只有加入独特的项目名单 [英] Only Add Unique Item To List
问题描述
我添加远程设备列表,因为他们宣布自己在网络上。我只想给设备添加到列表中,如果它没有previously被添加。
的通知按照跨越异步接口监听到来使code添加设备可以在多个线程运行。我不知道我在做什么错,但没有母校什么,我尝试我结束了重复。以下是我目前有.....
锁(_remoteDevicesLock)
{
远端设备rDevice =(从D在_remoteDevices
其中,d.UUID.Trim()。等于(notifyMessage.UUID.Trim(),StringComparison.OrdinalIgnoreCase)
选择D).FirstOrDefault();
如果(rDevice!= NULL)
{
//更新设备.....
}
其他
{
//创建一个新的远程设备
rDevice =新的远端设备(notifyMessage.UUID);
_remoteDevices.Add(rDevice);
}
}
如果您的要求是:有没有重复,你应该使用的 HashSet的。
HashSet.Add 将返回的错误时,该项目已存在(如果这甚至关系到你)。
您可以使用@pstrjds链接下面的构造函数(或这里)来定义相等运算符,否则你会需要实现远端设备
( GetHash code 平等方法code>&安培;
等于
)
I'm adding remote devices to a list as they announce themselves across the network. I only want to add the device to the list if it hasn't previously been added.
The announcements are coming across an async socket listener so the code to add a device can be run on multiple threads. I'm not sure what I'm doing wrong but no mater what I try I end up with duplications. Here is what I currently have.....
lock (_remoteDevicesLock)
{
RemoteDevice rDevice = (from d in _remoteDevices
where d.UUID.Trim().Equals(notifyMessage.UUID.Trim(), StringComparison.OrdinalIgnoreCase)
select d).FirstOrDefault();
if (rDevice != null)
{
//Update Device.....
}
else
{
//Create A New Remote Device
rDevice = new RemoteDevice(notifyMessage.UUID);
_remoteDevices.Add(rDevice);
}
}
If your requirements are to have no duplicates, you should be using a HashSet.
HashSet.Add will return false when the item already exists (if that even matters to you).
You can use the constructor that @pstrjds links to below (or here) to define the equality operator or you'll need to implement the equality methods in RemoteDevice
(GetHashCode
& Equals
).
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