[Qt-interest] QMetaObject and its method() function

[Louis Du Verdier]

Hello,

First, please excuse me if my english is bad, I am french.
I am writing an article about the Qt meta-objects system (in french), so I am creating an application which goal would be to illustrate that I say in it.
This application have to show all methods availible by the class. There it is the first code  version :

    QTableWidget *widget = new QTableWidget(this);
    widget->setColumnCount(3);
    widget->setHorizontalHeaderItem(0, new QTableWidgetItem("Type d'acces"));
    widget->setHorizontalHeaderItem(1, new QTableWidgetItem("Type de methode"));
    widget->setHorizontalHeaderItem(2, new QTableWidgetItem("Type de fonction"));

    widget->setRowCount(_metaObject->methodCount());
    for(int i = 0; i < _metaObject->methodCount(); i++)
    {
        widget->setVerticalHeaderItem(i, new QTableWidgetItem(_metaObject->method(i).signature()));
        if(_metaObject->method(i).access() == QMetaMethod::Private) widget->setItem(i, 0, new QTableWidgetItem("Prive"));
        if(_metaObject->method(i).access() == QMetaMethod::Protected) widget->setItem(i, 0, new QTableWidgetItem("Protege"));
        if(_metaObject->method(i).access() == QMetaMethod::Public) widget->setItem(i, 0, new QTableWidgetItem("Public"));
        if(_metaObject->method(i).methodType() == QMetaMethod::Signal) widget->setItem(i, 1, new QTableWidgetItem("Signal"));
        else if(_metaObject->method(i).methodType() == QMetaMethod::Slot) widget->setItem(i, 1, new QTableWidgetItem("Slot"));
        else widget->setItem(i, 1, new QTableWidgetItem("Autre"));
        widget->setItem(i, 2, new QTableWidgetItem(_metaObject->method(i).typeName()));
    }

The only problem is that the "else" condition, about the method type, is never called, and only the signals and slots are shown on the tableWidget.
If I create a private function (not a slot), this one is not shown on the window. I think it is because the meta-object system don't store this kind of function, but if someone can tell me if I am in the truth...
But if it is not wrong, this means that the enums QMetaMethod::Method and QMetaMethod::Constructor do not have any utility ?

Thank you in advance,
Louis.



      
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