavior.Concurrent)]
public IEnumerator<ITask> SubscribeHandler(Subscribe subscribe)
{
    SubscribeRequestType request = subscribe.Body;
    LogInfo("Subscribe request from: " + request.Subscriber);
    // Use the Subscription Manager to handle the subscribers
    yield return Arbiter.Choice(
        SubscribeHelper(_submgrPort, request, subscribe.ResponsePort),
        delegate(SuccessResult success)
        {
            // Send a notification on successful subscription so that the
            // subscriber can initialize its own state
            base.SendNotificationToTarget<Replace>(request.Subscriber, _submgrPort, _state);
        },
        delegate(Exception e)
        {
            LogError(null, "Subscribe failed", e);
        }
    );
    yield break;
}

Answer 1

虚拟处理程序用于您希望用作其他类基的类。非虚的处理程序要么在派生类中声明为覆盖，要么在不会派生的类中声明为覆盖(在示例代码中经常出现这种情况)。

参考以下文章:

http://msdn.microsoft.com/en-us/library/9fkccyh4 (v = VS.100) . aspx

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