Can't see widget, yet isVisible() returns true

**MattPhillips** · 2nd February 2010, 15:03

Hello,

This is happening in Windows 7 only, not Ubuntu, so there shouldn't be some fundamental problem with my code. My program has a QGLWidget subclass as its main window--as in, main.cpp looks like this:

Qt Code:

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int main(int argc, char *argv[])
{
    QApplication a(argc, argv);
    MyGLWidget w;
    w.show();
    return a.exec();
}

int main(int argc, char *argv[])
{
    QApplication a(argc, argv);
    MyGLWidget w;
    w.show();
    return a.exec();
}

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When I run this in Windows, the widget never appears on screen, nor does an icon for it appear in the taskbar (i.e. the bar on the bottom of the screen where program icons go). However, the executable *does* appear in the Processes list on the Task Manager; and within the program, I use a timer to check whether the widget has appeared yet with 'isVisible()', and this returns true. Does anybody know what could be going on? In general I am able to run Qt programs in Windows 7. Thanks--

Matt