I'm trying to create an exit button that correctly closes the GUI I have made in QT. I have tried doing this in the following way:
#include <QtGui/QApplication>
#include "mainwindow.h"
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
int window_width = QApplication::desktop()->width();
int window_height = QApplication::desktop()->height();
MainWindow w;
QPushButton * quit_btn = new QPushButton;
quit_btn->setParent(w.centralWidget());
quit_btn->setGeometry(window_width-50,12,32,32);
QObject::connect(quit_btn,SIGNAL(clicked()),qApp,SLOT(quit()));
w.resize(window_width,window_height);
w.show();
return a.exec();
}
Unfortunately when I push the button, the debugger gives an error:
Invalid address specified to RtlFreeHeap( 003E0000, 0028F950 )
Can anybody point me in the right direction?
Connect the button's clicked() signal to your main window's close() slot. That way things are cleaned up properly.
Related
It is very straightforward to connect to QMenu::triggered or QMenu::hovered signals by calling QObject::connect and pass the appropriate QAction.
However, I do not know how to use QMenu::aboutToHide signal, as there is no action passed to that signal.
How to use QMenu::aboutToHide and QMenu::aboutToShow signals or those are just virtual functions that can be overridden?
The signals in the world of Qt are not functions, never invoke them. The signals notify that something has happened with the QObject and send information if necessary.
In the case of triggered and hovered it is necessary to send the QAction because several QActions in a QMenu, then the developer thought that it is necessary to know with which QAction was interacting. On the other hand with aboutToShow and aboutToHide the signal does not send anything because it wants to notify is that if the QMenu was shown or hidden, respectively. Is there any need to know that QMenu was shown or hidden if he did it ? no, because the sender did it, I do not use other properties that we do not have at hand.
Example of use:
#include <QApplication>
#include <QMainWindow>
#include <QMenuBar>
#include <QDebug>
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
QMainWindow w;
QMenu *foo_menu = w.menuBar()->addMenu("Foo Menu");
for(const QString & name: {"action1", "action2", "action3"}){
foo_menu->addAction(name);
}
QObject::connect(foo_menu, &QMenu::aboutToShow, [](){
qDebug()<<"aboutToShow";
});
QObject::connect(foo_menu, &QMenu::aboutToHide, [](){
qDebug()<<"aboutToHide";
});
QObject::connect(foo_menu, &QMenu::triggered, [](QAction *action){
qDebug()<< "triggered: " <<action->text();
});
QObject::connect(foo_menu, &QMenu::hovered, [](QAction *action){
qDebug()<< "hovered: " <<action->text();
});
w.show();
return a.exec();
}
And what happens if you have several QMenu that connect to the same slot? How do I know QMenu was shown or hidden?
The solution is to use sender() which is a method that belongs to the QObject class that returns the object that emitted the signal, in this case the QMenu.
Example:
#include <QApplication>
#include <QMainWindow>
#include <QMenuBar>
#include <QDebug>
class MainWindow: public QMainWindow{
public:
MainWindow(QWidget *parent=nullptr):
QMainWindow(parent)
{
for(const QString & name_of_menubar: {"bar1", "bar2", "bar3"}){
QMenu *menu = menuBar()->addMenu(name_of_menubar);
connect(menu, &QMenu::aboutToShow, this, &MainWindow::on_aboutToShow);
connect(menu, &QMenu::aboutToHide, this, &MainWindow::on_aboutToHide);
for(const QString & name: {"action1", "action2", "action3"}){
menu->addAction(name);
}
}
}
private slots:
void on_aboutToShow(){
if(QMenu *menu = qobject_cast<QMenu *>(sender()))
qDebug()<<"aboutToShow" << menu->title();
}
void on_aboutToHide(){
if(QMenu *menu = qobject_cast<QMenu *>(sender()))
qDebug()<<"aboutToHide" << menu->title();
}
};
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
MainWindow w;
w.show();
return a.exec();
}
Got an error running the following problem:
#include <QMessageBox>
#include <QApplication>
int main() {
QApplication app();
QMessageBox msgBox(QMessageBox::Critical,
QObject::tr("text1"),
QObject::tr("text2"),
QMessageBox::Ok);
msgBox.exec();
return 0;
}
The error is:
The program breaks at QMessageBox msgBox(...);
The call stack is:
EDIT: Even after i have added QApplication instance in XTerm window named qtcreator_process_stub i see the following:
QWidget: Must construct QApplication before a QWidget
The line
QApplication app();
doesn't create a QApplication object - it actually declares a function taking no arguments and returning a QApplication! This is sometimes known as the "most vexing parse".
To actually construct the application object, you need to provide the program's arguments:
QApplication app(argc, argv);
The full program is then
#include <QMessageBox>
#include <QApplication>
int main(int argc, char **argv) {
QApplication app(argc, argv);
QMessageBox msgBox(QMessageBox::Critical,
QObject::tr("text1"),
QObject::tr("text2"),
QMessageBox::Ok);
msgBox.exec();
return 0;
}
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
QDialog dlg;
dlg.exec();
return a.exec();
}
That's all my code, but when I close the window, The process isn't exit, it seems that drop in the loop a.exec().
Generally speaking, calling any exec is a bad idea, other than QCoreApplication::exec() or QDrag::exec(). The presence of exec() and waitForXxx() methods is an enticing trap for the unwary. Those methods are "easy" to use, but that ease comes at a price of hard to track bugs. Don't use them.
You should simply show the dialog:
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
QMessageBox msg;
msg.setText("Hello");
msg.addButton(QMessageBox::Close);
msg.show();
return a.exec();
}
If you wish to wait for the dialog to be accepted or rejected, you should use the dialog's clickedButton slot. QMessageBox has a long-standing bug that makes the accepted and rejected signals useless :(
// https://github.com/KubaO/stackoverflown/tree/master/questions/messagebox-show-25545652
#include <QtGui>
#if QT_VERSION >= QT_VERSION_CHECK(5,0,0)
#include <QtWidgets>
#endif
#include <functional>
[...]
int main(int argc, char *argv[])
{
QApplication app(argc, argv);
QMessageBox msg;
msg.setText("Continue?");
msg.addButton(QMessageBox::Yes);
msg.addButton(QMessageBox::No);
auto onClick = [&msg]() {
auto role = msg.buttonRole(msg.clickedButton());
if (role == QMessageBox::NoRole)
QApplication::quit();
if (role == QMessageBox::YesRole) {
auto label = new QLabel("I'm running");
label->setAttribute(Qt::WA_DeleteOnClose);
label->show();
}
};
#if QT_VERSION >= QT_VERSION_CHECK(5,0,0)
QObject::connect(&msg, &QMessageBox::buttonClicked, onClick);
#else
QObject::connect(&msg, SIGNAL(buttonClicked(QAbstractButton*)),
new FunctorSlot{onClick, &msg}, SLOT(call()));
#endif
msg.show();
return app.exec();
}
#include "main.moc"
For Qt 4, you need the following helper:
// Qt 4 only
struct FunctorSlot : public QObject {
Q_OBJECT
public:
std::function<void()> callable;
template <typename Fun>
FunctorSlot(Fun && fun, QObject * parent = {}) :
QObject{parent}, callable{std::forward<Fun>(fun)} {}
Q_SLOT void call() {
callable();
}
};
Possible solution:
QApplication a(argc, argv);
QDialog dlg;
QTimer::singleShot( &dlg, 0, SLOT(exec()) );
return a.exec();
It will work well. First - application event loop will be started. Then dialog event loop will be executed. After closing of dialog, both dialog and application loop will be finished. Application loop will be terminated automatically (by default), when last window is closed.
But, as noted by #thuga - there are no reason to call exec(). It is enough to call show() method.
How do I have the user first login before getting to the main window in my QT app? I'm new to QT and have looked through the documentation and have not found anything very helpful. Thank you.
I would make this in the following way. Let's assume, that my login dialog is a QDialog:
class Login : public QDialog
{
[..]
};
In my application, I create an instance of my Login dialog and if it is accepted, i.e. closed with OK button, I open my main window:
int main(int argc, char *argv[])
{
[..]
QMainWindow mw;
Login login;
if (login.exec() == QDialog::Accepted) {
mw.show();
}
[..]
}
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
MainWindow w;
LoginWindow l; // create login-widget
connect(&l, SIGNAL(loggedIn), &w, SLOT(show()); // connect to mainwindow show()
l.show(); // show login-window instead of
// mainwindow at start
return a.exec();
}
You might want to :
1 - exec the loop only if the dialog is accepted, otherwise you application continues to run
2 - create the mainwindow only if accepted, cause it might be a heavy interface, which takes time to initialize
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
Dialog_Login dialog;
if(dialog.exec() == QDialog::Accepted){
MainWindow w;
w.setUser(dialog.GetUser());
w.show();
return a.exec();
}
else return 0;
}
You need to create the QApplication instance in parrallel with the login dialog.
The program name is GoToCell in the form I created a label and two push buttons and I wrote the code in main.cpp as follows:
#include <QtGui/QApplication>
#include<QDialog>
#include "ui_GoToCell.h"
#include "GoToCell.h"
int main(int argc, char *argv[]) {
QApplication a(argc, argv);
Ui::GoToCell ui;
QDialog *dialog=new QDialog;
ui.setupUi(dialog);
dialog->show();
return a.exec();
}
While running it I'm getting the following errors:
GoToCell is not a member of ui
What should I do?
I think you have misspelled object name in objectName property in the GoToCell.ui form.
Change it to GoToCell. It will then execute.