An interesting problem when using QMenu::aboutToShow() signal!!

[srazi]

I subclass QToolBar and create some QToolButtons and create and set menu for every QToolButton, all things work good until I connect 'QMenu::aboutToShow()' to a 'SLOT'!! And then I see that menus are showed in an incorrect position!!!
when I looked at 'qtoolbutton.cpp' I see this code:

void QToolButtonPrivate::popupTimerDone()
{
...
    QPoint p;
    QRect screen = QApplication::desktop()->availableGeometry(q);
    QSize sh = ((QToolButton*)(QMenu*)actualMenu)->receivers(SIGNAL(aboutToShow()))? QSize() : actualMenu->sizeHint();
    QRect rect = q->rect();
...
//computing 'p'! 'sh' is used in this computation
...
    actualMenu->exec(p);
...
}

void QToolButtonPrivate::popupTimerDone()
{
...
    QPoint p;
    QRect screen = QApplication::desktop()->availableGeometry(q);
    QSize sh = ((QToolButton*)(QMenu*)actualMenu)->receivers(SIGNAL(aboutToShow()))? QSize() : actualMenu->sizeHint();
    QRect rect = q->rect();
...
//computing 'p'! 'sh' is used in this computation
...
    actualMenu->exec(p);
...
}

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I can't understand using 'QObject::receivers(const char *signal)' in such a bad way!!!!! Why we should get different results just because of creating a connection!!!!?
It's unbelievable...

For solving the problem I use 'QEvent::Show' in replace of 'QMenu::aboutToShow()'!