我正在尝试制作一个可以扫描网络中的ARP请求并列出所有现有网络设备的应用程序.目前,我使用SharpPcap和PacketDoNet.

根据给定IP解析主机名时,解析“未知”主机时出现SocketException.因此,我将其放在了try / catch中.由于我认为这是不好的风格(忽略异常),因此我正在寻找其他解决方案.

这是一些代码：

// Button for scanning the network
private void btnStartScanningForClients_Click(object sender, RoutedEventArgs e)
{
    // Check for correct interface
    // [...]

    // Start scanning process
    if (!this.netWorkItOut.Startet)
    {
        // Dis-/Enable visual controls
        // [...]

        // Start scanning
        var index = this.cbNetworkInterface.SelectedIndex
        this.netWorkItOut.StartDevice(index);
        this.netWorkItOut.Scanner.StartScanningNetwork(resolveHostnames);
    }
}

这是控制对象,用于保存扫描程序,处理事件,接受数据包并将其放入队列

public void StartDevice(int deviceIndex)
{
    this.Startet = true;
    // [...]
    this.Device = WinPcapDeviceList.Instance[deviceIndex];

    // Activate Scanner
    this.Scanner = new Scanner(this.DeviceInfo);

    // Subscribe Events
    // [...]

    this.Device.Open(DeviceMode.Promiscuous, 1);
    this.Device.Filter = "(arp || ip || ip6)";

    this.Device.OnPacketArrival += device_OnPacketArrival;
    this.Device.StartCapture();
}

private void device_OnPacketArrival(object sender, CaptureEventArgs e)
{
    //PacketDoNet
    Packet packet;

    try
    { packet = Packet.ParsePacket(LinkLayers.Ethernet, e.Packet.Data); }
    catch (Exception)
    { return;  }

    if (packet is EthernetPacket)
    {
        var arp = ARPPacket.GetEncapsulated(packet);

        if (arp != null)
        {
            if (this.Scanner.Started)
            {
                lock (this.Scanner.PacketQueueARP)
                {
                    this.Scanner.PacketQueueARP.Add(arp);
                }
            }
        }
    }
}

这是Controling对象和Scanner类.扫描程序类处理ARP请求并解析主机名

public void StartScanningNetwork(bool resolveHostnames)
{
    // [...]
    this.ResolveHostnames = resolveHostnames;

    // start worker to listen for ARP packets
    this.workerARP = new Thread(WorkerARP);
    this.workerARP.Name = "Scanner thread (ARP)";
    this.workerARP.Start();

    this.Started = true;
}

private void WorkerARP()
{
    List<IPAddress> processedIps = new List<IPAddress>();

    // copy packets from storage queue to thread queue for processing
    while (Started)
    {
        // [...]

        if (this.threadQueueARP.Count > 0)
        {
            foreach (var packet in this.threadQueueARP)
            {
                // [...]

                if (!processedIps.Contains(ip))
                {
                    // [...]

                    if (this.ResolveHostnames)
                    {
                        var resolveHostnamesTask = Task.Factory.StartNew(ResolveHostnamesWorker, ip);
                    }
                }
                // [...]
            }

            // [...]
        }
        // [...]
    }

}

private void ResolveHostnamesWorker(object data)
{
    if (data is IPAddress)
    {
        var ip = (IPAddress)data;
        var hostname = "";

        try
        {
            hostname = Dns.GetHostEntry(ip).HostName;
        }
        catch { }

        // Raise Event for hostname resolved
    }
}

其所有有关行主机名= Dns.GetHostEntry(ip).HostName

因此：通过Dns.GetHostEntry()解决HostEntry时,如何避免使用try / catch？如果没有已知主机,是否有只返回null的函数？

提前致谢！

解决方法:

据我所知,没有像TryGetHostName()这样的方法不会引发异常.

但是我认为可以按您期望的方式捕获异常.因此,您应该将捕获的异常限制为您期望的异常：

private void ResolveHostnamesWorker(object data)
{
    if (data is IPAddress)
    {
        var ip = (IPAddress)data;
        var hostname = "";

        try
        {
            hostname = Dns.GetHostEntry(ip).HostName;
        }
        catch(SocketException socketException)
        {
            // maybe limit handling based on data in socketException and
            // call throw; to rethrow exception if not the expected one
        }

    // Raise Event for hostname resolved
}

}