I am trying to create a library(.so file), which will have methods
Create_widget (Which will create a widget in the library code)
Add/modify widget (QPixmap or QGraphicItem)
The caller code :
testFile.cpp will call the Create_widget of the library and the expectation is it will create the widget. Then add/modify(add_Image_to_widget) methods will called from testFile.cpp and expects a update on the widget.
TEST_F(SomeTests, Check1)
{
int argc = 0;
char* argv[0] = {};
QApplication qapp(argc, argv);
Create_widget();
sleep(5);
add_Image_to_widget();
sleep(5);
qapp.exec();
}
The problem :
Until the qpp.exec() is called the GUI will not show any update. This seems the behavior as explained in qt doc.
My expectation is as soon as Create_widget called widget needs to be created and after sleep of 5 seconds, image to be displayed. But in this case the event processing starts only after exec. (Blocking call).
Let me know how this can be achieved.
Tried to move the below operation to a thread
"
QApplication qapp(argc, argv); qpp.exec();
"
This made the errors of widget not created in QApplication.
Related
I am working on a CMAKE C++ project which uses the QT Libraries. (For me, 5.15.3, for others 5.12.x)
In this project, there is a class Vtk3DViewer : public QWidget. In its constructor, it tries to create one of its member variables, which is of type QVTKOpenGLNativeWidget. This is from the VTK libraries. (Located in include\vtk-9.2\QVTKOpenGLNativeWidget.h)
For me, this new QVTKOpenGLNativeWidget() call within the constructor of my QWidget fails with the following error:
"Must construct a QApplication before a QWidget"
But that's just it, we do create a QApplication in main() well before this point. And this only happens on Windows. Linux builds appear to not have any issue.
Switching from Debug to RelWithDebugInfo moves the error - making it happen much earlier and on creating a QToolBarExt instead.
Why is this happening, and how do I fix it?
Here is an example of main():
int main(int argc, char* argv[])
{
// Set info for settings & registry
QApplication::setOrganizationName(COMPANY_NAME);
QApplication::setOrganizationDomain(COMPANY_DOMAIN);
QApplication::setApplicationName(APP_NAME);
// Set up for software-based backend for VTK
QApplication::setAttribute(Qt::AA_UseOpenGLES);
QApplication a(argc, argv);
a.setWindowIcon(QIcon(":/main-window/favicon.ico"));
// Instantiate singletons
TaskExecutionManager::getInstance(); // Instantiate the task manager
DataDispatcher::getInstance();
// Create main window with default size
MainWindow w;
w.show();
// Start application event loop
return a.exec();
}
Then the main window's constructor calls:
void MainWindow::initializeMainWindow(Ui::MainWindow* ui)
{
this->setDockOptions(AnimatedDocks | AllowNestedDocks | AllowTabbedDocks | GroupedDragging);
// Main toolbar
m_topToolBar = new QToolBarExt(this); // This causes a "Must construct a QApplication before a QWidget" error
}
The issue was that VTK was built in RELEASE, while our project was built in DEBUG.
(I did not see this as an issue, since we would never need to step into/debug VTK's code)
It appears this cryptic/incorrect error message will appear under these circumstances. Ensuring VTK and the project it includes are compiled the same way fixes it.
I went throught suggested "questions" about my problem. However neither does not solve it.
I program two windows. The second window is opening from first window. I need active the both windows, however to start the first window(MainWindow) I use:
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
MainWindow w;
w.setWindowModality(Qt::NonModal);
return a.exec();
}
As was mentioned, the second window is started from pushButton, which is situated in first window(MainWindow) by same way.
void MainWindow::on_pushButton_2_clicked()
{
Graphics gr;
gr.setWindowModality(Qt::NonModal);
gr.exec();
}
I changed modality to NonModal,however the problem is without change. The Non-Modal mean:"The window is not modal and does not block input to other windows." <- from documentation
By documentation is recommended to avoid used .exec(). The alternatives are .show() and open(), which i tried. After the modification, the second window is shut down immediately after opening. after open immediately shut down.
Do you have any idea, how to solve that?
Graphics gr; defines a local variable, so the object is destructed as soon as it goes out of scope (at the end of your function).
In Qt, the typical approach is to work with pointers to Qt widgets (and, more generally, QObjects), but have a parent for each – the parent will clean it up.
Try this instead:
auto gr = new Graphics(this);
gr->setWindowModality(Qt::NonModal); // this is the default, no need for this call
gr->show();
This way the window will survive until your main window is destructed. In Qt there is also a mechanism for widgets/objects to self-destruct:
this->deleteLater();
So for example if you want to close and cleanup gr you can use that mechanism (you could also store gr as a member and call gr->deleteLater().
I use CAMA series (finger print module) for my application. In that document it said that recommended to use EnrollStandByMode command before exit the application. I create this command and call it in my destructor on QMainWindow in my application. (there is no problem here). But when I exit the application and again run it my finger print is confused and the response commands are incorrect. After trace my code ,I saw the destructor not run when the application close with close button.
This is my code to exit the qapp.
void MainWindow::on_pushButton_clicked()
{
qDebug()«"QApplication::quit()";
QApplication::quit();
}
I want to know how to call destructor before exit the application?
There are several ways to call something at the end of the application.
If your code does not require the Qt event loop to run, just call the function at the end of main()
int main(int argc, char **argv)
{
QApplication app(argc, argv); // or QGuiApplication/QCoreApplication
// application setup
const int resultCode = app.exec();
// call your shutdown code here
return resultCode;
}
Another option is to register a function with qAddPostRoutine() which is called by the Qt application object's destructor.
If you need the Qt event loop, connect a function to the Qt application object's aboutToQuit() signal.
I am trying to write an QtQuick program which works as an intelligent interface between user and a few CLI applications. I have implemented QtQuick + JavaScript application.
A QtQuick butten emits signal that is listened by C++ layer. So far everything works well.
However, in my C++ slot function I need to write to a certain Item in QtQuick application. This Item is an TextArea which serves as Log output of CLI applications. These CLI applications are run from the slot function in C++. I store their output into a variable, and I want to show output of this variable in this Log output TextArea.
I tried a lot of things, but I didn't find the right way to do that
I'd a similar problem.
This is how I solved it.
In C++ I created a class that handles the command with a QProcess (and I expose the class to QML), which attach the readyToRead signal to a C++ function in my exposed class, this function emits another signal showOutput with the output text.
With this information I just connect my new signal to a javascript function in qml:
cppExposed.showOutput.connect(jsFunction);
And in the javascript function I just append the text
function jsFunction(output) {
OutputTextArea.text += output;
}
To expose C++ properties to QML you can have a look at the documentation here: http://doc.qt.io/qt-5/qtqml-cppintegration-exposecppattributes.html
I think the easiest way for you is to create an object of your cpp class, and set as a context property in your main.cpp before load your main.qml:
Something like this:
QQmlApplicationEngine engine;
ProcessHandler procHandler;
engine.rootContext()->setContextProperty("procHandler", &procHandler);
Now you can access your object direct from QML, and you can connect signals
procHandler.showOutput.connect(jsFunction)
And in your C++ class don't forget to connect with the process ReadyToReady signal and emit your own signal with the data:
void readyToRead() {
emit showOutput(m_proc.readAllStandardOutput());
}
UPDATE:
Your property should be set before load the QML file by the engine:
AnalyzeSignal analyzeSignal;
engine.rootContext()->setContextProperty("analyzeSignal", &analyzeSignal);
engine.load(QUrl(QStringLiteral("qrc:/main.qml")));
And I don't think you need that writeToLogOutput to be a property (and it has some syntax error btw), it's a signal method, right? So it's automatically available to QML as a signal, not a property.
DON'T create a new QQMLContext.
In this line what you are doing is creating a new QQMLContext
QQmlContext *context = new QQmlContext(engine.rootContext());
context->setContextProperty("analyzeSignal", &analyzeSignal);
This won't work, as you are setting the property to the newly created context, not to the original root context.
My final solution of main function looks like this:
int main(int argc, char *argv[])
{
QApplication app(argc, argv);
QQmlApplicationEngine engine;
AnalyzeSignal analyzeSignal;
QQmlContext *context = engine.rootContext();
context->setContextProperty("analyzeSignal", &analyzeSignal);
engine.load(QUrl(QStringLiteral("qrc:/main.qml")));
// Registering slot onto a signal
QObject *win = engine.rootObjects()[0];
QObject *item = win->findChild<QObject*>("myButton");
QObject::connect(item, SIGNAL(doAnalyzeSignal(QString,QString,QString,QString)), &analyzeSignal, SLOT(cppSlot(QString,QString,QString,QString)));
return app.exec();
}
And this works!
I´ve beeing programming Java for some time right now...Now that I got into C++ and Qt I am a bit lost about GUI Thread (EDT Thread) and Worker Thread
I am trying to make the main window of my application open only when the configuration window is closed.
I dont want to put the code for creating the main window in the OK button of my configuration window.
I tryed to make them modal but the main window still opens.....
Afther configuration is complete I still have to see if there is an application update...So its something like
EDIT: This is my main:
ConfigurationWindow *cw = new ConfigurationWindow();
//if there is no text file - configuration
cw->show();
//**I need to stop here until user fills the configuration
MainWindow *mw = new MainWindow();
ApplicationUpdateThread *t = new ApplicationUpdateThread();
//connect app update thread with main window and starts it
mw->show();
Try something like this:
#include <QtGui>
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
QDialog *dialog = new QDialog;
QSlider *slider = new QSlider(dialog);
QHBoxLayout *layout = new QHBoxLayout(dialog);
layout->addWidget(slider);
dialog->setLayout(layout);
dialog->exec();
qDebug() << slider->value(); // prints the slider's value when dialog is closed
QMainWindow mw; // in your version this could be MainWindow mw(slider->value());
w.show();
return a.exec();
}
The idea is that your main window's constructor could accept parameters from the QDialog. In this contrived example I'm just using qDebug() to print the value of the slider in the QDialog when it's closed, not passing it as a parameter, but you get the point.
EDIT: You might also want to "delete" the dialog before creating the main window in order to save memory. In that case you would need to store the parameters for the main window constructor as separate variables before deleting the dialog.
You have to learn about signals and slots. The basic idea is that you would send a signal when you configuration is finished. You put your QMainWindow in a member variable and call mw->show() in a slot of your main programm that is connected with the configurationFinished signal.
If your ConfigurationWindow is a QDialog, you could connect the finished(int) signal to the MainWindow's show() slot (and omit the show() call from main).