C# check if a COM (Serial) port is already open

Question

Is there an easy way of programmatically checking if a serial COM port is already open/being used?

Normally I would use:

try
{
    // open port
}
cat         


        

Answer 1

For people that cannot use SerialPort.GetPortNames(); because they are not targeting .net framework (like in my case I am using .Net Core and NOT .Net Framework) here is what I ended up doing:

In command prompt if you type mode you get something like this:

mode is an executable located at C:\Windows\System32\mode.com. Just parse the results of that executable with a regex like this:

// Code that answers the question

var proc = new Process
{
    StartInfo = new ProcessStartInfo
    {
        FileName = @"C:\Windows\System32\mode.com",
        UseShellExecute = false,
        RedirectStandardOutput = true,
        CreateNoWindow = true
    }
};

proc.Start();
proc.WaitForExit(4000); // wait up to 4 seconds. It usually takes less than a second

// get ports being used
var output = proc.StandardOutput.ReadToEnd();

Now if you want to parse the output this is how I do it:

List<string> comPortsBeingUsed = new List<string>();
Regex.Replace(output, @"(?xi) status [\s\w]+? (COM\d) \b ", regexCapture =>
{
    comPortsBeingUsed.Add(regexCapture.Groups[1].Value);
    return null;
});

foreach(var item in comPortsBeingUsed)
{
    Console.WriteLine($"COM port {item} is in use");
}

Answer 2

You can try folloing code to check whether a port already open or not. I'm assumming you dont know specificaly which port you want to check.

foreach (var portName in Serial.GetPortNames()
{
  SerialPort port = new SerialPort(portName);
  if (port.IsOpen){
    /** do something **/
  }
  else {
    /** do something **/
  }
}

Answer 3

This is how I did it:

      [DllImport("kernel32.dll", CharSet = CharSet.Auto, SetLastError = true)]
    internal static extern SafeFileHandle CreateFile(string lpFileName, int dwDesiredAccess, int dwShareMode, IntPtr securityAttrs, int dwCreationDisposition, int dwFlagsAndAttributes, IntPtr hTemplateFile);

then later on

        int dwFlagsAndAttributes = 0x40000000;

        var portName = "COM5";

        var isValid = SerialPort.GetPortNames().Any(x => string.Compare(x, portName, true) == 0);
        if (!isValid)
            throw new System.IO.IOException(string.Format("{0} port was not found", portName));

        //Borrowed from Microsoft's Serial Port Open Method :)
        SafeFileHandle hFile = CreateFile(@"\\.\" + portName, -1073741824, 0, IntPtr.Zero, 3, dwFlagsAndAttributes, IntPtr.Zero);
        if (hFile.IsInvalid)
            throw new System.IO.IOException(string.Format("{0} port is already open", portName));

        hFile.Close();


        using (var serialPort = new SerialPort(portName, 115200, Parity.None, 8, StopBits.One))
        {
            serialPort.Open();
        }

Answer 4

Sharing what worked for me (a simple helper method):

private string portName { get; set; } = string.Empty;

    /// <summary>
    /// Returns SerialPort Port State (Open / Closed)
    /// </summary>
    /// <returns></returns>
    internal bool HasOpenPort()
    {
        bool portState = false;

        if (portName != string.Empty)
        {
            using (SerialPort serialPort = new SerialPort(portName))
            {
                foreach (var itm in SerialPort.GetPortNames())
                {
                    if (itm.Contains(serialPort.PortName))
                    {
                        if (serialPort.IsOpen) { portState = true; }
                        else { portState = false; }
                    }
                }
            }
        }

        else { System.Windows.Forms.MessageBox.Show("Error: No Port Specified."); }

        return portState;
}

Notes:
- For more advanced technique(s) I recommend using ManagementObjectSearcher Class.
More info Here.
- For Arduino devices I would leave the Port Open.
- Recommend using a Try Catch block if you need to catch exceptions.
- Check also: "TimeoutException"
- More information on how to get SerialPort (Open) Exceptions Here.

Answer 5

  public void MobileMessages(string ComNo, string MobileMessage, string MobileNo)
    {
        if (SerialPort.IsOpen )
            SerialPort.Close();
        try
        {
            SerialPort.PortName = ComNo;
            SerialPort.BaudRate = 9600;
            SerialPort.Parity = Parity.None;
            SerialPort.StopBits = StopBits.One;
            SerialPort.DataBits = 8;
            SerialPort.Handshake = Handshake.RequestToSend;
            SerialPort.DtrEnable = true;
            SerialPort.RtsEnable = true;
            SerialPort.NewLine = Constants.vbCrLf;
            string message;
            message = MobileMessage;

            SerialPort.Open();
            if (SerialPort.IsOpen  )
            {
                SerialPort.Write("AT" + Constants.vbCrLf);
                SerialPort.Write("AT+CMGF=1" + Constants.vbCrLf);
                SerialPort.Write("AT+CMGS=" + Strings.Chr(34) + MobileNo + Strings.Chr(34) + Constants.vbCrLf);
                SerialPort.Write(message + Strings.Chr(26));
            }
            else
                ("Port not available");
            SerialPort.Close();
            System.Threading.Thread.Sleep(5000);
        }
        catch (Exception ex)
        {

                message.show("The port " + ComNo + " does not exist, change port no ");
        }
    }

Answer 6

The SerialPort class has an Open method, which will throw a few exceptions. The reference above contains detailed examples.

See also, the IsOpen property.

A simple test:

using System;
using System.IO.Ports;
using System.Collections.Generic;
using System.Text;

namespace SerPort1
{
class Program
{
    static private SerialPort MyPort;
    static void Main(string[] args)
    {
        MyPort = new SerialPort("COM1");
        OpenMyPort();
        Console.WriteLine("BaudRate {0}", MyPort.BaudRate);
        OpenMyPort();
        MyPort.Close();
        Console.ReadLine();
    }

    private static void OpenMyPort()
    {
        try
        {
            MyPort.Open();
        }
        catch (Exception ex)
        {
            Console.WriteLine("Error opening my port: {0}", ex.Message);
        }
    }
  }
}