Unable to communicate with arduino using QSerialPort if the arduino-ide has not opened the connection before

Question

I am using the following code to open a serial port to communicate with the arduino.

  const qint32 baud = Global::getInstance().getSettings().value("Hardware/baudRate", 115200).toInt();
  const QString portName = Global::getInstance().getSettings().value("Hardware/port", "COM3").toString();
  port.setPortName(portName);

  if(!port.open(QIODevice::ReadWrite))
  {
    qFatal("Unable to open serial port");
    exit(1); 
  }
  if(!port.setParity(QSerialPort::NoParity) ||
     !port.setStopBits(QSerialPort::OneStop) ||
     !port.setDataBits(QSerialPort::Data8) ||
     !port.setFlowControl(QSerialPort::NoFlowControl ) ||
     !port.setBaudRate(baud))
  {
     qFatal("Unable to configure serial port");
     exit(1); 
  }
  if(port.error() != QSerialPort::NoError)
  {
    qFatal("some error occurred!");
    exit(1);
  }

Afterwards I try to read from the connection using the following code:

bool Light::waitForReady()
{
  char data[1];
  data[0] = -1;
  if(port.waitForReadyRead(10000))
  {
    const int numRead = port.read(&data[0], 1);
    if(numRead == 1)
    {
      return data[0] == (char)RDY;
    }
    else
    {
      qWarning("Read error, read %d bytes", numRead);
    }
  }
  else
  {
    qWarning("Read timeout");
    return false;
  }
  return false;
}

The read does not work, it times out after waiting for 10 seconds. However when I open and close the serial monitor inside the arduino-ide before running my QT code it works. My guess is that the ardunio-ide does something to the port that I am missing but as far as I can tell I am using the exact same settings as the arduino-ide to open the serial port.

This is the code that is running on the arduino:

void setup() 
{
  Serial.begin(115200);
  Serial.write(RDY);
}

This guy has had a similar problem and he compared the serial port settings of putty and QSerialPort. He found that the settings fDtrControl, fOutX and fIutX differed. I tried changing fDtrControl but it had no effect. I do not know how to set fOutX and fIutX using QT.

Any ideas what might be causing this bug?

edit: Solution: port.setDataTerminalReady(true) was missing

Hi Arne. Which QtSerialPort version are you using? Is it with Qt 4.8.5 and custom build, or the 5.2.1 binary package? Also, if you use the 5.3 beta or the code from master, you will have easier time because you can set the port up before opening. — László Papp, Apr 21 '14 at 13:19
@LaszloPapp I am using the QtSerialPort that came with Qt 5.2.0 (binary package) — Arne Böckmann, Apr 21 '14 at 13:21
Your failure test logic is wrong. Instead of `if (!a && !b && !c)` it should be `if (!a || !b || !c)`. — Kuba hasn't forgotten Monica, Apr 21 '14 at 13:22
@ArneBöckmann: do the command line example (sync or async) work for reading? — László Papp, Apr 21 '14 at 13:25
@KubaOber Ops, thanks. However that does not change the problem :) — Arne Böckmann, Apr 21 '14 at 13:27
Arne, it may not change the problem, but it won't work properly without you fixing it *either*. Often there are multiple problems that are layered, and presence of one problem lets you get away with other problems: since it doesn't work, you have no feedback when you add broken code to the mix - it still just doesn't work :) — Kuba hasn't forgotten Monica, Apr 21 '14 at 15:10

score 4 · Accepted Answer · edited May 23 '17 at 12:01

4

I do not know how to set fOutX and fIutX using QT.

Please read my answer is in here.

Also be convinced that the Putty (or your arduino-terminal) configuration (parity, flow control, baud rate) are same as QSerialPort configuration. In addition, you can try the Terminal example (from the QtSerialPort examples), e.g. instead of Putty (or your arduino-terminal).

UPD:

You can try to add:

setDataTerminalReady(true);

and

setRequestToSend(true);

after opening of port.

edited May 23 '17 at 12:01

Community

1
1

answered Apr 21 '14 at 14:26

Denis Shienkov

517
3
12

You have not read what the OP wrote in the comments, have you? The OP already mentioned he tried the examples. By the way, the terminal example is the worst example for testing. Keep that in mind. Also, please gain enough reputation for flagging a question duplicated if you think the answer is satisfactory elsewhere.. Answers are not the correct way to handle duplicates. It takes one only a little reputation to use the duplicate flagging system. – László Papp Apr 21 '14 at 14:34
> The OP already mentioned he tried the examples. Where? > By the way, the terminal example is the worst example for testing. Laszlo, please please stop your conjectures – Denis Shienkov Apr 21 '14 at 14:53
This is not "conjecture", but a somewhat "authoritative" answer from a contributor who wrote the simple examples and at least another one in the team thinking the same way. We wrote the command line examples because they are the simplest, and hence the most productive to test issues with. The terminal example is a left over with all the widget world inheritage. It is bloated, continously untested when doing changes, and so on. No one really cares about it. There is a QML example in the pipeline, but that has to wait. Even then, the cli examples are minimalistic. – László Papp Apr 21 '14 at 15:56

Unable to communicate with arduino using QSerialPort if the arduino-ide has not opened the connection before

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