GUI compile problem

[RSX]

If you are using designer then actionQuit you call through ui.

connect(ui->actionQuit, SIGNAL(triggered()), qApp, SLOT(quit()));

connect(ui->actionQuit, SIGNAL(triggered()), qApp, SLOT(quit()));

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[waynew]

Thanks RSX, that did it.

Just trying to get used to things in Qt.

[wysota]

Just trying to get used to things in Qt.

Your problem had nothing to do with Qt, it was strictly C++ related.

[squidge]

Indeed, if you want to refer to actionQuit as exactly that instead of ui->actionQuit, then you need to use multiple inheritance. An example is here.

[Lawand]

If you are using designer then actionQuit you call through ui.

connect(ui->actionQuit, SIGNAL(triggered()), qApp, SLOT(quit()));

connect(ui->actionQuit, SIGNAL(triggered()), qApp, SLOT(quit()));

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shoudn't it be ui.actionQuit ?

[john_god]

It's

ui.actionQuit

ui.actionQuit

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if ui was declared on the stack, and it's

ui->actionQuit

ui->actionQuit

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if ui was declare in heap, like a pointer

[Lawand]

Yes, but since it is auto-generated by Qt's UIC, then it must always be on the stack, right?

[wysota]

No, why...?

[john_god]

I just check in one project, it's in the heap

private:
    Ui::area_dlg *m_ui;

private:
    Ui::area_dlg *m_ui;

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[Lawand]

I see now, at first I thought that UIC always generates ui as an object rather than a pointer, but then I realized that it's the IDE that generated this, not UIC.

Sorry if I wasted your time

[wysota]

UIC generates classes, not objects.