SerialPort port.open “The port 'COM2' does not exist.”

Question

I\'m having a big problem with the SerialPort.Open();

I am communicating with an usb virtual com port (cdc), and it is listed as COM2.

It works fine

Answer 1

This error can be caused if the driver returns an unexpected "file type" for "COM2".

Try p/Invoking GetFileType and I believe you'll see the pattern. It has to be FILE_TYPE_CHAR or FILE_TYPE_UNKNOWN or else SerialPort will throw that exception.

class Program
{
  static void Main(string[] args)
  {
    string portName = @"COM2";
    IntPtr handle = CreateFile(portName, 0, 0, IntPtr.Zero, 3, 0x80, IntPtr.Zero);
    if (handle == (IntPtr)(-1))
    {
      Console.WriteLine("Could not open " + portName + ": " + new Win32Exception().Message);
      Console.ReadKey();
      return;
    }

    FileType type = GetFileType(handle);
    Console.WriteLine("File " + portName + " reports its type as: " + type);

    Console.ReadKey();
  }

  [DllImport("kernel32.dll", CharSet = CharSet.Auto, CallingConvention = CallingConvention.StdCall, SetLastError = true)]
  public static extern IntPtr CreateFile(string lpFileName, uint dwDesiredAccess, uint dwShareMode, IntPtr SecurityAttributes, uint dwCreationDisposition, uint dwFlagsAndAttributes, IntPtr hTemplateFile);

  [DllImport("kernel32.dll")]
  static extern FileType GetFileType(IntPtr hFile);

  enum FileType : uint
  {
    UNKNOWN = 0x0000,
    DISK = 0x0001,
    CHAR = 0x0002,
    PIPE = 0x0003,
    REMOTE = 0x8000,
  }
}

Also see this thread on MSDN forums.