[RESOLVED] [Migration] Serial port device is not responding to my VS 2008 code

[Daarklord]

I'm migrating a program from VB 6 to VS 2008. The program communicates with a microcontroller through a serial port... or at least, it's SUPPOSED to. DUN DUN DUN. Though I tried to keep the communication as similar to the VB6 version (which I didn't write) as possible, the microcontroller is not replying as it should.

This is the old code that works:

Function SendInstruction(w As String) As String
buffer$ = ""
buffera$ = ""
MSComm1.InputLen = 0
   
   MSComm1.Output = w & Chr$(13)
   Text3 = CStr(w)
   x = timeGetTime
   If Port > 2 Then delay (1)
   Timeout = False
   Do
      DoEvents
      signal$ = MSComm1.Input:
      buffer$ = buffer$ + signal$
      If signal$ = "" Then signal$ = Chr$(0)
      If (timeGetTime > (x + 200)) Then Timeout = True
   Loop Until Timeout Or (Asc(Right$(signal$, 1)) = 13)

Debug.Print timeGetTime - x

   Text1 = buffer$
   If Left$(buffer$, 1) = "?" Then:
End Function

This is the new code:

    Sub SendInstruction(ByVal w As String)
        My.Application.DoEvents()
        If SerialPort1.IsOpen Then
            Try
                'Send a command to the current COM port
                Dim reply As String
                buffer = "" 'clear buffer for more reply
                SerialPort1.WriteLine(w) 'Send instruction.
                TextBoxInstruction.Text = w 'CStr(w) 'display instruction in the instruction textbox
                'Read answer from the serial connection.
                reply = SerialPort1.ReadLine
                buffer = buffer & reply 'store reply

                TextBoxReply.Text = buffer 'display reply in the reply textbox
            Catch e As System.InvalidOperationException
                If e.Message = "The port is closed." Then
                    buffer = "<Disconnected.>"
                Else
                    Throw e
                End If
            Catch e As TimeoutException
                'A timeout will cause the command to be ignored
                buffer = "<Timed out.>"
                TextBoxReply.Text = buffer
            End Try
        Else
            buffer = "<Disconnected.>"
        End If
    End Sub

The reply goes into buffer. I send the same argument w in both cases, but the new code just times out a the SerialPort1.ReadLine. I'm pretty sure the port settings (port name, parity, baud rate, data bits, and stopbits) are the same too. Anything I might have missed? Also, what is the InputLen property and where did it go in VB 2008? Though I don't think that's the problem since it should just default to 0 in the new code.

Halp!

[Jenner]

Your code assumes that there's going to be data in the received buffer almost immediately after you send your command. Reading the serial port has to be done in a different way. Here's a quick example of serial communication in VB.NET:

    Private WithEvents port As New IO.Ports.SerialPort("COM1", 9600, IO.Ports.Parity.None, 8, IO.Ports.StopBits.None)

    Private Sub Button_Click(ByVal sender As Object, ByVal e As System.EventArgs) Handles btnChooseOrder.Click
        If Not port.IsOpen Then port.Open()
        'Send data.
        port.WriteLine("BLAH")
    End Sub

    Private Sub Received_Data(ByVal sender As Object, ByVal e As System.IO.Ports.SerialDataReceivedEventArgs) Handles port.DataReceived
        'This happens when the port receives data.  Read it then.
        MsgBox(port.ReadLine)
    End Sub

[Daarklord]

Greetings, Jenner, my partner in crime . All is good and the problem has been solved. It doesn't involve the DataReceived event, but you are right there is a delay. Good thing about ReadLine is that it's a blocking call and will keep running until a newline is received. But the real question is: what IS newline? *X-files music plays*

After I implemented your DataReceived suggestion and found it to not work I went berserk and nuked a small nation. That cooled my head down, then I made a simple program in VB6 which worked and suprised me to no extent. But then while typing it out I noticed the message ends in + Chr(13) which is carriage return. Ofcourse I considered the newline problem before and I tried to Write(w & Chr(13)) in VB 2008 some time ago before giving up and coming here.

But armed with new knowledge from the MSDN about System.IO.SerialPort, I tried setting SerialPort1.NewLine = Chr(13). It worked. Now the microcontroller returns text, the PROPER text, with a funny square character at the end which I assume to be a Line Feed. The main part of the problem is over. Thanks for the help.

[Jenner]

No problem. Glad you figured it out! Yea, I had the same problem talking to a Motor Controller on a stepper motor. Took me a day to realize I wasn't sending a newline (vbCrLf) after my commands.

I confess, I do a similar trick where after I sent the command, I did a DoEvents() loop until it either gave a response or my time-out value was hit.

[Daarklord]

Yes DoEvents() is supposed to be primitive and bad but I can't think of anything else to do short of multi-threading. However I don't put DoEvents() in a loop while waiting for a response because who knows the user might click on another read/write command and that leads to race conditions. I put it between read/write commands. So the program pauses while serial communication is taking place, but who cares, serial communication is the ONLY thing it's doing!

Peace,
Daarklord